phy 112 midterm
The electron will speed up.
An electron moving to the right at a constant speed moves into an electric field. The electric field is to the left. Which of the following correctly indicates how the field will affect the electron's motion? The electron will speed up. The electron will slow down. The electron's speed will remain the same, but it will deflect upwards. The electron's speed will remain the same, but it will deflect downwards.
b. Charge quantity
1. Which of the following is an example of a scalar measurement? a. Force b. Charge quantity c. Electric field strength
induction
2 step charge transfer process that involves both polarization and conduction
gained energy, lost energy, gained energy
A negatively charged particle moves through various electric fields. Which of the following correctly indicates whether the negatively charged particle gained or lost energy in each of the following examples? gained energy, lost energy, gained energy gained energy, gained energy, lost energy lost energy, lost energy, lost energy lost energy, gained energy, gained energy
1500 eV.
An electron moves through a potential difference of 1500 V. The magnitude of the change in potential energy experienced by the electron is 1500 J. 1500 eV. Not enough information is stated. The distance traveled needs to be known. Not enough information is stated. The energy depends on whether the field is uniform or nonuniform.
A > B > C
Based on the equipotentials shown in the example below, order the magnitude of the potentials at points A, B, and C (from greatest to least). C > B > A A > B > C B > A > C A > C > B
A > B > C
Based on the plot of equipotentials shown in the example below, order the magnitude of the electric field strength at points A, B, and C (from greatest to least). B = C > A C > B > A A > B > C A > B = C
30 V
Consider a uniform electric field of 50 N/C directed towards the east. If the potential measured at some point in the field is 80 V, what is the potential at a position 1.0 m directly east of that point? 50 V 130 V 30 V 80 V
Since A and B have the same resistance, they will initially (with the switch open) have the same voltage drop and the same bulb brightness. Bulb C is in parallel to the rest of the circuit. When the switch is closed, the total circuit resistance will decrease. Since the voltage source has not changed, this means the current through the battery will increase. Bulb A is in series with the battery, which means bulb A will also experience more current. Bulb A will get brighter. With the switch open, bulbs A and B have the same voltage drop. When C is added in parallel to B, the resistance decreases, which means the voltage drop across B will also decrease. Bulb B will get dimmer.
Consider the circuit below. All the bulbs are identical. What happens to the brightness of bulbs A and B when the switch is closed?
Answer: F = BILsinθ F = (4.0 T)(0.75 A)(2.0 m)(sin90) F = 6.0 N The force will be directed into the page (or computer screen).
Ex) A 2.0 m long horizontal wire has a current of 0.75 A directed to the right. The wire is in a uniform field of 4.0 T directed down the page (or computer screen). What are the magnitude and direction of the force on the wire?
Answer: Uc = ½ qV = ½ (CV)V Uc = ½ (25x10^-6 F)(12 V)2 Uc = 1.8x10^-3 J
Ex) A capacitor with a 25 μF capacitor is connected to a 12 V battery. How much energy is stored in the capacitor?
Answer: C = Q/V (25x10^-6 F) = Q / (9 V) Q = 2.25x10^-4 C
Ex) A capacitor with a capacitance of 25 μF is connected to a 9V battery. How much charge will be stored in the capacitor?
Answer: ΣF = ma qvBsinθ = mv2/r qB = mv/r (3.2x10^-19 C)(4.2 T)(sin90) = (6.64x10^-27 kg)(4x10^5 m/s) / r r = 0.00198 m or 1.98 mm
Ex) A helium nucleus with a charge of +3.2x10^-19 C and a mass of 6.64x10^-27 kg moving at a speed of 4x10^5 m/s enters a magnetic field with a strength of 4.2 T. What is the radius of the curved path taken by the helium nucleus?
Answer: F = qvBsinθ (2.0x10^-3 N) = (4.x10^-6 C)(700 m/s)(B)(sin90) B = 0.71 T Directed into the page (or computer screen); be sure to use your left hand.
Ex) A particle with a net charge of -4 μC moving to the right at a speed of 700 m/s enters a uniform magnetic field. The particle experiences a magnetic force of 2 mN directed down the page (or computer screen). What are the magnitude and direction of the magnetic field?
Answer: Draw lines that pass perpendicularly through each of the field lines.
Ex) Draw at least five equipotential lines for the following electric field.
Answer: Magnetic fields are directed away from the north and towards the south end of the magnet.
Ex) Draw the magnetic field that occurs near the open end of a horseshoe magnet.
Place a ferromagnetic material in a magnetic field, and either heat and cool the material or tap the material to force the domains into alignment with the external field.
How can you create a permanent magnet? How can you create an electromagnet?
Capacitors are made by alternating layers of conductors and insulators. They are used to store charge and energy.
How is a capacitor made? What is a capacitor used for?
The center of the atom is called the nucleus, and is composed of densely packed protons and neutrons. Electrons orbit the nucleus (although not in the same way that a planet orbits a sun). A neutral atom has an equal number of protons and electrons.
Give a basic description of the structure of an atom.
c. Energy an object has due to its position in a gravity field
Gravitational potential energy is nest described as: a. Energy an object has due to its motion b. Energy an object has due to a reversible deformation c. Energy an object has due to its position in a gravity field
a. The amount of push or pull between two masses
Gravity field strength, g, is best defined as: a. The amount of push or pull between two masses b. The ability of a single mass to apply gravity forces c. Always equal to a value of 9.8 N/kg d. None of the above
E is proportional to q. If q doubles, E will also double at all locations.
If the amount of charge on a particle doubles, what will happen to the field strength measured at some distance r from the particle?
repel as far away as possible
If you have two conductors touching when you bring the charged metal rod nearby=> electrons will ______. Some of electrons on metal bar transfer over to other metal bar
resistance
ability of a circuit component to impede current
permeability of free space
ability of a material to support a magnetic field
resistance
ability of a resistor to resist current. If it increases, flow rate decreases
right-hand curl rule
allows you to quickly identify the direction of the field near the wire. The field strength is directly proportional to the amount of current but inversely related to the distance from the wire
capacitance
amount of charge per volt that be stores in a capacitor
electron-volt
amount of energy that an electron would gain of lose when it moves through a potential difference of 1.0 volts
static electric force
amount of push or pull between two or more net charges
Dielectric
another name for insulating material in a capacitor. The specific effects of inserting a ____ between the capacitor plates depends on whether or not the battery is still connector to the capacitor · To summarize, inserting a ____ with no battery connected will cause the following changes: o εo will increase by a factor of k. o C will increase by a factor of k. o V will decrease by a factor of k. o E will decrease by a factor of k. o Q will not be affected.
permanent magnets
are always one; they are always magnetically attracted to and by one other ferromagnetic material to create a ___, an object made of ferromagnetic materials is placed in a strong magnetic field, and is either heater or tapped. can become demagnetized if they are repeatedly dropped (the domain pole direction shifts)
Magnetic effects
are created by movement of charge. The movement of charge is the source of all magnetism.
magnetic field
are created by the movement of charge. This is different from electric fields; electric fields can be created by stationary charge. For a ____ to be created, charge must be in motion
ammeters
are devices used to measure current through a circuit component. Have low resistance, and should be connected in series
voltmeters
are devices used to measure voltage differences across circuit components. Have high resistance, and should be connected in parallel
electromagnets
are temporary magnets created using electric current. Small electromagnets are relatively easy to make. You need magnet wire (sometimes called enameled wire) or insulated wire and a battery. create a coil of wire, connect it to the battery; the coil of wire will become magnetized. If you put an object made from ferromagnetic material (iron nails are common for this style of simple electromagnet) in the center coil, the electromagnetic effects strengthen
drift velocity
average velocity of charge moving in response to an electric field
semiconductors
class of materials that share properties of both conductors and insulators
atom
composed of protons, neutrons, and electrons Has equal number of protons and electrons=> have same amount of both types of charge, which = neutral (equal charge)
alternating current
current that fluctuate sinusoidally with time
capacitor
device that can be used to store charge composed of alternating layers of insulator and conductor
voltmeter
devices used to measure voltage differences between various circuit locations. Are always wired to a circuit in parallel. Have a high resistance so they do not interfere with the operation of a circuit; the high resistance will keep charge from branching through the ____. The circuit notation for a ___ is a circle or square with a V in the middle
voltage
difference in potential energy per unit charge between two different positions
net charge
differences in charge quantities that cause attracting and repelling forces is a fixed amt: o must be transferred from one object to another; if one object gains charge it is because another object lost an = amt of charge; the overall amt of charge was not changed
Gravitational potential energy:
energy an object has due to its position in a gravity field
coulomb's law
equation that relates the static electric force, charge quantity, and distance between charges
series
experience the same amount of current through each resistor. The sum of the voltage drops across each resistor in ___ must equal the voltage across the battery. Resistors wired together in ____ will result in an increase in overall ____ resistance
parallel
experience the same amount of voltage across each ____ branch. The sum of the currents through each branch must equal the total current through the battery. Resistors wired together in ____ will result in a decrease in overall ___ resistance
direct current
flow of charge in only one direction
current
flow rate of charge
Lorentz force
force on a charge moving through a magnetic field
charge
fundamental property of matter responsible for electrical interactions
Ferromagnetic materials
have regions or domains where atoms are collectively oriented to create north and south poles. These domains are oriented in random directions until the material becomes magnetized. The magnetization process will orient the domains in a fairly uniform direction. Permanent magnets gain their magnetic properties through the movement of charge at the atomic level iron, cobalt, nickel, and some alloys
dielectric
insulating material in capacitor
series combination
involves linking the bottom plate of one capacitor to the top plate of another whatever charge gets pushed off the bottom plate will get pushed onto the top of the next capacitor, which will cause an equal amount of charge to leave its bottom plate, etc. However, this process becomes much more difficult with each additional capacitor added; you are essentially increasing the overall separation distance between the top capacitor plate and the bottom capacitor plate. Increasing the plate separation distance results in a lower capacitance.
parallel combination
involves linking the top plate of one capacitor to the top plate of another while also linking their bottom plates together Linking capacitors in ____, however, has the overall effect of increasing the plate area of the capacitor. Charge can be spread out over multiple different top capacitor plates. Increasing plate area will result in an increase in overall capacitance
capacitance
is a function of its design features. Large plate areas and small plate separation distances will result in greater capacitances. The capacitance will also depend on the insulator used, which is called the dielectric. Capacitors wired in parallel will have a large total capacitance. Capacitors wired in series will have a small total capacitance.
Electric potential energy
is the amount of static electric energy a charged particle has due to its position in an electric field. It is possible for a particle to have a negative energy value since the energy is determined by whether work was done on or by the charged particle when it was moved from an initial location infinitely far away from the field to its final position in the field. It is scalar, and so changes in energy can be found by subtracting the initial energy possessed by a charge from the final energy possessed by the charge.
current
is the quantity of charge that flows through some area every second. It is caused by differences; current and voltage are directly proportional. Current is hindered by resistors; current and resistance are inversely proportional. Resistors can provide a constant resistance (ohmic) or changing resistance (nonconstant). The resistance of a wire increases with increasing cross sectional area but decreases with increasing wire length.
potential energy
is the specific amount of electric energy a particular charged particle has stored due to work done on or by the particle as it moves through an electric field
equipotenatial
lines along which the potential is constant
Equipotentials
lines drawn in electric fields that show where the positions of equal potential are located. If you have 2 oppositely charged parallel plates, equipotential lines would look like: When dealing with nonuniform fields, the ____ lines have uniform spacing but in nonuniform fields, the ____ are spaced increasingly far apart. In a uniform field, the voltage changes uniformly between the plates. However, in a nonuniform field, the voltage does not change uniformly; the voltage changes occur much more rapidly the closer you move towards the source. · Notice in both the uniform field and the nonuniform field the ___ lines always meet the electric field lines at right angles. You can use this property to draw ___ for unfamiliar sets of charges as long as you know the shape of the electric field.
conductor
materials that allow electrons to flow easily, usually metals
insulators
materials that do not allow electrons to flow easily, usually non-metals
insulators
materials that do not allow electrons to move freely through them Nonmetals
induction
method of charging that involves first inducing a dipole in a neutral object
triboelectric charging
method of charging that involves rubbing two objects together
conduction
method of charging that involves transferring charge between two objects that are in physical contact
magnetic fields
movement of charge creates Since current is the flow rate of charge, this means that current-bearing wires create · direction of the _____ created by the wire depends on the direction of the current in the wire. · interact with each other by applying forces. A north pole placed near a south pole will result in an attracting force. A north pole placed near a north pole or a south pole placed near a south pole will result in a repelling force
electromagnet
object that temporarily acts as a magnet when current is passed through it
drift velocity
overall averages of the component of motion through the field and is, in most cases, relatively slow (usually on the order of mm/s).
proton
particle responsible for carrying the + type of charge, not usually mobile due to its position in the nucleus
electron
particle responsible for carrying the - type of charge, one of the more mobile charge carrier particles
magnetic field lines
pictorial representation of strength and direction of a magnetic field
electric energy
potential energy a charge has due to its position in an electric field
electric potential
potential energy per unit charge
polarization
process that causes redistribution of charge within an object so that one end becomes more positive and the other end becomes more negative. the net charge does not change
charge
property of matter that causes electrical effects o Symbol: q o Units: Coulomb or C o Conserved value fundamental property of matter responsible for electrical effects; all atomic matter contains ____ o + or - o Protons: + o Electrons: - o Objects with net charge of same type will repel, different charges will attract
potential
provides info about various positions near a source charge; a way of describing relative amounts of energy a hypothetical test charge may have at various positions
power
rate at which energy is provided by a source or dissipated by a circuit component
power
rate of energy transfer
magnetic field
region near a magnet that has the potential to apply magnetic forces
domains
regions within a material that behave like small bar magnets
ohm's law
relationship between voltage and current
RMS current
root mean squared current or a measure of the average current in an AC circuit
right hand curl rule
rule use to determine the direction of magnetic field near a wire with current
right hand rule
rule used to show the relationship between he directions of force, charge motion, and magnetic field for a charge moving through a magnetic field
Maxwell's equations
set of equations that describe electromagnetic phenomena
triboelectric charging
transfer of charge caused by rubbing two objects against each other
ohm
unit of resistance, ratio of volts per amp
Tesla
unit used to measure the strength of a magnetic field
farad
units of capacitance
ampere
units of current, ratio of charge per second
Electric potential (voltage):
way of predicting what positions are most likely to result in large amounts of electric potential energy when a + charge is placed at that position. It's the ratio of static potential energy per amount of charge placed in the field
right-hand rule
· Charge moving through a wire in the presence of a magnetic field will also cause the wire to experience a magnetic force. Since the direction of the current is based on the direction of positive charge flow, the ____can also be used to determine the direction of the force on the wire. The external field can also be generated by another, nearby wire, in which case the wires will either attract or repel each other depending on the direction of current flow. If the wire in the filed forms a loop, the loop may experience a torque causing the loop to rotate
parallel
· This arrangement will result in a decrease in overall resistance. By adding new ___ resistors, you've also added entirely new pathways for charge flow. This is like the effects you see when you increase the area of a wire resistor; the larger the area, the lower the resistance o Since conventional current flows from positive to negative, each resistor will have charge moving downward through the resistor
Power
· When charge flows through a circuit, its potential energy is transformed into heat, light, work, and other possible forms of energy. The rate of energy transfer is called ___. ____ is directly proportional to both current and voltage.
current
· amount of charge that flows through the wire in one second measures the quantity of charge moving through a point each second. Current does not measure how fast the charges are moving. Speed can be related to current but they do not measure the same thing Charge is actually moving in a variety of directions through the wire but has an overall motion from high potential to low potential. rate of charge transfer; the definition does not indicate which type of charge must be transferred. It is possible to create a current with a flow of + or - charge. If all of these charges were in the same electric field, the direction of these two currents would be in opposite directions
magnetic effects
· are created by the movement of charge. All magnets have north and south poles. Like poles repel while unalike poles attract
capacitor
· are devices made by alternating layers of conductors and insulators. They are devices used to store charge. The ability of a capacitor to store charge is called its capacitance. Capacitance is the ratio of charge stored on a plate per potential used to push the charge on to the plate. The units of capacitance are C / V or farads (F).
Batteries
· are sources of constant voltage, and will create steady state currents (DC). Variable voltage sources can cause fluctuating voltage that, if the alternating voltage source is connected to a circuit, will result in alternating current (AC) through that circuit. U.S. households are designed to operate using AC electricity.
right-hand rule
· can be used to determine the direction of the force relative to the direction of charge motion and direction of the field. If the particle is negative, the left hand should be used. the amount of force is proportional to the speed of the charge, the strength of the field, and the amount of charge. Since the force is always perpendicular to the motion, the charged particle will be deflected in a circular path through the field
flow rate of charge through some point.
The definition of current is the potential difference across an electrical component. flow rate of charge through some point. speed of charge through some point. rate of energy dissipation by a resistor.
b. the rate of energy transfer.
The definition of power is best described as: a. the ability to do work. b. the rate of energy transfer. c. the energy associated with movement. d. the force per area.
the electric field near two positive point charges.
The following diagram is best described as the equipotential contour map near a positive and a negative point charge. the equipotential contour map near two positive point charges. the electric field near a positive and a negative point charge. the electric field near two positive point charges.
a. Repulsive
The force between two positively charges particles will be: a. Repulsive b. Attractive c. More info is required; direction of force depends on distance between particles d. More info is required; direction of force depends on amount of charge on both particles
The slope of the graph is the resistance. R = (12-0 V) / (0.8 - 0 A) R = 15 Ω R = ρL / A (15 Ω) = (2.65E-8 Ω-m)(0.50 m) / (A) A = 8.83E-10 m^2 A = πr^2 (8.83E-10 m^2) = (π)(r^2) r = 1.68E-5 m
Use the following voltage versus current graph to determine the radius of the wire resistor used in the circuit if the wire is 0.50 m of aluminum.
c. both the above.
Voltage is best defined as: a. the difference in potential between two points in an electric field. b. the difference in energy per charge ratio between two points in an electric field c. both the above.
Voltage changes equally from one equipotential line to the next (just like elevation changes equally from one line on a topographical map to the next). Closely spaced lines show that voltage changes very quickly relative to position. Widely spaced lines show that voltage changes slowly relative to position.
What is the significance of the spacing between two equipotential lines?
1.8 x 10^5 N/C
What is the strength of an electric field measured 0.5 m from a point source with a net charge of 5.0 μC? 9 x 10^4 N/C 2.5 x 10^-6 N/C 1.8 x 10^5 N/C 10 N/C
The symbol is B, and the units are usually a Tesla (T) or a Gauss (G).
What is the symbol for magnetic field strength? What are the units for measuring the strength of a magnetic field? (List two.)
The excess charge stored on one plate would be repelled back onto the other plate. Charge would move through the wire in order to rebalance the charge on each plate
What would happen if you charged the capacitor, disconnected the battery, and then connected the two plates with a wire?
circular
When current flows through a wire, the moving charge creates a ___ magnetic field around the wire
hills.
When drawing the equipotential 3D maps, positive point sources will form hills. valleys. planes (flat surface). positive point sources that don't have equipotential areas.
Use conservation of energy to solve for the final speed of the particle after it accelerates through the first electric field. Ue = K qEr = ½ mv^2 (3E-3 C)(4000 N/C)(0.12 m) = ½ (3E-6 kg)(v^2) v = 980 m/s Use Newton's first law to solve for the field strength. ΣF = 0 FB - Fe = 0 qvBsinθ - qE = 0 (980 m/s)(B) - (2000 N/C) = 0 B = 2.0 T The particle is moving to the right, and the magnetic force must be up the page to counter the electric force. This means the magnetic field must be directed into the page.
A +3 mC particle with a mass of 3x10^-3 g is accelerated to the right a distance of 12 cm from rest through a uniform electric field with a field strength of 4000 N/C. The particle then enters a new uniform field with a field strength of 2000 N/C directed downwards. What is the strength and direction of the magnetic field needed in order for the particle to pass through the electric field undeflected?
0.23 W.
A 1.5V battery connects to a 10 Ω resistor. The power dissipated by the resistor is 15 W. 0.15 W. 0.23 W. 22.5 W.
P = IV = V^2/R P = (9 V)^2 / (100 Ω) P = 0.81 W P = W / t (0.81 W) = W / (60 sec) W = 48 J The wires connecting the battery to the resistor will dissipate negligible energy (the voltage drop across both ends of a connecting wire tends to be very, very small. Since energy must be conserved, the power supplied by the battery is equal to the heat emitted by the resistor. W = 48 J
A 100 Ω resistor is connected to a 9 V battery. How much energy does the resistor dissipate in 1 minute? b. How much energy does the battery supply to the circuit in 1 min?
As usual, whenever forces are involved, start with a force diagram on the ball. Since the system is in equilibrium, the forces must cancel (Newton's first law). The vertical component of the tension force will equal the downward force of gravity, and the horizontal component of the tension force will equal the static electric force. Solve for the forces and then combine the components using the Pythagorean theorem. Fg = mg Fg = (0.005 kg)(9.8 N/kg) Fg = 0.049 N E = Fe / q (2x103 N/C) = F / (6x10-6 C) Fe = 0.012 N FT2 = Fg2 + Fe2 tanθ = (0.049 N) / (0.012 N) FT2 = (0.049 N)2 + (0.012 N)2 θ = tan-1 (4.08) FT = 0.0504 N θ = 76.2° above the -x axis
A Styrofoam ball suspended from a string is placed into an electric field with a field strength of 2x103N/C. The ball has a mass of 5.0 grams and a net charge of +6 μC. What is the tension in the string?
None of the above options are correct.
A battery connects to a resistor with resistance R, and the resulting current is measured. The resistor is removed, and a new resistor with a resistance of 2R is added. As a result, the voltage will double. the current will double. the charge speed will double. None of the above options are correct.
equal to 1.44 W.
A battery connects to a resistor. The rate of energy dissipated by the resistor is 1.44 W. The rate of energy supplied by the battery is greater than 1.44 W. less than 1.44 W. equal to 1.44 W. Not enough information is stated. The voltage of the battery must be known.
the current will double.
A battery with a potential difference of V connects to a resistor, and the resulting current is measured. You remove the battery, and a new battery with a potential difference of 2V is put in its place. As a result, the current will double. the resistance will double. both the current and the resistance will double. battery replacement will have no effect on the circuit.
0.54 Ω
A battery with an emf of 12 V connects in parallel to a 50 Ω and 25 Ω resistor. The current through the battery is 0.70 A. What is the internal resistance of the battery? 17.1 Ω 16.6 Ω 0.54 Ω 0 Ω
U = ½ qV U = ½ (CV)V U = ½ (15E-3 F)(12 V)^2 U = 1.08 J
A capacitor with a capacitance of 15 mF is connected to a 12 V battery. How much energy will be stored in the capacitor?
F = qvBsinθ (3E-3 N) = (q)(40 m/s)(1.5 T) q = 5E-5 C Using the right-hand rule, the charge must be positive.
A charged particle moving at a speed of 40 m/s to the right enters a 1.5 T magnetic field directed upwards (i.e., up the page). The particle experiences a force of 3.0 mN out of the page. What is the charge on the particle?
The speed of the particle will remain the same
A charged particle moving to the right enters a uniform magnetic field directed into the page. The charged particle experiences a magnetic force. Which of the following is a true statement? (disregard any gravitational effects)? The speed of the particle will increase The speed of the particle will decrease The speed of the particle will remain the same Not enough information is stated. The type of charge (positive or negative) must be known
The voltage difference between the plates would decrease. Inserting a dielectric causes the overall electric field strength to decrease because the total electric field between the two plates is the sum of all the electric fields. Looking at the equation V = Er, there is a directly proportional relationship between voltage and field strength. The distance has not changed so the r will be the same, but since the electric field strength decreased the voltage would decrease as well. Inserting a dielectric would cause the electric field strength to decrease because the total electric field between the two plates is the sum of all the electric fields. The total charge would remain unchanged because charge is not affected by a dielectric. Charge cannot transfer on or off the capacitor plates. Looking at the equation C = Q/V, decreasing the voltage would cause C to increase thus allowing the same amount of charge to be stored at a lower voltage. The energy stored in the capacitor would decrease. The dielectric causes the voltage to drop, which would also cause the energy stored in the capacitor to decrease as well. Looking at the equation Uc = ½ qV, energy and voltage have a directly proportional relationship, so since the voltage decreased, the energy stored in the capacitor will decrease as well.
A dielectric is inserted between the plates of a charged parallel plate air capacitor. Describe whether each of the following will increase or decrease and explain your reasoning. a) Voltage difference between the plates b) Electric field strength between the plates c) Total charge stored on the plates d) Energy stored in the capacitor
P = IV (32 W) = I (120 V) I = 0.27 A V = IR (120 V) = (0.27 A) R R = 444 Ω Using unit analysis: Cost = (32 W)(1 kW / 1000 W)($0.14 / kW-hr)(24 hours / 1 day)(30 days) Cost = $3.23
A florescent lightbulb is rated at 32 W. If there is a voltage difference of 120 V across the bulb, what is the current through the bulb? b. What is the operational resistance of the bulb? c. What is the cost of leaving the bulb on for 30 days if the cost of energy is $0.14 per kW-hr.?
b. A push or pull
A force is best described as: a. The ability to do work b. A push or pull c. The rate at which energy is transferred
Experience no change in brightness
A light bulb connects to a battery. A second, identical light bulb connects in parallel to the first light bulb (see below). The connecting wires have negligible resistance. When the switch is closed, the original light bulb will No longer light Become brighter Become dimmer Experience no change in brightness
Charge will conduct between the two metal spheres until they have the same total charge. Negative charge will continue to move off the negative sphere until they both reach a final charge of +1 μC.
A metal sphere with a net charge of +5 μC is brought into contact with a second, identical metal sphere that has a net charge of -3 μC. What is the final charge on both spheres after they are brought into contact?
Since the electroscope has a net positive charge, the gold foil leaves will already be separated as the net positive charge causes a repelling force. If a negatively charged rod is brought near the top of the electroscope, some of the negative charge still present in the electroscope will be repelled down into the leaves causing the leaves to become more neutral. The leaves will move closer together. If the rod had a net positive charge, then the remaining negative charge in the electroscope would be attracted to the rod, and the leaves would become even more positively charged causing them to move further apart.
A negatively charged rod is brought near the top of a positively charged electroscope. In response, the leaves of the electroscope will: a) separate further. b) move closer together. c) have no response. d) not enough information. Explain your answer.
The negatively charged sphere will polarize the pith ball pushing the negative charge to the left side of the pith ball. Coulomb's law shows the attractive static force between the negative sphere and the closer positive side of the pith ball will be larger than the repelling force between the negative sphere and the further away negative side of the pith ball. The pith ball will move into contact with the negative sphere. Once the pith ball is in physical contact with the metal sphere, negative charge will conduct onto the pith ball giving the pith ball a net negative charge. The negatively charged sphere and the negatively charged pith ball will now repel. The pith ball will repel into the neutral sphere, and will conduct its excess negative charge onto the neutral sphere. The pith ball will now be neutrally charged, and the process begins again. Unless the neutral sphere is grounded, the negative charge between the two metal spheres will eventually balance, and the pith ball will experience the same static electric forces from both sides causing it to remain in equilibrium in the middle between the two.
A neutral pith ball (a small Styrofoam ball) coated with graphite is suspended between a negatively charged metal sphere and a neutral metal sphere. The pith ball will bounce back and forth between the two metal spheres for a while until it eventually comes to a stop suspended midway between the two. Explain this behavior.
C = ε0 A/d C= ε0 A/2d Ui = ½ Q2/C1 Ui = ½ (50x10^-6 C^)2 / (25x10^-6 F) Ui = 50x10^-6 J C2 = ½ C1 C2 = ½ (25x10^-6 F) C2 = 12.5x10^-6 F Uf = ½ Q2/C2 Uf = ½ (50x10^-6 C)^2 / (12.5x10^-6 F) Uf = 100x10^-6 J W = Uf - Ui W = 100x10^-6 J - 50x10^-6 J W = 5.0x10^-5 J
A parallel plate air capacitor has a capacitance of C (25 µF). The charge on each plate is Q (50 µC). There are no batteries connected to the capacitor. How much work is required to double the separation distance between the plates?
C = εA/d C = (8.85E-12 F/m)(0.1 m^2) / (0.02 m) C = 4.4E-11 F This is a small capacitance. C = Q / V (4.4E-11 F) = Q / (12 V) Q = 5.3E-10 C This is a small amount of charge, which is not surprising given the small capacitance.
A parallel plate air capacitor has a plate area of 0.10 m2 and a plate separation distance of 0.02 m. What is the capacitance of the capacitor? b) How much charge can be stored in the capacitor if it is connected to a 12 V battery?
A = L x W A = (0.75 m) x (0.5 m) A = 0.375 m^2 C = εA/d C = (8.85x10^-12 F/m) (0.375m^2) / (0.003 m) C = 1.1x10^-9 F C = Q / V (1.1x10-9 F) = Q / (12V) Q = (1.1x10-9 F) (12V) Q = 1.3x10-8 C
A parallel plate air capacitor has a plate separation distance of d (3 mm), and the plate area measures L (0.75 m) by W (0.5 m). What is the capacitance of the capacitor? b) How much charge can this capacitor hold if connected to a 12V battery?
V = Er (12 V) = E(0.15 m) E = 80 N/C Adding a dielectric with no external constant voltage source will cause the field strength to decrease by a factor equal to the dielectric constant. Can you explain why? E = (80 N/C) / 3.4 E = 23.5 N/C The potential difference between the plates will remain equal to the battery voltage for as long as the battery is connected. Since the plate separation distance remained constant: V = Er The field strength also remains constant. The field strength will remain 80 N/C.
A parallel plate air capacitor has an initial voltage difference of 12 V and a plate separation distance of 15 cm. A dielectric is inserted between the two plates. If the dielectric constant is 3.4, what is the new electric field strength between the plates? There are no voltage sources connected to the plates. b) What would be the new field strength between the plates if the plates were connected to a 12 V battery while the dielectric is placed between the plates?
C = Q/V Q = CV Q = (0.02 F) (12V) Q = 0.24 C C' = kC C2 = (3.2) (0.02 F) C2 = 0.064 F U = ½ Q^2/C2 U = ½ (0.24 C)^2 / 0.064 F U = 0.45 J
A parallel plate air capacitor with a capacitance of C (0.02 F) is connected to a 12V battery and charged. The capacitor is then disconnected from the battery and a dielectric with a dielectric constant of k (3.2) is inserted between the plates. How much energy will be stored in the capacitor after inserting the dielectric?
100 m/s
A particle with a mass of 0.01 kg and a net charge of -0.05 C accelerates from rest through a uniform electric field. If the strength of the field is 2000 V/C, what is the speed of the particle after traveling for 0.5 m? 100 m/s 10000 m/s 20000 m/s 141 m/s
E = Fe / q Fe = E x q (1200 N/C) = Fe / (2µC) Fe = (1200 N/C) (2.0x10^-6 C) Fe = 2.4x10^-3 N
A particle with a net charge of +q (+2µC) is placed in a uniform electric field that has a field strength of E (1200 N/C). What is the force on the particle?
14.4 N
A particle with a net charge of -2 μC is halfway between two particles with a net charge of +4 μC and +6 μC, respectively. The two positively charged particles are 10 cm apart. What is the magnitude of the net force on the negative particle? 72 N 14.4 N 3.6 N 18 N
E = F / q (500 N/C) = F / (6E-9 C) F = 3E-6 N
A particle with a net charge of -6 nC is placed in a uniform electric field with a field strength of 500 N/C. What is the magnitude of the force on the particle?
equal to 10 Hz.
A particular alternating voltage supply completes 10 full voltage cycle every 1 sec. A resistor connects to the alternating voltage supply. The resulting current will cycle at a frequency greater than 10 Hz. less than 10 Hz. equal to 10 Hz. More information is required. The amount of resistance must be known.
Work is done on the particle so that the particle will end with more energy.
A positive particle begins at a position near a negative plate and ends at a position near a positively charged plate. Which statement is most accurate? Work is done on the particle so that the particle will end with more energy. Work is done by the particle so that the particle will end with more energy. Work is done on the particle so that the particle will end with less energy. Work is done by the particle so that the particle will end with less energy.
The potential has nothing to do with the test charge. Potential depends only on the distance from the source charge. Positions closer to the positive source have a greater potential than positions far away. Since both particles moved towards the source, both particles moved to positions with higher potential. The correct answer is C
A positive particle is moved towards a positive source. A negative particle is also moved towards a positive source. In each case, a) the positive particle moved to a higher potential while the negative moved to a lower potential. b) the positive particle moved to a lower potential while the negative moved to a higher potential. c) both particles moved to a higher potential. d) both particles moved to a lower potential. Explain your answer.
1.8 x 10^-5kg
A positively charged 2 mC particle moving in the upwards direction at a speed of 20 m/s enters a 3.0 T field directed into the page. The particle path curves with a radius of 6 cm. What is the mass of the particle? 0.22 kg 3 x 10^-4 kg 5 x 10^-3 kg 1.8 x 10^-5kg
ΣF = ma qvBsinθ = mv^2/r (4E-3 C)(3.5 T) = (m)(200 m/s) / (0.1 m) m = 7E-6 kg
A positively charged 4 mC particle moving upwards at a speed of 200 m/s enters a 3.5 T field directed into the page. The particle path curves with a radius of 10 cm. What is the mass of the particle?
c. Forces require two objects with mass but fields only require one object
A primary difference between gravity forces and gravity fields is that: a. Forces and fields require only a single object with mass b. Forces and fields both require the presence of at least two objects with mass c. Forces require two objects with mass but fields only require one object d. Forces require only one object with mass but fields require two objects
K = ΔUe 1/2mv^2 = qV ½ (9.11E-31 kg)(v^2) = (1.6-19 C)(1000 V) v = 1.87E7 m/s
A proton is accelerated from rest a distance of 0.5 m through a voltage difference of 1000 V. What is the final speed of the proton when it leaves the electric field?
Ue = K qEr = 1/2mv^2 (1.6x10^-19 C)(E)(0.01 m) = ½(1.67x10^-27 kg)(4x10^5 m/s) E = 0.21 N/C
A proton starts from rest, and is accelerated through a uniform electric field. If the proton reached a velocity of 4x10^5 m/s after traveling 1 cm in the field, what is the field strength?
a. Magnitude
A scalar measurement is one that has: a. Magnitude b. Direction c. Both A and B
in addition to producing light, the incandescent bulb also dissipates more heat than the CFL.
A student claims that a 60 W incandescent light bulb and a 15 W CFL light bulb are equally bright (i.e., produce the roughly the same amount of light each second). This is because the two bulbs dissipate about the same amount of total energy each second. in addition to producing light, the incandescent bulb also dissipates more heat than the CFL. the resistance of the CFL bulb is smaller than the resistance of the incandescent bulb. the student is wrong; these two bulbs cannot be equally bright.
c. Both the above are true
A student lifts a ball into the air, and releases it. Which of the following statements is/are accurate? a. The student performs work on the ball, which the ball stores as gravitational potential energy. b. When the ball falls, gravitational potential energy is converted to kinetic energy c. Both the above are true
Since current and resistance are inversely proportional, if the resistance doubles, the current will halve. Voltage and resistance are not related to each other. Changing out one resistor for another will have no effect on the voltage provided by the battery to the circuit. Similarly, changing the voltage will have no effect on an ohmic resistor.
A student replaces a resistor in a circuit with a new resistor that has double the resistance of the original. How will this affect the following: a) Current b) Voltage
1500 eV.
An electron moves through a potential difference of 1500 V. The magnitude of the change in potential energy experienced by the electron is 1500 J. 1500 eV. More information is needed. The distance traveled needs to be known. More information is needed. The energy depends on whether the field was uniform or nonuniform.
P = W / t (2000 W) = W / (600 sec) W = 1.2E6 J Q = mLv (1.2E6 J) = m(2257 kJ/kg)(1000 J / 1 kJ) m = 0.53 kg P = V^2 / R (2000 W) = (120 V)^2 / R R = 7.2 Ω V = IR (240 V) = I (7.2 Ω) I = 33.3 A The power dissipated by the blow dryer will now be: P = IV P = (33.3 A)(240 V) P = 8000 W The blow dryer was not designed for this amount of energy. It will likely overheat, and possibly catch on fire.
A student takes 10 minutes to blow dry her hair with a 2000 W blow dryer. How much energy was dissipated by the blow dryer during this time? Review Challenge: What mass of water can be phase changed by this amount of energy? b. If the blow dryer is designed to operate at 120 V, what current will the device draw if it is plugged into a 240 V European outlet? Why is this dangerous?
c. Both magnitude and direction
A vector measure is any measurement that has: a. Only magnitude b. Only direction c. Both magnitude and direction
c. Both A and B
A vector measurement is best described as: a. A measurement that has magnitude b. A measurement that has direction c. Both A and B
Solve for the field strength produced by the wire at the distance indicated. B = μI / (2πr) B = (4πE-7 N-m/A)(30 A) / (2π*2E-3 m) B = 0.003 T Since you're looking at a point above the wire, the field will be directed into the page. Solve for the force on the charged particle. F = qvBsinθ F = (4E-3 C)(1000 m/s)(0.003 T) F = 0.012 N The field is into the page, and the direction of motion is left so the force will be directed down the page (towards the wire).
A wire carries 30 A of current directed to the left. What is the initial magnitude and direction of the force on a +4.0 mC charged particle that is initially moving at a speed of 1000 m/s to the left at a distance 2.0 mm above the wire?
3.2 x 10^-3 N F = u0 i q v / (2 pi r)
A wire carries 8 A of current directed to the left. What is the magnitude of the initial force on a +6 mC charged particle that is moving at a speed of 2000 m/s to the left at a distance 6.0 mm from the wire? 2.7 x 10^-4 N 3.2 x 10^-3 N 4.32 N 0 N
Steady state current means the quantity of charge moving through an area each second remains a constant amount of charge. When the two oppositely charged capacitor plates are connected, large quantities of charge will move through the wire at first but will steadily decrease until there is no net flow of charge between the plates.
A wire is connected to the two plates of a charged capacitor. Explain why the charge movement through the connecting wire is not considered steady state current.
right
A wire oriented so that the current flows into the screen experiences a force that deflects downward. The direction of the magnetic field must be left right down up
One eV is the energy gained by moving one electron through a potential difference of 1 V. Since the electron moved through 12 V, the electron will have gained 12 eV. Ue = 12 eV To convert: (12 eV)(1.6E-19 J / 1 eV) = 1.92E-18 J Alternatively: V = Ue / q (12 V) = Ue (1.6E-19 C) Ue = 1.92E-18 J Which can then be converted to eV The answer does not depend on whether the field was uniform or nonuniform. (This is shown in the above math.)
An electron is accelerated from rest through a potential difference of 12 V. How much energy does the electron have when it leaves the field? Record the answer in units of J and eV. Does it matter if the field is uniform or nonuniform?
Energy start = energy end Uinitial = Ufinal + K 1/2mv^2 = (kqq/r)initial - (kqq/r)final ½(9.11x10^-31 kg)(v^2) = (9x10^9 N-m^2/C^2)(-1.6x10^-19 C)(-5x10^-91.62 C)(1/0.02 m - 1/0.03 m) v = 1.62x10^7 m/s
An electron is released from rest from a position 2 cm from a -5 nC source charge. What is the speed of the electron when it has reached a position 3 cm from the source charge?
The electron will lose energy and move to a higher potential.
An electron moves from a position next to a negatively charged source to a position next to a positively charged source (see below). Which statement is accurate? The electron will gain energy and move to a lower potential. The electron will lose energy and move to a lower potential. The electron will gain energy and move to a higher potential. The electron will lose energy and move to a higher potential.
negative
An electroscope is a device that has small gold foil strips suspended from a metal rod. A student brings a negatively charged rod near the top of a neutral electroscope. The student then briefly touches a hand to the top of the electroscope before removing the charged rod. The foil strips of the electroscope now spread apart. The student then brings a different rod near the top of the electroscope. The leaves of the electroscope move closer together (see below). The new rod is most likely to have a net charge that is positive. negative. neutral. not enough information.
separate further.
An electroscope is a device that has small gold foil strips suspended from a metal rod. If the electroscope is negatively charged, the foil strips will repel and separate from each other (see below). If you bring a negatively charged rod near the top of a negatively charged electroscope, the foil strips will separate further. move closer together. not move at all. not enough information.
· If the plates are isolated, then there is no way for charge to transfer on or off the plates; the total charge will not change. · Since oppositely charged plates attract each other, work must be done on the plates in order to separate them. This additional energy will be stored in the capacitor as potential energy. The stored energy will increase. · Voltage is the ratio of energy per test charge in a field. Since work was done on the system, the amount of energy available per test charge has increased, which means the potential difference between the plates will increase. · Mathematically, V = Er. Since voltage has increased, and distance has also increased, the field strength will remain constant. Conceptually, this is because the field is the vector result of all the forces applied by the source per quantity of charge on the test charge in the field. If you treat the bar like a collection of point source charges (see below), then the force diagram for the test charge would look like this: Remember the plate is infinitely long so there are a lot more forces - you just don't bother to include them since they get increasingly small as you move away from the test charge. The horizontal forces cancel out, and you add all the vertical components to solve for the net force. Now, imagine pulling the bar away from the test charge. As the source and test charge move apart, the force between them decreases; all the force arrows get shorter. However, the angle of the force arrows also changes. As the bar is pulled away, the horizontal component decreases, and the vertical component increases resulting in more overall vertical force. The two changes occurring at the same time mean the overall force on the test particle remains the same. To get the field strength, take the net force, and divide by the quantity of charge on the test particle. · Capacitance is determined by the design of the capacitor. As the distance between plates increases, the attraction between plates decreases, which means the same applied voltage will have a harder time pushing charge onto the plate. Capacitance will decrease.
An infinitely long parallel plate air capacitor has an initial voltage difference of 9 V. There are no voltage sources connected to the plates. The distance between the plates is increased. What effect does this have on each of the following? - Net charge on the top plate - Stored energy - Voltage - Field strength - Capacitance picture part of answer
· Since the battery is still connected, the potential difference between the two plates will remain equal to the potential difference of the battery. The voltage remains the same. · The attraction between plates decreases with increasing distance, which means the repelling forces between the excess charge on the top plate increase. The plate is not isolated, which means some of the excess charge will be pushed onto the bottom plate. The net charge on the top plate will decrease. · While you still have to perform work on the plates to separate them, you also have to provide energy to move the charge off the top plate onto the bottom plate. The energy stored in the capacitor will decrease. · Mathematically, V = Er. Since V remained constant, and r increased, E must decrease. Conceptually, everything you said in problem 2 is still accurate except that now there are fewer charges applying forces. Since there are fewer charges applying forces, the net force will decrease, which means the ratio of force per test charge will decrease. · As before, capacitance is determined by the design of the capacitor. As the distance between plates increases, the attraction between plates decreases, which means the same applied voltage will have a harder time pushing charge onto the plate. Capacitance will decrease.
An infinitely long parallel plate air capacitor is connected to a 9 V battery. With the battery still connected, the distance between the plates is doubled. What effect does this have on each of the following? - Voltage - Net charge on the top plate - Stored energy - Field strength - Capacitance
Protons in an atom are fixed in place; they are not mobile. Electrons, however, are mobile. Since electrons carry the negative type of charge, electrons must have been transferred off the material leaving behind more positive charge than negative. The amount of charge carried by one electron is 1.6x10^-19 C. Divide the total charge by the charge of one electron to solve for the number of electrons. (3.5x10^-6 C) / (1.6x10^-19 C) = 2.25x10^13 electrons were transferred off the object.
An object has a net charge of +3.5 μC. What does this mean? a) Electrons were transferred onto the object. b) Electrons were transferred off the object. c) Protons were transferred onto the object. d) Protons were transferred off the object. e) More than one of the above are possible. Explain your answer. Solve for the number of charge carrier particles (protons or electrons) transferred on or off the object.
a. A fundamental property of matter
Charge is best defined as: a. A fundamental property of matter b. The amount of material an object has c. The push or pull between two objects
c. both of the above.
Charge is best defined as: a. a fundamental property of matter. b. the property that is responsible for electrical effects. c. both of the above.
c. both of the above.
Charge movement is the result of: a. electrical forces. b. interaction between electrical fields. c. both of the above.
1/9 F.
Charged particles A and B experience a static electric force of F. If the distance between the charge particles triples, then the static electric force becomes 1/9 F. 1/3 F. 3 F. 9 F.
Polarization is a redistribution of charge rather than a transfer of charge; the total amount of charge on the polarized object does not change. Conduction is the transfer of charge between objects. Induction is a charge transfer process that involves polarizing an object and then allowing charge to transfer to another object through conduction. While still under the influence of the dipole, the charge transfer mechanism is removed (the two objects are moved apart). The two objects are now charged.
Compare and contrast conduction, induction, and polarization.
Electric field lines are drawn in the direction of the force the source charge would put on a positive test charge (i.e., field lines are drawn away from positive sources and towards negative sources). Equipotential lines are drawn perpendicular to the electric field lines.
Compare and contrast electric field lines and equipotential lines.
A static electric force is the amount of push or pull between two charges. If there is only one charge present, there can be no static electric force because there is nothing for that charge to push or pull. An electric field is a way of describing how well a single charge (or a collection of charges) might push or pull on a hypothetical charge depending on where that hypothetical charge was placed. Electric fields are a way of looking at the space surrounding a charge rather than the interaction between two charges.
Compare and contrast electric forces and electric fields.
Permanent magnets typically have field strengths from 5x10-3 T up to 2 T. The strength of a permanent magnet is fixed (unless the magnet becomes demagnetized), and is always active. Electromagnets have variable field strengths that can be much, much greater than 2 T. The field strengths depend on the amount of current used to create the magnet. Electromagnets can also be turned on and off.
Compare and contrast magnetic fields created by permanent magnets versus electromagnets.
Electrons carry negative charge, and so move away from negative charge sources and toward positive charge sources. Conventional current is the direction of positive charge flow, and so conventional current is always away from the positive source and towards the negative. It's useful to note that, in both cases, charge is moving away from regions of high potential and towards regions of low potential.
Compare and contrast the direction of electron flow and the direction of conventional current.
a. positive charge.
Conventional current is based on the movement of: a. positive charge. b. negative charge. c. any charge.
voltage differences.
Current is caused by voltage differences. resistance. capacitance. none of the above.
Current is the flow rate of charge or the amount of charge that flows through a given area in one second. Velocity is the distance that charge travels in one second. Drift velocity is the average distance along the length of a wire traveled by the moving charge carriers every second. Drift velocity tends to be relatively slow.
Define current. How is current different from velocity?
Any positive point sources will act like hills or peaks, while any negative point sources will act like valleys or sink-holes. The uniform field will slope uniformly down from the positive plate to the negative plate (like an embankment). If you have trouble visualizing these, there are several 3D equipotential rendering programs you can operate from your browser.
Describe how each of the previous 2D contour maps would look in 3D.
A uniform field has the same field strength at all points in the field, and will apply the same force to a charge placed at any location in the field. Uniform fields are also unidirectional. The field created by a point source is nonuniform, and becomes weaker as you move away from the source. The amount of force on a charge placed in a nonuniform field depends on the distance between the charge and the source. Nonuniform fields are not unidirectional.
Describe the differences between a uniform field and a field created by a point source.
lost energy no chage in energy gained gained gained lost
Describe whether the positively charged particle gains or loses electrical potential energy in each of the following processes. b. Describe whether the negatively charged particle gains or loses electrical potential energy in each of the following processes.
conventional current
Direction of current described in terms of + charge flow § If discussing current, it is almost always conventional current § If discussing the direction of electron flow, they will generally not refer to this as current but specifically as electron flow
An electric field has a positive test charge. The field and positive charge will experience a repelling force due to the positively charged field and positive charge. Two of the same types of charges will repel each other. An electric field has a positive test charge. The field and positive charge will experience an attractive force due to two opposite charges. Two different types of charges will be attracted to each other. An electric field has a positive test charge. This means the negative charge will experience attracting forces with the field since two different charges will attract and the positive charge will experience repelling forces with the field since two of the same types of charges will repel. If the positively charged field is placed by the positive plate, the positive plate will apply a large repelling force while the negative plate will apply a small attracting force. If it was placed near the negative plate the negative plate would apply a large attracting force while the positive plate would apply a small repelling force. The total force on the field will be the same no matter where it is placed because this would be a uniform field.
Draw the electric fields for each of the following combinations of point charges and charged parallel plates (2 points each). Case A (defined by your instructor): field near a positive charge Case B (defined by your instructor): field near a negative charge Case C (defined by your instructor): field near a positive and a negative charge Case D (defined by your instructor): field between two oppositely charged parallel plates picture part of answer
b. The presence of charge
Electric fields are created by: a. The push or pull between two or more charges b. The presence of charge c. The movement of charge
b. towards negatively charged sources.
Electric fields are in a direction: a. towards positively charged sources. b. towards negatively charged sources. c. both of the above.
he ratio of energy per amount of charge on a particle in an electric field.
Electric potential is work either done on or by a charged particle. the ratio of energy per amount of charge on a particle in an electric field. the force between two charged particles. the ratio of force per amount of charge on a particle in an electric field.
a. presence of charge.
Electrical fields are created by: a. presence of charge. b. interaction between two or more charges. c. presence of mass.
a. the presence of charge.
Electrical fields are created by: a. the presence of charge. b. the movement of charge. c. the presence of mass.
Answer: Since the object has extra of the positive type of charge, electrons must have been transferred off the material. Since electrons carry the negative type of charge, removing electrons would have left behind more positive than negative. The total amount of charge moved off the material was 6.5 μC. Each electron moved off carries 1.6x10^-19 C. Convert μC into C and divide. 6.5x10^-6 C / 1.6x10^-19 C = 4.06x10^13 electrons were removed.
Ex) A particular object has a net 6.5 μC of the positive type of charge. Were electrons transferred onto or off the object? How many?
Answer: I = Q / t I = (15 C) / (20 seconds) I = 0.75 A V = IR (12 V) = (0.75 A)(R) R = 16 Ohms
Ex) A particular resistor has 15 C of charge pass through it every 20 seconds. If the potential difference across the resistor is 12 V, what is the resistance of the resistor?
Answer: I = Q/t (0.5 A) = Q / (60 sec) Q = 30 C Number electrons = total charge / charge on 1 electron (30 C) / (1.6x10^-19 C) = 1.88x10^20 electrons
Ex) A particular wire has 0.5 A of current. How many electrons will pass through the wire in one minute?
Answer: E = F / q (100 N/C) = F / (1.6x10^-19 C)F = 1.6x10^-17 N
Ex) A proton is placed into a uniform electric field with a field strength of 100 N/C. What is the electric force on the proton
Answer: F = qvBsinθ F = (1.6x10^-19 C)(4x10^5 m/s)(0.5 T)(sin90) F = 3.2x10^-14 N Directed up the page (or computer screen)
Ex) A proton moves to the left at a speed of 4x10^5 m/s through a magnetic field with a strength of 0.5 T that is directed out of the page (or computer screen). What are the magnitude and direction of the Lorentz force on the proton?
Draw a best fit trend line (since the resistor is ohmic, you know the graph is linear) and solve for the slope (rise / run). R = (4.5-0 V) / (0.75 - 0 A) R = 6 Ω V = IR V = (2 A)(6 Ω) V = 12 V
Ex) A student applies various voltages to an ohmic resistor, and measures the resulting current. The data results in the following graph. What is the resistance of the resistor? How much voltage would it take to run a current of 2 A through the resistor?
Answer: F = BILsinθ F = (2.0T)(12 A)(0.10 m)(sin90) F = 2.4 N Current will move down through the bar; the field is out of the page so the force will be left. Obviously, this particular design would be more of a novelty projectile launcher than something that would do any real damage.
Ex) A titanium alloy bar with a mass of 0.10 kg is placed across two oppositely charged parallel rails. The distance between the rails is 10 cm. A current of 12 A runs through the bar. The whole system is in the presence of a uniform magnetic field with a strength of 2 T. The diagram below shows this with a top-down view. What are the magnitude and direction of the force on the bar?
Answer: B = uI/(2πr) B = (4πx10^-7 T-m/A)(0.2A) / (2π*0.05 m) B = 8x10^-7 T directed out of the page (or +z axis)
Ex) A wire has 0.2 A of current running horizontally to the left. What is the direction and magnitude of its magnetic field at a point 5 cm below the wire?
Answer: The electron starts with electric potential energy, which is converted into kinetic energy as the particle moves through the electric field. Ue = KVq = 1/2mv^2 (12 V) (1.6x10^-19 C) = ½ (9.11x10^-31 kg)v^2 v =2.05x106 m/s
Ex) An electron accelerates from rest through a uniform electric field with a voltage difference of 12 V. What is the speed of the electron as it leaves the electric field?
Answer: The magnetic and electric forces must cancel out. ΣF = 0 Fe - FB = 0 (Can you explain why the electric force is positive and the magnetic is negative?) qE - qvBsinθ = 0 q(V/r) - qvB = 0 V/r - vB = 0 (12 V) / (0.04 m) - (7.3x10^6 m/s)(B) = 0 B = 4.1x10^-5 T Since the magnetic force is down, and the negative charge is moving left (use your left hand for the negative particle), the field must be out of the page (or screen).
Ex) An electron has a velocity of 7.3x10^6 m/s directed towards the left. The electron will move between two oppositely charged horizontal parallel plates. The positive plate is on top. The plates have a voltage difference of 12 V and are 4.0 cm apart. What is the strength and direction of the magnetic field required to keep the electron moving between the plates in a straight line?
Answer: C = Q/V = εoA/d (0.25x10^-3 C) / (25 V) = (8.85x10^-12 F/m)(A) / (0.12) A = 1.36x105 m2 This is an extremely large area so this is not very reasonable. One way the designer can reduce this required area is to replace the insulator between the plates with some other material.
Ex) An electronics designer needs a parallel plate air capacitor that can store 0.25 mC of charge using a voltage of 25 V. If the distance between the plates is 12 cm, what must the area of the plates be? How reasonable is this design?
Answer: The negative charge on the rod repels the negative charge already present in the electroscope. Since the electroscope is made from metals, electrons will be pushed down into the gold strips. The electroscope is now polarized (the top is more positive, and the bottom is more negative). However, the relocated negative charge now in the thin gold strips will repel each other causing the gold strips to separate.
Ex) An electroscope is a device that has small gold foil strips suspended from a metal rod. If a negatively charged rod is brought near the top of a neutral electroscope, the gold foil strips will separate. (See below.) Explain why this happens.
Answer: If the field is directed to the east, then the positive plate is on the left and the negative plate is on the right. Since you're looking at a point to the right of the given point, you should expect the voltage to be smaller (the position is closer to zero). Solve for the position of the 80 V potential reading. V = Ue / q = Er (80 V) = (50 N/C)r r = 1.6 m Solve for the voltage with an additional 1.0 m subtracted from that position. V = ErV = (50 N/C)(1.6 - 1.0) V = 30 V
Ex) Consider a uniform electric field of 50 N/C directed towards the right. If the potential measured at some point in the field is 80 V, what is the potential at a position 1.0 m directly to the right of that point?
Answer: Point your right thumb to the right, and curl your fingers. When your fingers are positioned above your thumb, they point towards you (out of the page). The magnetic field above the wire will be out of the page. Point your right thumb to the right, and curl your fingers. When your fingers are below your thumb, they point away from you (into the page). The magnetic field below the wire will be into the page
Ex) Current moves to the right through a wire (see below). What is the direction of the magnetic field created by the wire at a position a) above the wire and b) below the wire?
Answer: Particle A has a positive charge. To move a positive charge from infinitely far away to its starting position near the negative source, the particle would have to lose electric potential energy. Particle A starts out with Ue as a negative value. To pull the positive particle away from the negative source, work would have to be performed on the particle (the movement requires an outside force). Particle A will gain electric potential energy during this process. Particle B has a negative charge. To move a negative charge from infinitely far away to its starting position near the negative source, the particle would have to gain energy (an outside force must perform work on the charged particle). Particle B starts out with Ue as a positive value. In order to pull the negative particle away from the negative source, no work needs to be done on the particle; the two particles repel each other. Particle B will lose electric potential energy during this process.
Ex) Examine the following system that contains a negative source charge and two charged particles in the field created by the source charge. Particle A is positive, and particle B is negative. Describe whether the particles start with positive or negative energy. Determine whether the particles will gain or lose energy when they are moved to the new indicated positions.
Answer: The distance between A and the positive source is: r^2 = (0.30 m^)2 + (0.15 m)^2 r = 0.34 m ΣUe = kqq/r + kqq/r Ue = (9x10^9 N-m^2/C^2)(9x10^-6 C)(-8x10^-3 C) / (0.15 m) + (9x10^9 N-m^2/C^2)(9x10^-6 C)(8x10^-3 C) / (0.34 m) Ue = -2414 J The negative on the final answer should make sense since the positive charge is closer to the negative source so the energy lost putting the charge near the negative source would be larger than the energy gained putting the charge near the positive source. Answer: You already have Uinitial but you still need to solve for Ufinal. Ufinal = (9x109 N-m2/C2)(9x10-6 C)(-8x10-3 C) / (0.34 m) + (9x109 N-m2/C2)(9x10-6 C)(8x10-3 C) / (0.15 m) Ufinal = +2414 J (Hopefully, this wasn't too surprising.) ΔUe = Ufinal - UinitialΔUe = (+2414 J) - (-2414 J)ΔUe = 4828
Ex) Examine the following system. Solve for the static electric energy of particle A if it has a net charge of +9 μC and q has a value of 8 mC. Ex) Using the same system from the previous example, solve for the change in electric potential energy required to move particle A horizontally to a position below the positive source charge.
Answer: Solve for the energy required to put particle A at a position relative to the negative source on the left. Ue = kqq/r Ue = (9x10^9 N-m^2/C^2)(2x10^-6 C)(5x10^-3 C) / (0.10 m) Ue = 900 J In order to put the negatively charged particle near the negative source from a position infinitely far away, work would need to be done on the particle so this term should be positive. You can then solve for the energy relative to the positive source charge. Ue = kqq/r Ue = (9x10^9 N-m^2/C^2)(2x10^-6 C)(5x10^-3 C) / (0.20 m) Ue = -450 J In order to put the negatively charged particle near the positive source from a position infinitely far away, work would need to be done by the particle so this term should be negative. The particle's total energy is thenUe = 900 J + -450 J = 450 J The positive sign should make sense since the negative particle is closer to the negative source than the positive source; placing the particle would require more work be done on the particle than by the particle.
Ex) Examine the system below. Determine the amount of electric potential energy stored in particle A if the value for q is 5.0 mC, and A has a net charge of -2 μC.
Answer: The brightest bulb is the one that emits the most energy per second (has the largest power rating). P = IV = V2/R As resistance increases, power decreases. The brightest bulb is the one with the smallest resistance - the 144 Ω bulb. P = (120 V)^2 / (144 Ω) P = 100 W P = (120 V)^2 / (240 Ω) P = 60 W
Ex) Houses in the United States are wired to provide 120 volts of electric pressure. Which lightbulb would be brighter: a bulb with an operational resistance of 144 Ω or a bulb with an operational resistance of 240 Ω?
Answer: C = Q/V = kεoA/d Q / (12 V) = 5 (8.85x10^-12 F/m)(0.05 m2) / (3x10^-3 m) Q = 8.85 x 10^-9 C
Ex) How much excess charge could be stored in a capacitor if it has a plate area of 0.05 m2, a plate separation distance of 3.0 mm, has a dielectric with a dielectric constant of 5, and is connected to a 12 V battery?
Answer: W = Ue W = qEr W = (3x10-6)(900 N/C)(0.25 m) W = 6.75x10-4 J
Ex) How much work is required to move a 1.0 g particle with a net charge of +3 μC to a point halfway between two oppositely charged parallel plates separated by a distance of 0.5 m? The electric field between the plates is a uniform 900 N/C.
Answer: Vrms = Vo /√2 (120 V) = Vo / (√2) Vo = 170 V
Ex) If the rms voltage of a standard U.S. wall outlet is 120 V, what is the peak voltage?
Answer: Sketch the scenario, and draw a vector diagram. Remember the test charges used to determine the direction of electric fields are, by convention, positively charged. Particle A will repel a positive test charge, while particle B will attract a positive test charge. E = kq / r^2 EA = (9x10^9 N-m^2/C^2)(15x10^-9 C) / (0.25 m)^2 EA = 2160 N/C EB = (9x10^9 N-m^2/C^2)(10x10^-9 C) / (1.25 m)^2 EB = 57.6 N/C There are no y components so the net field strength will be the sum of the x components. Note the field created by particle A points to the left so it should have a negative value. ΣE = ΣEx = EA + EBΣE = (-2160 N/C) + (57.6 N/C)ΣE = -2102 N/C
Ex) Particle A has a net charge of +15 nC, and particle B has a net charge of -10 nC. The two particles are placed 1.0 m apart with particle A on the left and B on the right. What is the net field strength measured 0.25 m to the left of particle A?
Answer: Sketch the scenario, and then draw a free body diagram for particle B Solve for the two static electric forces. F = kqq/r^2 F a on b = (9x10^9 N-m^2/C^2)(4x10^-3 C)(2x10^-3 C) / (2 m)^2 F a on b = 18,000 N F c on b = (9x10^9 N-m^2/C^2)(6x10^-3 C)(2x10^-3 C) / (3 m)^2 F c on b = 12,000 N Since these forces don't have any angles, you don't need to solve for their x and y components. Solve for the net force and the direction of the net force. ∑F^2 = Fy^2 + Fx^2 ∑F2 = (18,000 N)^2 + (12,000)^2 ∑F = 21,633 N θ = tan-1 (1.5) tan θ = (18, 000 N) / (12,000 N) θ = 56.6˚
Ex) Particle A has a net charge of +4 mC, and is located at position (0, 2) m.Particle B has a net charge of -2 mC, and is located at coordinate position (0, 0) m.Particle C has a net charge of +6 mC, and is located at position (3, 0) m.What is the net charge on particle B? Ignore any gravity effects. picture is part of answer
The magnetic force is always perpendicular to the direction of motion. This will cause a centripetal acceleration. The velocity of the particle will change but the speed will remain constant. Note if the field is not a uniform field then the particle will also change speed.
Explain why a charged particle moving through a uniform magnetic field will change direction but will not change speed.
Answer: Draw the system and the free body diagram. Solve for the forces. F = kqq/r^2F A on C = (9x10^9 N-m^2/C^2)(6x10^-6 C)(6x10^-6 C) / (4 m)^2 FA on C = 0.0203 N (this is the same magnitude as FD on C) FB on C = (9x10^9 N-m^2/C2)(6x10^-6 C)(6x10^-6 C) / (5.66 m)^2 FB on C = 0.0101 N Fx = (0.0101 N)(cos 45) = 0.00716 N (This is the same as Fy.) ΣFx = (0.00716 N) + (-0.0203 N) = - 0.0131 N ΣFy = (0.00716 N) + (0.0203 N) = 0.0275 N ΣF^2 = Fy^2 + Fx^2 ΣF^2 = (0.0275 N)^2 + (-0.0131 N)^2 ΣF = 0.0305 N tanθ = (0.0275 N) / (-0.0131 N)θ = 115.5°
Ex) Particles A and B have charges of +6 μC, and are located at positions (0, 4) m and (4, 4) m respectively. Particles C and D have charges of -6 μC, and are located at positions (0, 0) m and (4, 0) m respectively. Solve for the net force on particle C. picture is part of answer
Answer: Electric potential (voltage) is not dependent on the charge of the particle; voltage is strictly about what position is likely to lead to large energies, and not the energy value once a particle is actually placed there. This question is trying to be tricky by providing enough information to address energy itself since energy and potential are often confused. V = Er The most potential will be at a position that is "high up" or far away from the negative plate. Since electric fields are directed from positive source charges to negative source charges, the plate on the left must have a net negative charge. This means that A is at a position of low potential, B and C are tied, and both D and E are at positions with high potential: D = E > B = C > A
Ex) Rank the following particles in order from greatest to least electric potential. Particles A and B have charges of +1 μC, C and D have charges of +2 μC, and particle E has a charge of -2 μC.
Answer: C = F > A = B > D > E Recall that particles in a uniform magnetic field have Ue = qEr. So ΔUe = qEΔr. E is at the end because negative charges moving with electric fields will lose energy, and the question asked about energy gained. C and F both start with no Ue but have the same large q, and move the same large Δr. A and B both start with no Ue but have the same small q, and move the same large Δr. The horizontal element to B's motion does not affect its electric potential energy at all. D starts with Ue but has a small q and a small Δr. Note that D will end with the same final amount of Ue as B and C but the question asked about changes in potential energy not final quantities.
Ex) Rank the following particles in order from most energy gained to least energy gained. Particles A, B, and D have a charge of +1 μC, particle C has a charge of +2 μC, and particles E and F have a charge of -2 μC.
Answer: Sketch the circuit diagram. Since you know the voltage across the 20 Ω resistor is the same as the voltage at the battery, you can use Ohm's law to solve for the current through the 20 Ω resistor. V = IR (12 V) = I (20 Ω) I = 0.6 A
Ex) Resistors of 10 Ω, 20 Ω, and 40 Ω are all connected in parallel to a 12 V battery. What is the current through the 20 Ω resistor? picture part of answer
Answer: Sketch the circuit diagram, and solve for the total resistance. R = (10 + 20 + 40) = 70 Ω Solve for the current through the circuit. V = IR (12 V) = I (70 Ω) I = 0.17 A This current is constant throughout the entire series circuit. This means the current through the 20 Ω resistor is also 0.17 A. You can use the current through the resistor along with the resistance of the resistor to solve for the voltage drop across the resistor. V = IR V = (0.17 A)(20 Ω) V = 3.4 V
Ex) Resistors of 10 Ω, 20 Ω, and 40 Ω are all connected in series to a 12 V battery. What is the voltage drop across the 20 Ω resistor? picture part of answer
Answer: Solve for the total resistance: 1/Rp = 1/R + 1/R + 1/R 1/Rp = 1/(100 Ω) + 1/(200 Ω) + 1/(400 Ω) Rp = 57.1 Ω Note that this resistance is smaller than any one of the individual resistors; the total resistance has decreased. P = IV = V2/R P = (9 V)2 / (57.1 Ω) P = 1.42 W
Ex) Resistors of 100 Ω, 200 Ω, and 400 Ω are all connected in parallel to a 9 V battery. What is the power supplied by the battery?
Answer: Solve for the total circuit resistance. Rs = (100 Ω) + (200 Ω) + (400 Ω) Rs = 700 Ω V = IR (9 V) = I (700 Ω) I = 0.013 A
Ex) Resistors of 100 Ω, 200 Ω, and 400 Ω are all connected in series to a 9 V battery. What current flows through the battery?
Answer: Solve for the total circuit resistance and then the current through the battery. Series combination: R = 3+4 = 7 Ω Parallel combination: 1/R = ½ + 1/7 R = 1.6 Ω Series combination: R = 8 + 1.6 = 9.6 Ω V = IR (9 V) = I (9.6 Ω) I = 0.94 A Start to "unwind" the circuit back into its original form. The 1.6 Ω equivalent resistor shares the same beginning and ending connecting wires as the 2 Ω resistor (compare the below diagram with the initial diagram). The voltage difference must be the same. V = IR V = (0.94 A)(1.6 Ω) V = 1.5 V Further unwind the circuit. You have the voltage drop across the 2 Ω (which is also equal to the voltage drop across the equivalent 7 Ω resistor). Solve for the current using Ohm's law. V = IR (1.5 V) = I (2 Ω) I = 0.75 A
Ex) Solve for the current through the 2 Ω resistor.
Answer: The 4 and 1 Ω resistors are in series (no branches split off between them). R = 1 + 4 = 5 Ω The 2 and 5 Ω resistors are in parallel (they share start and end connections). 1/R = ½ + 1/5 R = 1.43 Ω The 6 and 1.43 Ω resistors are in series (no branches split off between them). R = 6 + 1.43 R = 7.43 Ω V = IR (9 V) = I (7.43 Ω) I = 1.21 A
Ex) Solve for the current through the battery.
Answer: The distance between A and the positive source is: r^2 = (0.30 m)^2 + (0.15 m)^2 r = 0.34 m ΣV = kq/r + kq/r ΣV = (9x10^9 N-m^2/C^2)(-0.2x10^-6 C) / (0.15 m) + (9x10^9 N-m^2/C^2)(0.2x10^-6 C) / (0.34 m) ΣV = -12000 V + 5294 V ΣV = -6706 V
Ex) Solve for the voltage at the location of particle A if q has a value of 0.2 μC.
Answer: Force has an inverse-squared relationship with distance. If the distance is doubled, then the force will change by a factor of (1/2)2 = ¼. In other words, the new force will be four times smaller. Since the two particles are both positive, the force will be a repelling force.
Ex) The distance d between two positively charged particles is measured. The distance between the two particles is increased to 2d. If the original electrostatic force between the objects was F, what is the new force?
When the dielectric is inserted, the field between the plates will cause the dielectric to polarize, which will cause weak electric fields to form in the dielectric. These electric fields will be in the direction opposite the field between the two capacitor plates. The superposition of fields will result in an overall weaker field.
Explain why a parallel plate capacitor with a dielectric will have a smaller field strength between the plates than a similarly designed air based parallel plate capacitor.
Answer: Each parallel branch will experience a voltage drop equal to the voltage gain across the battery, which means all the resistors have equal voltage drops. Ohm's law says V = IR. Since all the resistors have the same voltage, the resistor with the largest resistance will experience the smallest flow rate of charge. The 4R resistor will have half the current the 2R resistor has and a quarter of the current the R resistor has. Power can be determined by P = IV. Since the voltage across each resistor is the same, the resistor with the largest current will dissipate the most power. The resistor with the largest current is the one with the smallest resistance. The R resistor will dissipate twice the power as the 2R resistor and four times the power as the 4R resistor.
Ex) Three resistors are connected in parallel to a battery as shown below. Which resistor experiences the largest voltage drop? Which resistor has the largest current? Which resistor dissipates the most power?
Answer: Since the resistors are connected in series, they will all have the same current. This is due to conservation of charge; charge is not created or destroyed within the loop. According to Ohm's law, V = IR. Since all the resistors have the same current, the resistor with the largest voltage drop will be the one with the largest resistance. The 4R resistor will have the largest voltage drop. The 4R resistor will have a voltage drop that is four times larger than the voltage drop across R and twice the voltage drop of the 2R resistor. Power can be found by P = IV. Since the current is the same through all resistors, the resistor with the largest voltage drop will dissipate the largest power. The 4R resistor will give off four times more energy each second than the R resistor and two times more energy each second than the 2R resistor
Ex) Three resistors are connected in series to a battery as shown below. Which resistor experiences the greatest current? Which resistor has the largest voltage drop? Which resistor dissipates the most power?
Answer: Solve for the magnetic field strength created by the top wire at a distance where the bottom wire is located. B = uI/(2πr) B = (4πx10^-7 T-m/A)(5.0 A) / (2π*0.2 m) B = 1.6x10^-6 T Solve for the force acting on the current-bearing bottom wire in the magnetic field created by the top wire. F = BILsinθ F = (1.6x10^-6 T)(10.0 A)(2.0 m)(sin90) F = 3.2x10^-5 N Use the right-hand curl rule to determine the direction of the field below the top wire. The field is directed into the page. Use the right-hand rule to determine the direction of the force. The force will be directed down the page.
Ex) Two horizontal wires are spaced 20 cm from each other. The top wire has 5.0 A of current directed to the right. The bottom wire has 10 A of current directed to the left. The wires are 2.0 m long. What are the magnitude and direction of the force of the top wire on the bottom?
Answer: Draw arrows to show the force between each source and the test charge (remember test charges are positive). The positive source will repel strongly (the x position is relatively close to the positive source), while the negative source will attract weakly (the x position is relatively far from the negative source). Add the field arrows together to find the overall field arrow (in red below).
Ex) Two oppositely charged particles have the same quantity of charge. Draw an arrow indicating the direction and strength of the net electric field at the position marked by the x below.
Answer: B = uI/(2πr) B = (4πx10^-7 T-m/A)(1.0 A) / (2π*0.25 m) B = 8x10^-7 T This is the field strength for both wires since the point of interest is equidistant, and the wires have the same current. However, the wire on the left will produce a field directed into the page while the wire on the right will produce a field that is directed out of the page. The two fields will cancel each other out. ΣB = (-8x10^-7 T) + (+8x10^-7 T) = 0
Ex) Two wires oriented vertically along the page (or computer screen) each have a current of 1.0 A directed upwards. If the wires are 0.5 m apart, what is the strength of the magnetic field halfway between them?
Answer: Positions closer to positive sources are high potential while positions closer to negative sources are low potential (again, think of these as 2D drawings of 3D electric elevation maps). Since D and E are on the same equipotential, they will have the same voltage. Since B is on the equipotential line closest to -, it will have the lowest voltage. D = E > C > A > B
Ex) Use the following equipotential map to rank the particles in order from largest potential to smallest potential. Explain your reasoning.
Answer: P = IV P = (2.37 A)(19 V) P = 45 W Using unit analysis: Cost = (45 W) (1 kW / 1000 W) ($0.12 / 1 kW-hr.)(672 hr.) Cost = $3.62
Ex) Use the following product label to determine the approximate cost of running the device for 4 weeks straight if the cost of energy is 12 cents per kW-hr.
Answer: if the test charge is placed near the + sheet, + sheet would apply a large repelling force and - sheet apply small attracting force. If test charge is placed near - sheet, - sheet applies large attracting force, + applies small repelling force. Total force on test charge will be same no matter where particle is placed: this is a uniform field
Ex) What if you had two oppositely charged metal sheets placed near each other and placed a test charge in between? picture is part of answer
Answer: Series: 1/C = 1/C1 + 1/C2 + 1/C3 1/C = 1/(5 μF) + 1/(5 μF) + 1/(5 μF) 1/C = 0.6 C = 1.67 μF Parallel: C = C1 + C2 + C3 C = (5 μF) + (5 μF) + (5 μF) C = 15 μF The parallel combination has 15, which is ≈ 9 times more overall capacitance than the series combination at 1.67.
Ex) What is the difference in overall capacitance if three 5 μF capacitors are wired together in series versus parallel?
Answer: E = kq/r^2 E = (9x10^9 N-m^2/C^2)(8x10^-9 C) / (0.5 m)^2 E = 288 N/C To solve for the field strength at the new position, you can resolve the entire problem after replacing r = 1.0 m. An easier way to solve the question would be to consider the relationship between field strength and position. Field strength has an inverse-square relationship with distance. Since you've doubled the distance, the field strength will decrease by a factor of 4. E = (288 N/C) / 4 = 72 N/C
Ex) What is the electric field strength of a -8 nC charged particle at a position measured 0.5 meters away? How does this compare to the strength of the field measured 1.0 m away from the source charge?
Answer: Ue = kqq/r Ue = (9x10^9 N-m^2/C^2)(1.5x10^-6 C)(3x10^-3 C) / (0.2 m) Ue = 202.5 J
Ex) What is the electric potential energy of a particle with a net charge of +1.5 μC located at a position 20 cm from an object with a net charge of +3 mC?
Answer: F = kqq/r^2 F = (9x10^9 N-m^2/C^2)(3x10^6 C)(6x10^6 C) / (0.02 m)^2 F = 405 N Remember the signs on the charges have no mathematical meaning so you should not include them in the math. The different types of net charge do mean the force is an attracting force. Newton's Third Law notes any two objects pushing or pulling on each other will push/pull with equal but opposite forces. Each applies the same 405 N of attracting force on the other.
Ex) What is the static electric force between a -3.0 μC charge and a +6 μC charge if they are 2.0 cm apart? Also, which is larger, the force of the negative particle on the positive or the force of the positive particle on the negative?
Answer: T = NIABsinθ T = (15)(3.2 A)(π)(0.035 m)2(0.5 T)(sin 30) T = 0.046 N-m
Ex) What is the torque on 15 circular loops with a radius of 3.5 cm carrying a current of 3.2 A in a magnetic field with a strength of 0.5 T when the loop is at an angle of 30° relative to the field?
Answer: Ammeters should be placed in series so that eliminates the first choice. The second choice places the ammeter in series with the first bulb so it will read current through the first bulb. The third choice places the ammeter in series with the battery so it will read current through the battery. The final choice places the ammeter in series with bulb 2 so is the correct choice.
Ex) Which of the following arrangements shows the correct way to connect an ammeter to measure the current through bulb 2?
Answer: The first arrangement correctly places the voltmeter in parallel across bulb 2.
Ex) Which of the following diagrams shows the correct way to connect a voltmeter to a circuit in order to read the voltage drop across bulb 2?
Answer: Current moves down the wires in the front and up the wires in the back. Curl your right-hand fingers to match, and your thumb should point to the right. Since magnetic fields point away from the north and towards the south, the right side of the bar will be magnetic north.
Ex) Which side of the iron bar will be magnetic north?
Since R3 and R4 are in series, they must have the same amount of current regardless of their individual resistance values (conservation of charge). That said, the sum of R3 and R4 compared to R2 will determine how large or small that current will be. Parallel branches share the same voltage drop. This means the voltage across R2 will be the same voltage across the entire branch that holds R3 and R4. In other words, the voltage across R3 and the voltage across R4 must add to equal the voltage across R2. This means the voltage across R2 must be larger than the voltage across R3. Since R1 and R5 are in series, they must have the same amount of current regardless of their individual resistance values (conservation of charge). The current though R5 is the same current through the battery. This current splits at the junction between the two parallel branches; some of the current goes through R2, and some of the current goes through the line that holds R3 and R4. This means the current through R5 must be larger than the current through R2. As noted earlier, the two resistors are in series so they must have the same current. V = IR If R1 > R5 and current is the same, then the voltage across R1 must be larger than the voltage across R5.
Examine the following circuit, and answer the questions. Compare the current through R3 and R4 if R3 > R4. Compare the voltage across R2 to the voltage across R3. Compare the current through R1 to the current through R5. Compare the current through R5 to the current through R2. Compare the voltage across R1 and R5 if R1 > R5.
CFL bulbs emit more energy in the form of light but far less energy in the form of heat. Incandescent bulbs emit far more heat than light; most of the power rating of an incandescent bulb is due to heat emission rather than light emission.
Explain how CFL lightbulbs can be brighter than incandescent lightbulbs but still have lower power ratings than incandescent bulbs.
A negative charge would lose energy moving from a position infinitely far away to a position near a positive source charge. As a result, the negative charge would have negative energy. However, since potential is a function of the source charge and not the test charge, the test charge near a positive source charge would be at a positive potential.
Explain how a negative charge can have negative energy but a positive potential.
Two oppositely charged parallel plates will allow a current to flow through any conductor placed across them. If the conductor has current flowing through it, and is also in the presence of a magnetic field, then the conductor will experience a magnetic force. The force should accelerate the conductor off the rails turning it into a projectile. Even novelty rail guns require very strong magnetic fields and very large currents through the conductor. Solve for the current. V = IR (12 V) = I (2 Ω) I = 6 A Solve for the force. F = BILsinθ F = (1.25 T)(6 A)(0.06 m) F = 0.45 N Unless the aluminum bar has very little mass, or the system has a very small coefficient of friction, this is unlikely to be enough force to overcome the friction force between the bar and the rails.
Explain how a rail gun works. b. A student decides to build a rail gun. She connects two parallel bars spaced 6 cm apart to a 12 V source, and sets an aluminum bar across the two bars. The student uses a series of neodymium magnets to create a reasonably uniform 1.25 T magnetic field. If the internal resistance of the system is 2 Ω, what is the magnetic force on the aluminum bar?
Capacitors are connected to voltage sources like batteries. The positive end of the battery pushes charge onto the top plate of the capacitor. This causes charge to be pushed off the bottom plate of the capacitor through static electric forces. Charge does not travel through the insulator separating the two plates.
Explain how capacitors are charged.
The balloon will polarize the ceiling. If the balloon is negatively charged, then the ceiling will polarize such that the positive pole of the ceiling is closer to the balloon than the negative pole. The attractive static electric force will then be larger than the repelling static electric force. As long as the net static force of attraction is equal to or larger than the gravity force acting on the balloon, the balloon will stick to the ceiling.
Explain why a charged balloon will electrostatically stick to a neutral ceiling.
The same amount of extra charge pushed into one plate will be pushed off the other plate so the total amount of net charge remains zero.
Explain why a charged capacitor still has an overall net charge of zero.
Linking capacitors in series is similar to increasing the distance between the top plate and bottom plate of a single capacitor. As the distance between plates increases, the attraction between plates decreases, which means the repelling forces between the excess charge particles will become more significant and the same applied voltage will have a harder time pushing charge onto the plate. Capacitance will decrease. Linking capacitors in parallel is similar to increasing the total area of the top and bottom capacitor plates. With more area, it's possible to collect more excess charge on a plate since they have more room to spread out. The repelling forces between the excess charges don't become a problem as quickly as they would on plates with smaller areas.
Explain why combining capacitors together in series results in an overall smaller capacitance. Explain why combining capacitors together in parallel results in an overall larger capacitance.
Since the types of charge are equally balanced on both plates, there is no net attracting or repelling forces acting on any of the charges. Without any net electrical forces, charge will not be driven from one plate to another. Charge will only flow through a wire from one plate to another if the capacitor is charged (there is a difference in net charge on both plates creating a voltage difference across the two plates).
Explain why connecting a wire to an uncharged capacitor would not create a flow of charge through the connecting wire.
Only one type of mass can interact in one way (attract). There are two types of charge and the different combinations of charge type lead to both attracting and repelling forces. The specific nature of these will be explored more in Lesson 3.4 on electric fields.
Explain why gravity forces are always attractive but static electric forces can be either attractive or repulsive.
Any lightning that hits the car will conduct through the frame of the car to the ground. Any passengers inside will be fine.
Explain why it is safe to sit in a car during a lightning storm.
Large plate areas allow charge to spread out minimizing the repelling static electric forces. Since the repelling forces are reduced, more charge can be added. Small separation distances allow for greater static electric forces between the two charged plates. The attracting force between the oppositely charged plates allows more charge to be added since the attracting force can be used to counter some of the repelling force experienced between the charges on the same plate.
Explain why large plate areas and small plate separation distances result in larger capacitances.
V = IR Since R is almost always larger than 1 Ω, the current will usually be smaller than the voltage that causes it.
Explain why the amplitude on an alternating voltage versus time graph is usually larger than the amplitude on an alternating current versus time graph.
A = D > B > C
Four equal positive point charges move through a uniform electric field as shown below. Which of the following correctly ranks the energy GAINED by the particles from greatest to least? A = D = C > B D > A = C > B A = D > B > C C > B > A = D
25.4 V
Four point charges form a square 1.0 m on each side (see below). What is the potential at a point in the center of the square of the value of q is 2 nC? 0 V 18 V 12.8 V 25.4 V
Similarities: Changes in gravitational potential energy are caused by moving masses through gravity fields. Changes in electric potential energy are caused by moving charges through electric fields. For uniform fields: Ug = mgh Ue = qEr For nonuniform fields: Ug = Gmm/r Ue = kqq/r Differences: There is only one type of mass, which means that work must always be done on a mass to move it away from a second mass. (i.e., if you lift a book up away from the Earth, you are doing work on the book.) Since there are two types of charge, sometimes the work done to separate charges is positive, and sometimes the work is negative. For example, work must be done on a +q to move it away from a -q, but work must be done by a +q to move it away from a second +q.
How is electric potential energy similar to gravitational potential energy? How are they different?
Electric potential is the ratio of energy per charge. Electric field strength is the ratio of force per charge. This means both potential and field strength depend only on the source charge, while energy and force depend on the charges of both the source and the test charge.
How is the relationship between electric potential energy and electric potential similar to the relationship between electric forces and electric field strength?
3.8 x 10^20 electrons
How many electrons will flow through a wire in 2.0 min if the current is 0.5 amps? 6.3 x 10^18 electrons 7.5 x 10^20 electrons 3.8 x 10^20 electrons 2.5 x 10^19 electrons
0.27 J
How much energy does it take to move a particle with a charge of +2 nC, from an initial position infinitely far from a second particle with a charge of +3 mC, to a final position 0.20 m from the second particle? 90 J 0.27 J 1.35 J 450 J
2.1x10^-3 J
How much energy does it take to move a particle with a charge of +6.0 μC a distance 0.5 m through a uniform electric field with a strength of 700 N/C? 350 J 2.16x10^5 J 2.1x10^-3 J 8.4x10^-3 J
First, solve for the amount of charge carried by 1 mole of electrons: Q = (6.02E23 e)(1.6E-19 C) Q = 96320 C I = Q / t (2.0 A) = (96320 C) / t t = 48160 seconds = 13.3 hours
How much time would it take for 1 mole of electrons (6.02E23 electrons) to pass through a wire if the flow rate of charge is 2.0 A?
No work is done moving a charged particle along an equipotential line. There is no change in voltage, which means the energy per charge ratio (definition of voltage) remains constant. Voltage is a determined by the source charge and not the charge of the particle in the electric field so it does not matter if the test charge is positive or negative.
How much work is done on a charged particle if it is moved along an equipotential line? Does it matter if the particle has a net positive or negative charge? Explain.
Resistance is directly proportional to the length of the wire. Doubling the wire length will double the resistance. Resistance is inversely proportional the area of the wire but area is proportional to the radius squared. If the length of the wire doubles, the area will increase by a factor of 2^2 = 4. If the area increases by a factor of 4, then the resistance will decrease by a factor 4. Half of the radius is (1/2)^2 = ¼ times the area. If the area decreases by a factor of 4 (i.e., ¼ the area) then the resistance will increase by a factor of 4.
How will each of the following changes affect the resistance of a particular wire resistor? a) Double the length of the wire b) Double the radius of the wire c) Halve the radius but triple the length of the wire
out of page to right down and left
Identify the direction of the magnetic force on each wire.
electrons
If a material gains _____, material will have - net charge, if loses ______ it will have + net charge · mobile charge carriers, move easily through conductors but not insulators
Power values are usually given as average power values. Pavg = ½ Ppeak (2000 W) = ½ (P) P = 4000 W Solve for peak voltage Vrms = V / √2 (120 V) = V / (1.41) V = 169 V P = IV (4000 W) = I (169 V) I = 23.7 A
If a particular blow dryer is rated at 2000 W, what is its peak power usage? b. If the blow dryer is plugged into a U.S. outlet with a Vrms of 120 V, what is the peak current through the dryer?
An electric field has a positive charge. If the test charge is kept in a uniform field, the positive charge will be attracted by a negative plate and will move in the same direction as the field. This is because both the field and positive test charge have the same type of charge, so both of them would be attracted to a negative charged plate. The negative test charge would be repelled by a negative plate and attracted to a positive plate, so the negative test charge would move against the field.
If a positive test charge is placed in a uniform field, the test charge will move in the same direction as the field. If a negative test charged is placed in a uniform field, the test charge will move against the field. Explain why this happens.
By agreement of physicists, electric field directions are based off the direction of the force applied by a source on a positive test charge. As a result, a negative charge placed in the field will experience a force in the direction opposite the field lines.
If a proton is placed in a uniform electric field, the proton will experience a force in the direction of the field. Explain why an electron experiences a force in the opposite direction of the electric field.
The movement of charge at the atomic level create regions or domains where there are randomly aligned N and S regions. When the ferromagnetic material becomes magnetized, the magnetic direction of the domains are forced into alignment. Basically, the magnetic field of a permanent magnet is still created by the movement of charge but it is movement at the atomic level.
If moving charge is the source of all magnetism, how do permanent magnets work?
F - protons have a net charge of 1.6x10^-19 C of the positive type of charge. T T F - if an object has a neutral charge, it contains equal amounts of positive and negative T
Label each of the following statements as true or false. If the statement is false, explain why. Protons have a charge of +1 C. The total amount of charge present must remain constant for any process. Charge is the fundamental property of matter that is responsible for electrical effects. If an object has a neutral charge, it contains no positive or negative charge. charge. It is not possible for an object to have a net charge of +5.6x10^-19 C.
The smallest resistance will occur in the circuit that is entirely parallel (circuit 4). The largest resistance will occur in the circuit that is entirely series (circuit 1). Solving circuit 2: Rs = R + R = 2R 1/Rtotal = 1/2R + 1/R = 3 / (2R) Rtotal = 2/3 Solving circuit 3: 1/Rp= 1/R + 1/R = 2/R Rp = ½ R R Rtotal = ½ R + R = 3/2 R Circuit 4 < Circuit 2 < Circuit 3 < Circuit 1
List the schematic diagrams below in order from smallest to largest equivalent resistance if all of the bulbs have resistance R.
Vrms = V / √2 V = (230 V)(1.41) V = 324.3 V The graph should be a sine (or cosine) wave that peaks at +/- 324 V. There should be 50 complete voltage cycles in 1 second. This is the same as showing 5 waves in 0.1 second. The peak V values would be lower. Vrms = V / √2 V = 120 (1.41) = 169 V The number of voltage cycles that occur each second would increase. There would be 60 waves in 1 second or 6 waves in 0.1 second.
Outlets in France have a Vrms of 230 V that alternates at a frequency of 50 Hz. Sketch an accurate graph of voltage versus time for an outlet in France. (Hint: how can you determine the peak V values?) picture part of answer b. Describe how your graph would change for the U.S. standard values. (Vrms = 120 V and f = 60 Hz)
The distance between particles C and A is: r^2 = 2^2 + 2^2 r = 2.83 m Due to symmetry, the force of C on A is the same force as the force of B on A. F = kqq/r^2 F = (9E9 N-m^2/C^2)(2E-6 C)(5.5-6 C) / (2.83 m)^2 F = 0.0124 N The x components of the forces cancel out so the net force will be equal to the sum of the two y-component forces. ΣF = ΣFy = (0.0124 N)(sin 45) + (0.0124 N)(sin 45) ΣF = 0.0175 N at 90°
Particle A has a net charge of +5.5 μC, and is located at (0,0) m. Particle B has a net charge of -2 μC, and is located at (2, 2) m. Particle C has a charge of -2 μC, and is located at (-2, 2) m. What is the net charge on particle A?
F = kqq/r^2 F = (9E9 N-m^2/C^2)(8.2E-6 C)(25E-9 C) / (0.5 m)^2 F = 7.38E-3 N attractive Equal. Newton's third law notes that the force of object 1 on object 2 is always equal but in the opposite direction to the force of object 2 on object 1.
Particle A has a net charge of -8.2 μC, and is placed 0.5 m to the left of Particle B, which has a net charge of +25 nC. What is the static electric force between the two? b. Which applies the largest force, Particle A on B or Particle B on A?
Particles with the same type of net charge repel, which means particle B cannot have a positive net charge. Particles with the opposite types of net charge attract, which means particle B must also have a net negative charge.
Particle A has a positive net charge. Particle B has a net charge, and is attracted to particle A. What type of net charge does particle B have? Explain.
The distance between any of the particles and the origin is: r^2 = 2^2 + 2^2 r = 2.83 m The electric fields for Ed and Eb will have the same strength. The electric fields for Ec and Ea will have the same strength. The field strength for Ea is half the field strength of Ed. E = kq / r^2 E = (9E9 N-m^2/C^2)(10E-9 C) / (2.83 m)^2 ED = EB = 11.25 N/C EA = EC = ½ ED EA = EC = ½ (11.25 N/C) EA = EC = 5.63 N/C All of the y components cancel out due to symmetry so the total field strength will be equal to the sum all four x components. Note that all four components are in the negative direction. ΣE = ΣEx = 2 (-5.63 N/C)(cos 45) + 2 (-11.25 N/C)(cos 45) ΣE = -23.9 N/C at 180°
Particle A is placed at position (-2, 2) m, particle B is placed at (2, 2) m, particle C is placed at (-2, -2) m, and particle D is placed at (2, -2) m. Solve for the net electric field strength at position (0, 0) if particles A and C have a charge of -5 nC, and particles B and D have a charge of +10 nC. picture part of answer
B = uI/(2πr) B = (4πx10^-7 T-m/A)(2.0 A) / (2π*0.05 m) B = 8.0x10^-6 T Bearth = 5x10-5 T Earth's magnetic field is roughly 6 times stronger than the field calculated in part a.
Solve for the magnitude of the magnetic field strength produced by a wire carrying 2.0 A of current measured at a position 5.0 cm away from the wire. b) How does this field strength compare to Earth's field strength?
c) r^2 = 3^2 + 3^2 r = 4.24 m E = kq / r^2 EA = EB = (9x10^9 N-m^2/C^2) (5x10^-6 C) / (4.24 m)^2 EA = EB = 2503.1 N/C EC = ED = (9x10^9 N-m^2/C^2) (10x10^-6 C) / (4.24 m)^2 EC = ED = 5006.2 N/C ∑E = ∑EY = (2503.1 N/C + 2503.1 N/C + 5006.2 N/C + 5006.2 N/C) (sin45˚) ∑E = 1.06x10^4 N/C at 180˚
Particle A is placed at position (3, 3) m, particle B is placed at (-3, 3) m, particle C is placed at (-3, -3) m, and particle D is placed at (3, -3) m. Particles A and B have a charge of -q(q= -5µC) and particles C and D have a charge of +2q (q= +10µC). a) Draw a properly labeled coordinate plane with correctly placed and labeled charges. b) Draw and label a vector diagram showing the electric field vectors at position (0, 0) m (3 points). c) Solve for the magnitude and direction of the net electric field strength at position (0, 0) m (7 points). picture part of answer
F is proportional to the inverse square of distance. 1/3^2 = 1/9 F The force will decrease by a factor of 9. F is proportional to q. The force will double or 2F. F and q are directly proportional, so an x3 increase in q will result in an x3 increase in F. F and r have an inverse square relationship, so a 1/3 change in r will result in an x9 increase in F. Combined, this means that F will increase by a factor of 3 x 9 = x27 above the original system force.
Particles A and B both have a net charge of +q, and are separated by distance d. Explain how each of the following changes will affect the static electric force Fe. a) The distance between the two is increased to 3d b) The charge on particle A is increased to +2q c) The charge on particle B is increased to +3q and the distance is decreased to 1/3 d
Rubber suits are insulators, which means it is difficult for charge to transfer through the suit to the person inside. The fine chain mesh is metallic, and will conduct the charge to the ground leaving the person inside electrocution free. In both cases, the wearer should be extremely cautious to make sure there are no openings or holes in the suit that would allow charge to transfer to the skin.
People who play with / perform with large tesla coils (a type of static electricity generator) wear full body suits that are either a) entirely made of rubber-like materials or b) fine chain mesh. How do each of these materials keep the wearer safe?
up
Point x is equidistant from two particles with net charges of +q and -q of equal magnitude. If a negative point charge is at the position marked by the x, the direction of the net electrical force on the negative point charge is down. up. left. right.
a. position in an electric field.
Potential differences are most closely related to: a. position in an electric field. b. the amount of charge on a particle in an electric field. c. both the above.
The field arrows point down, which means the negative or zero potential plate is the bottom plate. (Can you explain why?) The potential is related to the position between the plates, and not at all related to the amount of charge between the plates. A = C = E > D > B > F V = Er ΔV = EΔr · A, C, and D all move from low potential to high potential with the greatest Δr. · D has the same final potential as A, C, and D but the question asked about the change and not the final value. D experienced a smaller Δr. · B and F both show a decrease in potential. Since B has a smaller Δr, it has a smaller decrease.
Rank the following particles in order from largest increase in potential to smallest increase in potential if the particle charges are: A = C = +1 μC, B = D = -1 μC, E = +2 μC, and F = -2 μC.
E = F > A = C > B > D Ue = qEr E is positive moving against the field, and F is negative moving with the field so, in both cases, the particles gain energy. They both have large q values, and travel the same r. A and C travel the same r as E and F but have smaller q values. Both A and C are positive moving against the field. The motion of A perpendicular to the field has no effect on its Ue. B is negative moving with the field, and has the same charge has A and C but has a smaller r. D has the same q and r value as B but is negative moving against the field, which means it is losing energy rather than gaining.
Rank the following particles in order from most energy gained to least energy gained if they have the following net charges: A = C = +1 μC, B = D = -1 μC, E = +2 μC, and F = -2 μC.
τ = NIABsinθ τ = (20)(1.0 A)(0.02 m)^2(2 T) τ = 0.016 N-m This is a very small torque. Increase the number of loops. Increase the current. Increase the size of the loops. Increase the strength of the magnetic field.
Solve for the maximum torque on a set of 20 square loops that are 2 cm on each side if they experience a current of 1.0 A in a field of 2 T. b. What adjustments could be made to the system to increase the maximum torque?
Rs = R3 + R4 Rs = 4.0Ω + 4.0Ω Rs = 8.0Ω 1/Rp = 1/R2 + 1/R3+4 1/Rp = 1 / 4.0Ω + 1 / 8.0Ω 1/Rp = 0.375Ω Rp = 2.67Ω R = Rp + R1 + R5 R = 2.67Ω + 4.0Ω + 4.0Ω R = 10.67Ω V = IR (9.0V) = I x (10.67Ω) I = (9.0V) / (10.67Ω) I = 0.843 A V1 = V5 = IR1 V1 = V5 = (0.843 A) (4.0Ω) V1 =V5 = 3.37V V2 = V - V1 - V5 V2 = 9.0V - 3.37V - 3.37V V = 2.26V P = IV = I^2 R = V^2/R R2 = R3 + R4 2.26V = 2R R4 = 1.13V P = V^2/R P = 1.13^2 / 4.0Ω P = 0.319 W
Several resistors are connected to a battery as shown below R1 (4.0Ω), R2 (4.0Ω), R3 (4.0Ω), R4 (4.0Ω), R5(4.0Ω), V (9.0V)a) Solve for the total circuit resistance. b) Solve for the current through the battery. c) Solve for the voltage drop across R2. d) Solve for the power dissipated by R4.
Note that in conductors the negative charge is very mobile, and the repelling forces between the negative rod and the negative charge carriers (electrons) will force the electrons to the opposite end of the bar. The charge carriers will also repel each other so the electrons will move to the surface of the material. (The protons didn't move; the diagram above shows the net charge distribution and not a total charge distribution.) Insulators are a lot better at holding on to their electrons, which means the electrons will move to the opposite side of the molecule but not to the opposite end of the material.
Sketch pictures that indicate the charge distribution in a) a plastic bar and b) a metal bar when a negatively charged rod is brought near each.
Solve for the total resistance: R = 10 + 15 = 25 Ω Use Ohm's law to find the total current V = IR (20 V) = I (25 Ω) I = 0.80 A The current is the same everywhere through the series circuit so the current is 0.80 A through each resistor. V = IR V = (0.80 A)(10 Ω) V = 8.0 V The total voltage drop across all resistors is the same as the voltage across the battery. Either use Ohm's law again or just subtract: Vbattery = Vresistor 1 + Vresistor 2 (20 V) = (8.0 V) + V V = 12.0 V P = IV = I^2R P = (0.80 A)^2 (15 Ω) P = 9.6 W
Solve for the current through the battery. b) Solve for the current through each resistor. c) Solve for the voltage across each resistor. d) Solve for the power rating of the 15 Ω resistor.
Solve for the total resistance. 1/R = 1/10 + 1/15 R = 6 Ω Solve for the current using Ohm's law. V = IR (20 V) = I (6 Ω) I = 3.3 A Parallel branches share the same voltage drop values. In this case, the voltage across each resistor is the same as the voltage across the battery: 20 V. V = IR (20 V) = I (10 Ω) I = 2 A The total current through the battery is equal to the sum of the currents through each resistor. Either use Ohm's law again or subtract: I battery = I resistor 1 + I resistor 2 (3.3 A) = (2 A) - I I = 1.3 A P = IV = V^2/ R P = (20 V)^2 /(15 Ω) P = 26.7 W
Solve for the current through the battery. b) Solve for the voltage across each resistor. c) Solve for the current through each resistor. d) Solve for the power dissipated by the 15 Ω resistor.
Solve for the distance between the particle and any one of the source charges: r^2 = (0.25 m)^2 + (0.25 m)^2 r = 0.35 m Electric potential energy of a charged particle in a nonuniform field: ΣUe = Σ(kqq/r) Since there are two positive and two negative energy terms in the first system all with equal values, the net potential energy of the central charge is zero. In the second system: ΣUe = Σ(kqq/r) ΣUe = 3 (9x10^9 N-m^2/C^2)(+3x10^-9 C)(+5x10-3 C) / (0.35 m) + (9x10^9 N-m^2/C^2)(+3x10^-9 C)(-5x10^-3 C) / (0.35 m) ΣUe = + 0.77 J The third system will be the same as the second except that final value will be negative. ΣUe = - 0.77 J
Solve for the electric potential energy stored by the charged particle in the center of each system if it has a net charge of +3 nC, and the source particles have charge values of 5 mC of the type indicated in the diagram.
Field due to wire on the left B = uI/(2πr) B = (4πx10^-7 T-m/A)(10 A) / (2π*0.02 m) B = 1E-4 T directed out of the page Field due to wire on the right B = uI/(2πr) B = (4πx10^-7 T-m/A)(4 A) / (2π*0.02 m) B = 4E-5 T directed out of the page ΣB = (1E-4 T) + (4E-5 T) = 1.4E-4 T
Solve for the field strength located halfway between two parallel wires if the wire on the left has a current of 10 A directed downwards and the wire on the right has a current of 4 A directed upwards. The wires are 4 cm apart.
Solve for the distance between the particle and any one of the source charges. r^2 = (0.25 m)^2 + (0.25 m)^2 r = 0.35 m Electric potential energy of a charged particle in a nonuniform field ΣV = Σ(kq/r) Since there are two positive and two negative energy terms in the first system all with equal values, the net potential at the center of the first system is zero. In the second system: ΣV = Σ(kq/r) ΣV = 3 (9E9 N-m^2/C^2)(+5E-3C) / (0.35 m) + (9xE9N-m^2/C^2)(-5E-3C) / (0.35 m) ΣV = 2.57E7 V The third system will be the same as the second except that final value will be negative. ΣV = -2.57E7 V
Solve for the potential at the position marked by the dot in each of the following systems. Each source particle has a net charge of 5 mC of the type indicated in the diagrams.
The arrows represent the direction of the electric field, the circles represent positions with equal potential, and the source charge is negative.
The diagram below shows both the electric field lines and the equipotentials near a point source. Which of the following statements accurately describes the diagram? The arrows represent the direction of the electric field, the circles represent positions with equal potential, and the source charge is negative. The arrows represent the direction of the electric field, the circles represent positions with equal potential, and the source charge is positive. The arrows represent positions with equal potential, the circles represent the electric field, and the source charge is negative. The arrows represent positions with equal potential, the circles represent the electric field, and the source charge is positive.
E is inversely proportional to r^2. E will decrease by a factor of 4. The field at the new position will be ¼ E.
The electric field strength measured a distance r from a point source is determined to be some value E. By what factor will the field strength change if the field is measured at a point 2r?
c. both the above.
The electric potential energy stored by a charged particle in an electric field is most closely related to: a. the position of the particle in the field. b. the amount of charge on the particle. c. both the above.
c. 22 N
The magnitude of a force that has an x component of 10 N and a y component of 20 N is: a. 30 N b. 500 N c. 22 N
charge.
The property of matter that is responsible for static electric effects is charge. mass. momentum. density.
C) energy.
The purpose of a battery in a circuit is to act as a source of A) resistance. B) charge. C) energy. D) both b and c.
a. To act as an energy source
What is the purpose of a battery in a circuit? a. To act as an energy source b. To act as a charge source c. Both the above
b. Charge
The static electric force is caused by: a. Mass b. Charge c. Volume d. More than one of the above
a. <2,6>
The sum of vectors A and B where A = <3,2> and B = <-1,4> is: a. <2,6> b. <4,6> c. <7,1>
Coulomb.
The unit used to measure quantity of charge is the kilogram. Newton. Coulomb. joule.
Since the voltage cycles 60 times in 1 second, the current through the lightbulb will also cycle 60 times in 1 second. The voltage will cycle through zero twice each cycle. (Can you explain why?) Since voltage drives current (they cycle together), every time the voltage drops to zero, the current will also drop to zero. A lightbulb plugged into an AC source (such as a standard household outlet) is actually cycling on and off very rapidly. The reason people don't notice is because their eyes can't detect those rapid changes. (Similar to watching a movie from film, if you move a series of stationary images rapidly enough, your eyes can't detect the individual images.) That said, most fluorescent bulbs light as a result of the high temperatures of the filament. When current stops flowing through the bulb, it still takes a moment for the filament temperature to drop and for light production to cease. The bulb brightness will still cycle at a rate double the voltage cycle frequency (can you explain why?) but the light doesn't usually completely turn off during this process. Since fluorescent bulbs don't generally rely on filaments, the flickering may be more noticeable, which is why some florescent lighting systems are designed with additional electrical components that will increase the frequency of the voltage cycle through the bulb.
The voltage for a U.S. outlet cycles at a frequency of 60 Hz. Describe what happens to the current through a lightbulb plugged into a U.S. outlet during this time. How does this affect the lightbulb?
a. 17.3 N
The x component of a force that is 20 N at 30˚ is: a. 17.3 N b. 20 N c. 10 N
1/C = 1/C + 1/C + 1/C 1/C = 1/(40 μF)x3 C = 13.3 μF C = C + C + C C = (40 μF)x3 C = 120 μF
Three capacitors, each with a capacitance of 40 μF, are wired together in series. What is the overall capacitance? Three capacitors, each with a capacitance of 40 μF, are wired together in parallel. What is the overall capacitance?
135°
Three charged particles are held in place. Particle A has a net charge of +2 μC and is at (0, 0) m. Particle B has a net charge of +4 μC and is placed a (0.5, 0) m. Particle C has a net charge of -4 μC and is placed at (0, 0.5) m. What is the angle of direction of the net force acting on particle A? 45° 135° 225° 315°
0.41 N
Three charged particles are held in place. Particle A has a net charge of +2 μC and is at (0, 0) m. Particle B has a net charge of -4 μC and is placed a (0.5, 0) m. Particle C has a net charge of -4 μC and is at (0, 0.5) m. What is the magnitude of the net force on particle A? 0.41 N 0.29 N 0.58 N 0.17 N
1/Rp = 1/R1 + 1/R2 + 1/R3 +... 1/Rp = 1 / 2.0Ω + 1 / 4.0Ω + 1 / 6.0Ω 1/Rp = 0.9167Ω Rp = 1.091Ω V= IR (9.0V) = I (1.091Ω) I = (9.0V) / (1.091Ω) I = 8.25 A V = IR (9.0V) = I (4.0Ω) I = (9.0V) / (4.0Ω) I = 2.25 A
Three resistors (R1 2.0Ω, R2 4.0Ω, and R3 6.0Ω) are connected in parallel to a battery with a voltage of V 9.0V. a) Solve for the current through the battery. b) Solve for the current through R2.
R = 2.0Ω + 4.0Ω + 6.0Ω R = 12.0Ω V = IR (9.0V) = I (12.0Ω) I = (9.0V) / (12.0Ω) I = 0.75 A V = IR V = (0.75 A) (4.0Ω) V = 3.0V
Three resistors (R1 2.0Ω, R2 4.0Ω, and R3 6.0Ω) are connected in series to a battery with a voltage of V 9.0V. a) Solve for the current through the battery. b) Solve for the voltage across R2.
Each branch has the same voltage difference across the branch. Since each branch has the same electric pressure, the flow rate will be largest where the resistance in the branch is smallest. The 1 Ω resistor will experience the largest flow rate of charge (current). The voltage across the branch is equal to the voltage across the battery. Use Ohm's law. V = IR (12 V) = I (1 Ω) I = 12 A Voltages across parallel branches are equal. Since the resistors are wired in parallel, each resistor will experience a voltage drop equal to the battery's voltage. Voltage battery = 12 V so the voltage across each of the resistors is 12 V.
Three resistors, (R1 = 1 Ω, R2 = 2 Ω, R3 = 4 Ω), are wired in parallel, and then connected to a 12 V battery. a) Draw the circuit diagram. picture part of answer b) Which resistor will have the most current? Explain. c) Solve for the current. d) Which resistor will have the greatest voltage drop across the resistor? Explain. e) Solve for the voltage.
The current is constant throughout the entire series circuit; the resistors will all experience the same current. Solve for the equivalent resistance first. R = 1 + 2 + 4 = 7 Ω Use Ohm's law. V = IR (12 V) = I (7Ω) I = 1.7 A V = IR Each resistor experiences the same flow rate of charge (current) because charge must be conserved throughout the loop. In order for this to happen, resistors with high resistance will have a larger voltage across the resistor (a greater electric pressure difference is required to produce the same flow through a large resistance). The 4Ω resistor will have the largest voltage drop. V = IR V = (1.7 A)(4 Ω) V = 6.9 V
Three resistors, (R1 = 1 Ω, R2 = 2 Ω, R3 = 4 Ω), are wired in series, and then connected to a 12 V battery. a) Draw the circuit diagram. picture part of answer b) Which resistor will have the most current? Explain. c) Solve for the current. d) Which resistor will have the greatest voltage drop across the resistor? Explain. e) Solve for the voltage.
Solve for the magnetic field strength created by the bottom wire at the top wire's location. B = μI / (2πr) B = (4πE-7 T-m/A)(8 A) / (2π*0.02 cm) B = 8E-5 T directed into the page Solve for the force. F = BILsinθ F = (8E-5 T)(4 A)(2.0 m) F = 8E-4 N directed upwards, the wires will repel. (The bottom wire will be repelled downwards with the same amount of force; you can check the math if you like.)
Two 2.0 m long horizontal parallel wires are placed 2 cm apart. The top wire carries 4 A of current to the right, and the bottom wire carries 8 A of current to the left. Do the wires attract or repel? By how much force?
C = Kε0A/d C1 = Kε1A/d C1 = 5(8.85x10^-12 F/m) (0.75 m^2) / (0.005 mm) C1 = 6.64x10^-9 F C2 = Kε2A/d C2 = 7(8.85x10^-12 F/m) (0.75 m^2) / (0.005 mm) C2 = 9.29x10^-9 F 1/Cs = 1/C1 + 1/C2 1/Cs = 1/6.64x10^-9 F + 1/9.29x10^-9 F 1/C = 2.58x10^8 F C = 3.9x10^-9 F Uc = ½ qV. U = ½ (CV)V U= ½ (3.9x10^-9 F) (12V)^2 U = 2.8x10^-7 J
Two capacitors are connected in a series. They both have a plate separation distance of d (5 mm) and a plate area of A (0.75 m^2). They each have a different dielectric with dielectric constants of k1 (5) and k2 (7), respectively. What is the overall capacitance of the system? b) If this system of capacitors is connected to a 12V battery, how much total energy will be stored in the capacitors?
The particles will repel. As they get further apart, the repelling force will decrease.
Two charged particles both have positive net charges. They are near each other and released from rest. Which of the following statements best describes the interaction of the particles? The particles will attract. As they get closer together, the force of attraction will increase. The particles will attract. As they get closer together, the force of attraction will decrease. The particles will repel. As they get further apart, the repelling force will increase. The particles will repel. As they get further apart, the repelling force will decrease.
Since the charges are opposite, they will attract, and move towards each other. Since the static electric force between the two will increase as the particles get closer together, the acceleration will also increase according to Newton's second law (ΣF = ma). The correct answer is B.
Two charged particles have opposite net charges. They are placed near each other, and released from rest. Which of the following statements best describes the motion of the particles? a) The particles will accelerate towards each other at a constant acceleration. b) The particles will accelerate towards each other with an increasing acceleration. c) The particles will accelerate away from each other at a constant acceleration. d) The particles will accelerate away from each other with a decreasing acceleration. Explain your answer.
0.34 m
Two charged particles, each with a charge of +8.0 μC are at some position near each other. What is the distance between the particles if the static electric force between them is 5.0 N? 0.12 m 0.34 m 8.7 m 2.9 m
Both a and b are possible.
Two charged spheres are suspended from strings. The charged spheres repel each other. Which statement correctly explains this behavior? Both spheres have a negative net charge. Both spheres have a positive net charge. One sphere has a positive net charge, and the other has a negative net charge. Both a and b are possible.
2V
Two identical batteries, each with a voltage of V, connect in a series to a resistor (see below). The voltage across the resistor is V 2V ½ V 0V
V
Two identical batteries, each with a voltage of V, connect to each other in parallel. This arrangement then connects to a resistor (see below). The voltage across the resistor is V 2V ½ V 0V
D
Two identical metal blocks are in contact side-by-side. A positively charged rod moves near the left block. Which of the following most accurately shows the resulting distribution of charge throughout the metal blocks?
C = Q / V (10 μF) = Q / (9.0 V) Q = 90 μC 1/C = 1/C + 1/C 1/(10 μF) = 2 (1/C) C = 20 μF C = εA/d (20E-6 F) = (8.85E-12 F/m)(A) / (0.005 m) A = 11299 m2 Given the required plate area, it is not likely.
Two identical parallel plate capacitors are wired together in series. The overall capacitance is 10 μF. How much charge can be stored in the capacitors if they are connected to a 9.0 V battery? b) What is the individual capacitance of one of the capacitors? c) If the capacitor has a plate separation distance of 0.5 cm, how likely is the capacitor to use air as the insulating material? Explain.
positive.
Two neutral metal blocks are in physical contact side-by-side. Block A is on the left, and block B is on the right. A positively charged rod is near the left side of block A (see below). With the positive rod still present, the blocks separate. As a result, block B will have a net charge that is positive. negative. neutral. not enough information.
What is the electric field strength between the plates? V = Er (9 V) = E (0.2 m)E = 45 N/C V = Er V = (45 N/C)(0.1 cm) V = 4.5 V
Two parallel plates separated by a distance of 20 cm are connected to a 9 V battery. What is the electric field strength between the plates? b) What is the potential at a point halfway between the two plates?
Some point above the top wire
Two parallel wires are horizontally oriented. The top wire has current of some value I directed to the left, and the bottom wire has current of 2 x I directed to the right. At what point will the magnetic fields cancel out? Some point between the two wires Some point above the top wire Some point below the bottom wire No position cancels the magnetic fields
3F
Two positively charged particles experience a static electric force of F. One of the particles is replaced with a new charged particle that has three times more charge than the original. The new static electric force will be F. 3F 1/3F 9F
Resistor B has the greatest resistance.
Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which resistor has the greatest resistance? Resistor A has the greatest resistance. Resistor B has the greatest resistance. The two resistors have the same resistance. Not enough information is stated. The current needs to be measured to determine
16 Ω
Use the following voltage versus current data to determine the resistance of a particular resistor. 16 Ω 36 Ω 0.063 Ω 1 Ω
· Series: has a large resistance, will have small power output, which means small power input. downside is small power output so dim bulbs. Another downside is if any connecting wires breaks or if one bulb burns out, the entire circuit is broken, and none of the bulbs will light · Parallel: more material to construct, and is a bit more expensive to operate. Has a small resistance, will have large power output, which means large power input. The upside is the larger power output also means brighter bulbs. Another upside is, since the bulbs are independently connected to the battery, if one bulb burns out, the others will remain lit · A combination shares the advantages and disadvantages of both wiring methods. It's cheaper than an all-parallel combination but more expensive than an all-series combination. In this mixed combination, if one bulb burns out, all the lights on the same branch will also go out but the other branches will remain lit.
What are some limitations and benefits of series and parallel combinations?
Electric potential differences drive the movement of charge. Steady, unchanging voltages such as those provided by batteries will create direct current. Fluctuating voltages will create alternating current.
What causes current to flow in a circuit? b. What causes steady state (direct) current? What causes alternating current?
Static electric forces are created by the interaction of charge. Two positive charges or two negative charges will always apply repelling forces on each other while positive and negative charges will attract each other.
What causes static electric forces?
A dielectric is the insulator placed between the two conducting plates of a capacitor. Dielectrics will increase the overall capacitance of the capacitor. They also allow the capacitor components to be rolled up, and put into smaller containers rather than relying on parallel plate designs.
What is a dielectric? What is the purpose of a dielectric?
A component in a circuit that hinders the flow of charge through the circuit
What is an electric resistor?
d. The flow rate of charge in a circuit
What is current? a. The speed of charge moving in a circuit b. The impedance to charge flow rate in a circuit c. The amount of charge in a circuit d. The flow rate of charge in a circuit
d. More than one of the above
What is resistance? a. The potential difference between two points in a circuit b. The impedance to charge flow rate in a circuit c. The ratio of voltage per current in a circuit d. More than one of the above
0.21 A
What is the current in a wire if 4 x 10^18 electrons flow through the wire every 3 sec? 1.3 x 10^19 A 0.64 A 1.92 A 0.21 A
First, you need to solve for the amount of charge carried by 5E26 electrons: Q = (5E18 e)(1.6E-19 C) Q = 0.8 C I = Q / t I = (0.8 C) / (2 s) I = 1.6 A
What is the current in a wire if 5E18 electrons pass through the wire every 2 seconds?
R = ρL / A R = (10.6E-8 Ω-m)(0.3 m) / (π (2E-5 m /2)^2) R = 101.3 Ω V = IR (12 V) = I (101.3 Ω) I = 0.12 A
What is the current through a platinum resistor with a length of 30 cm and a diameter of 2E-5 m when it is connected to a 12 V battery?
Out of the screen
What is the direction of the magnetic field at the position marked by the X ? Up Down Into the screen Out of the screen
Down
What is the direction of the magnetic field at the position marked by the X? Left Right Down Up
Right
What is the direction of the magnetic field at the position marked by the X? Left Right Into the screen Out of the screen
right
What is the direction of the magnetic force on the wire? Left right Up No force
No force
What is the direction of the magnetic force on the wire? (The current in the wire is flowing to the right .) Up Down Into the screen No force
E = kq / r^2 E = (9x10^9 N-m2/C2) (4.0x10^-3 C) / (2.5 m)^2 E = 5.76x10^6 N/C E = Fe / q Fe = E x q Fe = (5.76x10^6 N/C) (6x10^-9 C) Fe = 3.4x10^-2 N
What is the electric field strength at a position measured at r (2.5 m) from a 4.0 mC point source charge? b) If a particle with a charge of -q (-6 nC) is placed at the position indicated in the previous question, what is the magnitude of the force on the particle?
F = BILsinθ F/L = (0.1 T)(4 A) F/L = 0.4 N/m
What is the force per length ratio on a wire that has 4.0 A of current perpendicularly oriented to a uniform magnetic field of 0.1 T?
no force into page to right to left (- particle so use left hand) down, into page to left, out of page
What is the initial direction of the force on each of the following moving particles? last 2
3.75 N
What is the magnitude of the force on a +3 mC charge moving at a speed of 1000 m/s through a 1.25 T magnetic field? 0.42 N 3.75 N 2.4 N 0.0013 N
6x 10^-5 T
What is the magnitude of the magnetic field strength at a position 1.0 cm away from a wire with 3.0 A of current? 6 x 10^-7 T 3.8 x 10^-4 T 9.4 x 10^-9 T 6x 10^-5 T
-1.3 x 10^-5 C
What is the net charge on an object that has gained 8 x 10^13 electrons? -8 x 10^13 C -5 x 10^32 C -1.3 x 10^-5 C -2 x 10^-33 C
Batteries create a voltage difference in a circuit that then creates an electric field through the circuit causing charge everywhere in a wire to begin moving. Batteries provide energy (through the creation of electric fields) and potential differences. The correct answer is D.
What is the purpose of a battery in a circuit? a) Source of charge b) Source of energy c) Cause of potential difference d) B and C only e) A, B, and C Explain your answer.
A television that uses a 120 V potential difference to draw 1.25 A of current
Which device would cost more to run continuously for 1 hr? 100 W light bulb A clock radio with an total resistance of 200 Ω that operates with a 120 V potential difference A hand mixer with a total resistance of 120 Ω that draws 1 A of current A television that uses a 120 V potential difference to draw 1.25 A of current
The one with the largest power rating will have the greatest cost P = I^2R P = (3 A)^2(80 Ω) P = 720 W P = 100 W P = iV P = (120 V)(1 A) P = 120 W The toaster has the largest power rating so it will cost the most to run for one hour.
Which device would cost more to run for one hour? a) A toaster that draws 3 A of current and has a resistance of 80 Ω b) A 100 W lightbulb c) A TV that has a potential difference of 120 V and draws a current of 1.0 A Explain.
c. Both the above
Which of the following correctly defines power in a circuit? a. The amount of energy provided to a circuit each second b. The amount of energy dissipated by a circuit each second c. Both the above
Right
Which of the following correctly identifies the initial direction of the force on the moving negatively charged particle? Left Right Into the screen Out of the screen
No force
Which of the following correctly identifies the initial direction of the force on the moving negatively charged particle? Up Down Into the page No force
Right
Which of the following correctly identifies the initial direction of the force on the moving positively charged particle? Into the screen Out of the screen Right Left
Each half will have both north and south poles.
Which of the following correctly indicates the result of cutting the following permanent magnet in half along the dotted line as shown below? The left half will be a south magnet and right half will be a north magnet. The left half will be a north magnet and the right half will be a south magnet. Each half will have both north and south poles. The halves will no longer act as magnets and will have no magnetic poles.
C
Which of the following correctly shows the equipotentials near a negative point source? A B C D
a. If voltage increases, current increases.
Which of the following is a true statement? a. If voltage increases, current increases. b. If resistance increases, current increases. c. Both the above
All of the above are examples of insulators
Which of the following is an example of an insulator? A piece of paper A block of wax A block of glass All of the above are examples of insulators
Conductors are metal and allow charge to move freely through them.
Which of the following statements best describes conductors? Conductors are nonmetal and do not allow charge to move freely through them. Conductors are nonmetal and allow charge to move freely through them. Conductors are metal and do not allow charge to move freely through them. Conductors are metal and allow charge to move freely through them.
The total charge present in the system and surroundings is constant in any process.
Which of the following statements concerning charge is most accurate? In order for an object to have zero net charge, it must contain zero positive or negative charge. One electron has a charge of -1.0 C. The total charge present in the system and surroundings is constant in any process. More than one of the above is accurate.
Conventional current is the movement of positive charge and flows through a circuit in a direction from positive to negative.
Which of the following statements concerning conventional current is correct? Conventional current is the movement of positive charge and flows through a circuit in a direction from positive to negative. Conventional current is the movement of positive charge and flows through a circuit in a direction from negative to positive. Conventional current is the movement of negative charge and flows through a circuit in a direction from positive to negative. Conventional current is the movement of negative charge and flows through a circuit in a direction from negative to positive.
Objects with like charges repel, while objects with opposite charges attract.
Which of the following statements regarding charge is most accurate? One type of charge acts like an invisible fluid that can move between objects. Objects with like charges repel, while objects with opposite charges attract. Objects with like charges attract, and objects with opposite charges also attract. Static electric forces, like gravity forces, always attract and never repel.
Electric field lines are perpendicular to equipotential lines.
Which of the following statements regarding equipotentials and electric fields is accurate? Electric field lines are parallel to equipotential lines. Electric field lines are perpendicular to equipotential lines. Electric field lines can be any angle relative to equipotential lines. No equipotential lines are possible if there are electric field lines.
a. Two oppositely charged parallel plates
Which of the following will create a uniform electric field? a. Two oppositely charged parallel plates b. Two oppositely charged point sources c. Two point sources with the same type of charge d. More than one of these
Electric potential depends on the amount of charge on a particle in an electric field.
Which statement regarding electric potential is NOT accurate? Electric potential is measured in units of volts. Electric potential depends on the position in an electric field. Electric potential depends on the amount of charge on a particle in an electric field. Electric potential depends on the strength of the electric field at a given position.
Wire A has the same resistance as wire B.
Wire resistor A has twice the length and twice the cross sectional area of wire resistor B. Which of the following accurately compares the resistances of wire resistors A and B? Wire A has twice the resistance of wire B. Wire A has half the resistance of wire B. Wire A has the same resistance as wire B. None of the above
Capacitor
· device used to store electrical charge o Made by alternating layers of conductor and insulator. For, example you could make one by layering aluminum foil and wax paper. Most basic one is two parallel metal plates with air as an insulator o Stores charge by having some kind of electrical pump (like a battery) push charge onto one of the capacitor plates. As a result, charge already on the other capacitor plate will be pushed off. Of course, electrical capacitors do this through electrical forces between charges A good one is one that can hold a lot of excess charge with very little electrical push. This ratio of quantity of excess stored charge per applied volt is call
capacitors
· devices used to store charge and electrical energy · commonly used electrical component · used for camera flashes, to start up some motors, to charge defibrillators, as sensors for keyboards, as components in signal processing, along with many other uses
circuit diagram
· diagram that shows how various circuit elements are all linked together in a circuit. You use specific symbols to represent various circuit components o Symbol for a battery: § Long horizontal line represents the + end of the battery and the short horizontal line represents the -. This symbol can be rotated to match the direction of the actual battery in a circuit o Symbol for a resistor: o Circuit notation for a capacitor: § You can link these components with lines that represent wires in the circuit
Electric potential energy
· energy a charge has due to its position in an electric field o Ue = qEr, q is net charge on particle in electric field, E is strength of electric field, and r is distance between object relative to determined zero positon o When dealing with uniform electric fields, use one of the charged plates as the zero position because its functionally the same as setting ground as zero height in gravity fields, set the - plate as the zero position o + particle placed near - plate will remain at rest just like a box placed on the floor will remain at rest; neither object has potential energy. To lift box off ground against field, work must be performed on the box. To move the + charge towards the + plate against the electric field, work must be performed on the charge. If particle in electric field has a - net charge, - charge will be at rest near a + plate, and work must be applied to push it towards the - negative plate. o Work being done on the particle is stored by the particle as electric potential energy o According to law of energy conservation: total energy at the start + net work done on the system = total energy at the end § 0 + W = Ue
electric potential
· is a measure of the energy a charged particle could potentially have if placed at a given position in an electric field; it is an energy per charge ratio. is location specific, and depends only on the source charge that is creating the field. Potential has units of J/C or volts (V). is scalar, and so changes in potential can be found by subtracting the potential of one location in a field from the potential at a second location in a field. Equipotential surfaces or equipotentials are maps of all the locations in a field that have the same potential. Equipotential lines are always perpendicular to the electric field lines.
electric force
· is a push or pull between two charges whereas an electric field is just a way to visualize the space near a single charge o Direction of electric fields based on direction of the force the source would apply to a + charge § + source charges always create fields directed away § - source charges always create fields directed towards charge o Fields created by sources are not uniform; direction and size of field depend on how far away from the source you measure § Possible to create a uniform field where the ability of the source to push or pull on a test charge is the same direction and strength at all positions
electric fields
· maps of electric force applied by a source charge on a nearby positive test charge o Created by charge o Way of conceptualizing the region of space near a charge and the ability of that charge to apply forces to other nearby objects o Any object that has mass has its own gravity field § The more mass an object has, the stronger its gravity field, the greater the ability of the object to apply forces: this is also true for ______ · a way of conceptualizing the region of space around a charged object and the ability of that object to apply electric force to other nearby objects o Based on direction of force a source charge would apply to a nearby + test charge o Are always directed away from + source charges and towards - source charges o If multiple exist in an area, they will superimpose on each other to create a net electric field o Can determine the value of net field strength through vector addition
circuit
· means "go around" and charge pushed by the battery, is moving around in a circle. As long as the battery can continue to push charge, the flow rate of the charge will remain steady. o Most basic ____ is a battery connected to a loop of wire. o Need to be created with conducting materials (insulators will not allow charge to go through the material) with no breaks
conduction
· movement of charge from one object to another o Easy to move charge from metal to metal, but not insulator to insulator
static electric force
· proportional to the quantity of net charge on both objects and proportional to the inverse-square of the distance between two charged objects o Fe α q1q2 / r2 § Fe is force § q1 and q2 are the net charges on the object § r is distance between the two objects · between charged objects can be quantitatively determined using Coulomb's law where force is directly proportional to charge but proportional to the inverse square of the distance between the charges. If multiple charges in a system, the net force can be determined through vector addition
polarization
· redistribution of charge within an object; one end becomes + charge and the other - without changing total amt of charge in the object
series connection
· resistors linked together one after another in a line o Connecting resistors in ___ will result in an increase in total resistance § Each resistor in a ___ increases the restriction on charge flow § This is like the effects you see when increasing the length of a wire resistor; the longer the wire resistor, the larger the resistance
Ammeters
·devices used to measure current through a particular wire circuit component. Are always wired to a circuit in series. Have very low resistance so they do not interfere with the operation of the circuit; the low resistance allows the current to flow through the ___ without changing the flow rate through the circuit. The circuit notation for an ammeter is a circle or square with an A in the middle
conductors
·materials that allow electrons to move easily through them Metals
