Physics 112 Midterm Exam

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?

blocks with - all on one side and + on the other side

A positive particle is moved towards a positive source. A negative particle is also moved towards a positive source. In each case,

both particles moved to a higher potential. 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.

What is the direction of the magnetic field at the position marked by the X ?

Out of the screen

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?

P = IV (32 W) = I (120 V) I = 0.27 A

The property of matter that is responsible for static electric effects is

charge

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.

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.

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.

An object has a positive net charge. The most likely reason for the net charge is that

electrons were transferred off the object.

The electronvolt is a unit that measures

energy

The purpose of a battery in a circuit is to act as a source of

energy.

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

equal to 1.44 W.

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

equal to 10 Hz.

The definition of current is the

flow rate of charge through some point.

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?

kqq/r^2 - kqq/r^2 (9x10^9)(2x10^-6)(4x10^-6)/(0.05)^2=28.8 (9x10^9)(2x10^-6)(6x10^-6)/(0.05)^2=43.2 43.2-28.8= 14.4 N

Potential differences are most closely related to:

position in an electric field.

Electrical fields are created by:

presence of charge.

The purpose of a resistor in a circuit is to

restrict the flow of charge.

Which of the following correctly identifies the initial direction of the force on the moving positively charged particle?

right

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

up

What is the net charge on an object that has gained 8 x 10^13 electrons?

(8*10^13)(-1.6*10^-19)= -1.28*10^-5 Coulombs. -1.3 x 10-5 C

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.

Another way to connect resistors together is in parallel

1/Rp=1/R1+1/R2+1/R3+... Where Rp is the total resistance, and the other R values are the individual resistances of the resistors wired together in parallel. This matches the earlier observation that connecting resistors in parallel should decrease the overall resistance This arrangement will result in a decrease in overall resistance

Two identical batteries, each with a voltage of V, connect in a series to a resistor (see below). The voltage across the resistor is

2V

electric force

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.

Which device would cost more to run continuously for 1 hr?

A television that uses a 120 V potential difference to draw 1.25 A of current

Which of the following correctly shows the equipotentials near a negative point source?

C

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.

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.

A capacitor with a capacitance of 25 μF is connected to a 9V battery. How much charge will be stored in the capacitor?

C = Q/V (25x10-6 F) = Q / (9 V) Q = 2.25x10-4 C

a particular capacitor has plate area of 0.25 m2, a plate separation distance of 2 mm, and needs to be able to store 8.89E-8 C of charge with 12 V of potential difference.

C = Q/V = KεA/d (8.89E-8 C) / (12 V) = K (8.85E-12 F/m)(0.25 m^2) / (0.002 m) K = 6.7

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?

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

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?

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.

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?

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.

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.

How will each of the following changes affect the resistance of a particular wire resistor?

Double the length of the wire Resistance is directly proportional to the length of the wire. Doubling the wire length will double the resistance. Double the radius of the wire 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. Halve the radius but triple the length of the wire 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. Tripling the length of the wire will triple the resistance. Combined, the overall resistance will increase by 4x3 = 12 times greater.

What is the direction of the magnetic field at the position marked by the X?

Down

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?

E = F / q (100 N/C) = F / (1.6x10-19 C) F = 1.6x10-17 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?

E = F / q (500 N/C) = F / (6E-9 C) F = 3E-6 N

Uniform electric field

E = Fe/q Electric fields that have the same strength everywhere are called uniform fields. 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. Where E is the strength of the electric field created by a source charge, Fe is the amount of static electric force the source will apply on the test charge, and q is the test particle's charge. It's worth reiterating that q in this equation is the amount of charge on a particle that is in a field created by some source charge. It is not the charge of the source creating the field.

What is the strength of an electric field measured 0.5 m from a point source with a net charge of 5.0 μC?

E = K*q/r^2 = (9*10^9)*(5.0*10^-6) / (0.5)^2 = 1.8*10^5 N/C

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?

E = kq / r2 EA = (9x109 N-m2/C2)(15x10-9 C) / (0.25 m)2 EA = 2160 N/C EB = (9x109 N-m2/C2)(10x10-9 C) / (1.25 m)2 EB = 57.6 N/C ΣE = ΣEx = EA + EB ΣE = (-2160 N/C) + (57.6 N/C)ΣE = -2102 N/C

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?

E = kq/r2 E = (9x109 N-m2/C2)(8x10-9 C) / (0.5 m)2 E = 288 N/C 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

nonuniform field

E = kq/r2 the amount of force applied by a point source on a test charge does depend on the distance between charges. 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 Where E is the electric field strength created by the source charge, k is the Coulomb's law constant, q is the net charge on the source particle, and r is the distance away from the source particle. Notice that electric field strength also has an inverse-square relationship with distance. It's worth reiterating the q value in this field strength equation is the net charge on the particle creating the field.

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?

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.

If the amount of charge on a particle doubles, what will happen to the field strength measured at some distance r from the particle?

E is proportional to q. If q doubles, E will also double at all locations.

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 I= 2.37 V=19

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

d) Solve for the power rating of the 15 Ω resistor.

P = IV = I2R P = (0.80 A)2 (15 Ω) P = 9.6 W

d) Solve for the power dissipated by the 15 Ω resistor.

P = IV = V2/ R P = (20 V)2 /(15 Ω) P = 26.7 W

A 100 Ω resistor is connected to a 9 V battery. How much energy does the resistor dissipate in 1 minute?

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

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?

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?

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

Power

P=IV is the rate of energy transfer. When you studied mechanical forms of energy in PHY 111, power was defined as work over time In this case, work is equal to the electric potential energy lost by the charged particles as they move around the circuit. Where P is power (the rate of energy transfer), I is the current, and V is the potential difference (voltage). Power has units of A-V, which is the same as J/s. As usual, this ratio is also called a watt (W). The power equation is often combined with Ohm's law: V = IR to create other forms of the power equation: P = IV = I^2R = V^2/R

Which of the following correctly indicates the result of cutting the following permanent magnet in half along the dotted line as shown below?

Each half will have both north and south poles.

Three resistors are connected in series 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? 2R, R, 4R

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.

Which of the following statements regarding equipotentials and electric fields is accurate?

Electric field lines are perpendicular to equipotential lines.

b) Solve for the voltage across each resistor.

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.

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?

I = Q / t I = (15 C) / (20 seconds) I = 0.75 A V = IR (12 V) = (0.75 A)(R) R = 16 Ohms

A particular wire has 0.5 A of current. How many electrons will pass through the wire in one minute?

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) = 1.88x1020 electrons

current.

I=ΔQ/Δt current 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. The amount of charge that flows through the wire in one second Where I is the current, ΔQ is the amount of charge that moves through a given point, and Δt is the time it takes the charge to move through the given point.

Wall outlets in Australia provide a peak voltage of 338 V. What is the RMS current through a device that has an average power rating of 800 W?

Pavg = Vrms Irms 800=(338/1.41)(Irms)

Which of the following processes will NOT result in a change in net charge on an object?

Polarization

Which statement regarding voltage is most accurate?

Positive voltages occur near positive point sources, and negative voltages occur near negative point sources.

static electric force

Proportional to the quantity of net charge on both objects, and proportional to the inverse-square of the distance between the two charged objects. In other words: Fe α q1q2 / r2 Where Fe is the force, q1 and q2 are the net charges on the object, and r is the distance between the two objects.

What is the current in a wire if 4 x 1018 electrons flow through the wire every 3 sec?

Q = (4E18 e)(1.6E-19 C) Q = 0.64 C I = Q / t I = (0.64 C) / (3 s) I = 0.21 A

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?

R = (4.5-0 V) / (0.75 - 0 A) R = 6 Ω V = IR V = (2 A)(6 Ω) V = 12 V

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?

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

Which of the following resistors will experience the largest flow rate of charge? All the resistors have identical resistance.

R4

series connection

Resistors linked together one after another in a line Connecting resistors together in series will result in an increase in total resistance

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

Right

Two +2 C charges are placed at two of the vertices of an equilateral triangle. What is the direction of the net electric field at point P?

Right

What is the direction of the magnetic force on the wire?

Right

Which of the following correctly identifies the initial direction of the force on the moving negatively charged particle?

Right

in a series circuit, the current that flows through the battery and each resistor remains constant.

Rs=R1+R2+R3+... Where Rs is the total series resistance, and the other R values are the resistances of each resistor in series. This matches the earlier observation that adding resistors in series should increase the total resistance.

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

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?

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 = 4.06x1013 electrons were removed.

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?

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.

Explain why a charged balloon will electrostatically stick to a neutral ceiling.

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.

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?

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

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

Which device would cost more to run for one hour? 1. A toaster that draws 3 A of current and has a resistance of 80 Ω 2. A 100 W lightbulb 3. A TV that has a potential difference of 120 V and draws a current of 1.0 A

1. P = I^2R P = (3 A)^2(80 Ω) P = 720 W 2. p=100W 3. 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.

How much energy will an electron gain if it moves through a potential difference of 1.0 V?

1.0 eV

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?

12=0.70(16.67+0.06) v=(I)(req+r) 0.54 Ω

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?

135°

An electron moves through a potential difference of 1500 V. The magnitude of the change in potential energy experienced by the electron is

1500 eV.

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

3F

If each of the resistors has an identical resistance of R, what is the total circuit resistance?

5/3 R

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 > 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).

A > B > C

Which of the following statements regarding charge is most accurate?

All of the above are accurate.

Which of the following is an example of an insulator?

All of the above are examples of insulators

Which of the following arrangements shows the correct way to connect an ammeter to measure the current through bulb 2?

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.

Electric fields

Are created by charge. Electric fields are a way of conceptualizing the region of space near a charge and the ability of that charge to apply forces to other nearby charges. It is just a way of visualizing the space near a single charge. 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.

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?

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)

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?

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

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?

B = uI/(2πr) B = (4πx10-7 T-m/A)(2.0 A) / (2π*0.05 m) B = 8.0x10-6 T b. Bearth = 5x10-5 T Earth's magnetic field is roughly 6 times stronger than the field calculated in part a.

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?

B = uI/(2πr) B = (4πx10-7 T-m/A)(3.0A) / (2π*0.01 m) 6x 10-5 T

The value of the magnetic constant is a material property.

B=μoI/2πr In this course, you'll only focus on wires whose resulting fields form in air. The magnetic constant for air uses the symbol μo and has a value of 4πx10-7 T-m/A. B α μ Where B is the magnetic field strength created by a long, straight current-bearing wire, μo is the magnetic constant, I is current in the wire, and r is the distance away from the wire. It's worth reiterating this equation only works for long straight wires.

Two charged spheres are suspended from strings. The charged spheres repel each other. Which statement correctly explains this behavior?

Both a and b are possible.

Which of the following statements regarding energy and potential is accurate?

Both energy and potential are scalar measurements.

capacitor

C=Q/V Where C is capacitance, Q is the amount of excess charge stored in the capacitor, and V is the voltage required to push the charge onto the capacitor plate. is a device used to store electrical charge. A simple capacitor is made by alternating layers of conductor and insulator. A good capacitor 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 called capacitance. If the capacitor plate area increases, there is more room for charge to spread out. Less interaction between repelling charges is a good thing; larger area means it's easier to store charge. Larger plate areas will result in greater capacitances

electric permittivity

C=ϵA/d constant of proportionality this new equation describes how to design a capacitor to achieve a particular capacitance. is a measure of the resistance encountered when forming electric fields in the insulator. It is represented using the symbol ε, and has units of N/m. The permittivity of a vacuum (sometimes called the permittivity of free space) uses the symbol εo, and has a value of 8.85x10-12 F/m.

Explain how CFL lightbulbs can be brighter than incandescent lightbulbs but still have lower power ratings than incandescent bulbs.

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 capacitors are charged.

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.

Which of the following statements best describes conductors?

Conductors are metal and allow charge to move freely through them.

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.

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

Which statement regarding electric potential is NOT accurate?

Electric potential depends on the amount of charge on a particle in an electric field.

What causes current to flow in a circuit?

Electric potential differences drive the movement of charge.

Compare and contrast the direction of electron flow and the direction of conventional current.

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.

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?

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

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?

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).

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?

F = BILsinθ F/L = (0.1 T)(4 A) F/L = 0.4 N/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?

F = k q1 q2 / d^2 5 = 9*10^9 * 8*10^-6 * 8 * 10^-6 / d^2 d^2 = 0.576/5 = 0.1152 d= 0.34 m

Particle A has a net charge of +4 mC, and is located at coordinate 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.

F = kqq/r2F a on b = (9x109 N-m2/C2)(4x10-3 C)(2x10-3 C) / (2 m)2F a on b = 18,000 N F c on b = (9x109 N-m2/C2)(6x10-3 C)(2x10-3 C) / (3 m)2F 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. ΣF2 = Fy2 + Fx2 ΣF2 = (18,000 N)2 + (12,000 N)2 ΣF = 21,633 N tanθ = (18,000 N) / (12,000 N) θ = tan-1 (1.5) θ = 56.3°

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?

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.

What is the static electric force between a -3.0 μC charge and a +6 μC charge if they are 2.0 cm apart?

F = kqq/r^2 F = (9x10^9 N-m2/C2)(3x10^6 C)(6x10^6 C) / (0.02 m)^2 F = 405 N

large collection of charge - for instance, current moving through a wire?

F = q (Δx / t) Bsinθ = (q / t)(Δx)Bsinθ Where Δx is the distance the charge travels or, in this case, the length of the wire in the field. Replace Δx with L for length and you have: F=BILsinθ

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?

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.

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?

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 proton moves to the left at a speed of 4x105 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?

F = qvBsinθ F = (1.6x10-19 C)(4x105 m/s)(0.5 T)(sin90) F = 3.2x10-14 N Directed up the page (or computer screen)

Experiments have shown when a charged particle moves through a magnetic field, the particle will experience a deflecting force. This is sometimes referred to as a magnetic force or a Lorentz force

F=qvBsinθ The amount of force experienced by the charged particle depends on the amount of net charge, the velocity of the particle, and the strength of the field through which it is moving. Where F is the force, q is the net charge on the particle, v is the velocity of the particle, and Bsinθ is the component of the field perpendicular to the motion of the particle. Any parallel components will not contribute to the force on the charged particle.

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.

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

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?

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

What is the current in a wire if 5E18 electrons pass through the wire every 2 seconds?

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 magnitude of the force on a +3 mC charge moving at a speed of 1000 m/s through a 1.25 T magnetic field?

Fm = Bvq 1.25*1000*3e-3 = 3.75 N

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?

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.

Which of the following statements regarding static electric charge is most accurate?

Force is proportional to the inverse square distance between the charged particles.

Qs

If moving charge is the source of all magnetism, how do permanent magnets work? 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. How can you create a permanent magnet? How can you create an electromagnet? 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. Compare and contrast magnetic fields created by permanent magnets versus electromagnets. 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.

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?

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 = Er V = (50 N/C)(1.6 - 1.0) V = 30 V

b. What adjustments could be made to the system to increase the maximum torque?

Increase the number of loops. Increase the current. Increase the size of the loops. Increase the strength of the magnetic field.

What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the left and the bottom wire has a current of 2 A to the right?

Out of the page

Explain why large plate areas and small plate separation distances result in larger capacitances.

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.

conductors

Materials that allow electrons to move easily through them. Metals are good conductors

insulators

Materials that do not allow electrons to move freely through them. Nonmetals are good insulators

Magnetic effects are caused by

Moving charge

Which of the following statements concerning the structure of an atom is accurate?

Neutrally charged atoms contain equal amounts of protons and electrons.

The units used to measure the strength of an electric field are

Newtons per Coulomb (N/C).

Which of the following correctly identifies the initial direction of the force on the moving negatively charged particle?

No force

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,

None of the above options are correct.

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?

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 FT2 = (0.049 N)2 + (0.012 N)2 FT = 0.0504 N tanθ = (0.049 N) / (0.012 N) θ = tan-1 (4.08) θ = 76.2° above the -x axis

Explain why connecting a wire to an uncharged capacitor would not create a flow of charge through the connecting wire.

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).

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?

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

What if you have more than two charged particles?

Sketch the system if a drawing is not provided. Draw and label a free body diagram. Calculate all the forces. Solve for any x and y force components. Add all the x direction forces together, and then add all the y direction forces together. Use the Pythagorean theorem to combine the components into their vector form. Use arctangent to solve for the direction of the final vector form.

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?

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.

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

Solve for the energy required to put particle A at a position relative to the negative source on the left. Ue = kqq/r Ue = (9x109 N-m2/C2)(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 = (9x109 N-m2/C2)(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 then Ue = 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.

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?

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 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?

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.

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?

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

Solve for the current through the battery.

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

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?

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 = V^2/R P = (9 V)^2 / (57.1 Ω) P = 1.42 W

Solve for the current through the battery.

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

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 above the top wire

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.

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.

b. What causes steady state (direct) current? What causes alternating current?

Steady, unchanging voltages such as those provided by batteries will create direct current. Fluctuating voltages will create alternating current.

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?

T = NIABsinθ T = (15)(3.2 A)(π)(0.035 m)2(0.5 T)(sin 30) T = 0.046 N-m

Solve for the number of charge carrier particles (protons or electrons) transferred on or off the object.

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) = 2.25x1013 electrons were transferred off the object.

Which of the following correctly defines power in a circuit?

The amount of energy provided to a circuit each second The amount of energy dissipated by a circuit each second

b) Solve for the current through each resistor.

The current is the same everywhere through the series circuit so the current is 0.80 A through each resistor.

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.

The distance between A and the positive source is: r2 = (0.30 m)2 + (0.15 m)2 r = 0.34 m ΣUe = kqq/r + kqq/r Ue = (9x109 N-m2/C2)(9x10-6 C)(-8x10-3 C) / (0.15 m) + (9x109 N-m2/C2)(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.

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.

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 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?

The distance between particles C and A is: r2 = 22 + 22 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°

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?

The electron starts with electric potential energy, which is converted into kinetic energy as the particle moves through the electric field. Ue = K Vq = 1/2mv2 (12 V) (1.6x10-19 C) = ½ (9.11x10-31 kg)v2 v =2.05x106 m/s

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 lose energy and move to a higher potential.

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.

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?

The first arrangement correctly places the voltmeter in parallel across bulb 2.

Which of the following correctly identifies the initial direction of the loop's rotation?

The left side of the loop will move out of the screen, while the right side of the loop will move into the screen.

An electron has a velocity of 7.3x106 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?

The magnetic and electric forces must cancel out. ΣF = 0Fe - 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) - (7.3x106 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).

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.

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.

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.

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 particle with a charge of +2 μC is at the position marked by the x. Which statement concerning the particle's energy at this position is most accurate?

The particle will have a positive energy term.

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?

The particles will accelerate towards each other with an increasing acceleration. 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).

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?

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.

Explain why a charged capacitor still has an overall net charge of zero.

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.

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.

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 m2 A = πr2 (8.83E-10 m2) = (π)(r2) r = 1.68E-5 m

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 remain the same.

symbol B represents magnetic field strength

The unit used to measure the strength of a magnetic field is the Tesla (T). Magnetic fields measured in units of Teslas tend to be very small. Another common unit for magnetic field strength is Gauss (G) where 1 G = 10-4 T. Experiments have shown the strength of a magnetic field created by current in a wire is inversely proportional the distance from the wire. The further from the wire, the weaker the field. B α 1 / r Unsurprisingly, experiments have also shown the larger the current, the stronger the field. Magnetic field strength and current in a wire are directly proportional. B α I

b. How much energy does the battery supply to the circuit in 1 min?

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

Particle A has a net charge of -5 μC. Particle B has a net charge of +10 μC. Which of the following is the largest force?

They apply the same force on each other.

Fe = kq1q2/r2

This constant is called the Coulomb's law constant, and has a value of 9x10^9 N-m2/C2.

Magnetic effects are created by the movement of charge

This is worth repeating: the movement of charge is the source of all magnetism Permanent magnets are always on; they are always magnetically attracted to and by other ferromagnetic materials. Examples of ferromagnetic materials are iron, cobalt, nickel, and some alloys.

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?

Ue = kqq/r Ue = (9x109 N-m2/C2)(1.5x10-6 C)(3x10-3 C) / (0.2 m) Ue = 202.5 J

How much electric potential energy does a -7.3 nC charged particle have at a position 2.0 m from a +0.05 C source?

Ue = kqq/r Ue = (9x109 N-m2/C2)(7.3x10-9 C)(0.05 C) / (2.0 m) Ue = 1.64 J However, when a negative charge moves towards a positive source, the charge will lose electric potential energy. To move the negatively charged particle from infinitely far away to the given position, the particle would have to lose 1.64 J of energy. The final answer is then: Ue = -1.64 J

Electric potential energy in nonuniform field

Ue=kq1q2/r If the charged particle loses energy getting into position, then it has negative electric potential energy. If the charged particle gains energy getting into position, then it has positive energy.

electric potential energy in uniform field

Ue=qEr is the energy a charge has due to its position in an electric field. The equation for electric potential energy is very similar to the equation for gravitational potential energy.

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?

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 mv2 (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.

c. What is the cost of leaving the bulb on for 30 days if the cost of energy is $0.14 per kW-hr.?

Using unit analysis: Cost = (32 W)(1 kW / 1000 W)($0.14 / kW-hr)(24 hours / 1 day)(30 days) Cost = $3.23

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.

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

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?

V = IR (12 V) = I (20 Ω) I = 0.6 A

b. What is the operational resistance of the bulb?

V = IR (120 V) = (0.27 A) R R = 444 Ω

c) Solve for the current through each resistor.

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

Explain why the amplitude on an alternating voltage versus time graph is usually larger than the amplitude on an alternating current versus time graph.

V = IR Since R is almost always larger than 1 Ω, the current will usually be smaller than the voltage that causes it.

c) Solve for the voltage across 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

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?

V1 -V2 = Er 80v - V2 = 50*1 V2 = 30 V

Two charged parallel plates have a uniform field strength of 1200 N/C between them. If the separation distance between the plates is 0.15 m, what is the potential difference between the plates?

V=Er V=(1200)(0.15) V= 180V

resistors

V=IR The ability of a resistor to resist current is called its resistance. If resistance increases, flow rate decrease. This relationship is called Ohm's law. If you were to create a graph of voltage versus current, the slope would be equal to the resistance of the circuit. A component in a circuit that hinders the flow of charge through the circuit The constant slope relationship between voltage and current indicates a constant circuit resistance. Resistors that maintain a constant resistance value are called ohmic. There are some nonohmic resistors that do not maintain a constant value.

voltages (electric potential) in uniform fields.

V=Ue/q=Er 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. Where V is electric potential (voltage), Ue is electric potential energy, q is the charge of the particle in the electric field, E is the strength of the electric field, and r is the distance relative to some zero position. is a way of predicting what positions are most likely to result in large amounts of electric potential energy when a positive charge is placed at that position.

voltages (electric potential) in nonuniform fields

V=kq/r

The relationship between the rms value and the peak value is:

Vrms=Vo/√2 Irms=Io/√2 Alternating current and alternating voltage have maximum values (called peak values), and they also have average values. You use a specific statistical type of average called the root-mean-squared (rms) average for these values. Where the naught subscript indicates the peak value. The average power rating is the product of the two rms values. Pavg = Vrms Irms Pavg = (Vo / √2)(Io / √2) Pavg = ½ Vo Io

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.

W = Ue W = qEr W = (3x10-6)(900 N/C)(0.25 m) W = 6.75x10-4 J

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.

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.

Three resistors, (R1 = 1 Ω, R2 = 2 Ω, R3 = 4 Ω), are wired in parallel, and then connected to a 12 V battery.

Which resistor will have the most current? Explain. 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). Solve for the 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 Which resistor will have the greatest voltage drop across the resistor? Explain. 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. Solve for the voltage. Voltage battery = 12 V so the voltage across each of the resistors is 12 V.

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 the same resistance as wire B.

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.

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 - Uinitial ΔUe = (+2414 J) - (-2414 J) ΔUe = 4828 J

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?

a) Gained energy, lost energy ,gained energy

Which of the following is a true statement?

a) If voltage increases, current increases. b) If resistance increases, current increases. c) Both the above

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

a. F is proportional to the inverse square of distance. 1/32 = 1/9 F The force will decrease by a factor of 9. b. F is proportional to q. The force will double or 2F. c. 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.

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?

a. Power values are usually given as average power values. Pavg = ½ Ppeak (2000 W) = ½ (P) P = 4000 W b. 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

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

a. Since current and resistance are inversely proportional, if the resistance doubles, the current will halve. b. 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.

Three resistors, (R1 = 1 Ω, R2 = 2 Ω, R3 = 4 Ω), are wired in series, and then connected to a 12 V battery. a. Which resistor will have the most current? Explain. b. Solve for the current. c. Which resistor will have the greatest voltage drop across the resistor? Explain. d. Solve for the voltage.

a. The current is constant throughout the entire series circuit; the resistors will all experience the same current. b. 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 c. 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. d. V = IR V = (1.7 A)(4 Ω) V = 6.9 V

a. 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?) b. Describe how your graph would change for the U.S. standard values. (Vrms = 120 V and f = 60 Hz)

a. 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. b. 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.

Voltmeters

are devices used to measure voltage differences between various circuit locations. Voltmeters are always wired to a circuit in parallel. Voltmeters 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 voltmeter. The circuit notation for a voltmeter is a circle or square with a V in the middle.

Semiconductors

are a class of materials that, as you might expect, share properties of both.

Batteries

are a source of voltage. Source of energy and cause of potential difference 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.

Ammeters

are devices used to measure current through a particular wire or circuit component. Ammeters are always wired to a circuit in series. Ammeters 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 ammeter 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.

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

in addition to producing light, the incandescent bulb also dissipates more heat than the CFL.

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.

Polarization

is a redistribution of charge rather than a transfer of charge; the total amount of charge on the polarized object does not change.

dielectric

is another name for the insulating material in a capacitor. The specific effects of inserting a dielectric between the capacitor plates depends on whether or not the battery is still connected to the capacitor.

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. Velocity is the distance that charge travels in one second

electronvolt

is the energy an electron would gain (or lose depending on direction) moving through one volt of potential difference. Since V = Ue/q, then Ue = Vq so 1 eV = (1 V)(1.6x10-19 C). In other words, 1 eV = 1.6x10-19 J

Conduction

is the movement of charge from one object to another. It is easy to move charge from metal to metal (conductor to conductor) but much more difficult to move charge from insulator to insulator. The transfer of charge caused by rubbing two objects against each other is a special type of conduction called triboelectric charging.

What would happen if you inserted the dielectric between the capacitor plates while the battery was still connected?

the battery will keep the voltage difference between the capacitor plates at a constant value equal to the voltage of the battery

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 following diagram is best described as

the electric field near two positive point charges.

If a material gains electrons

the material will have a negative net charge

If a material loses electrons

the material will have a positive net charge.

The U.S. has an alternating current frequency of 60 Hz. China has an alternating current frequency of 50 Hz. This means that

the number of times voltage cycles in a minute in the U.S. is greater than the number of times voltage cycles in China.

The definition of power is best described as:

the rate of energy transfer.

Electric potential is

the ratio of energy per amount of charge on a particle in an electric field.

If one object has more of the positive type of charge and one object has more of the negative type of charge

then the objects will attract each other.

If two objects both have more of the negative type of charge than positive (referred to as a negative net charge)

then the objects will repel each other.

If two objects both have more of the positive type of charge than negative (usually referred to as a positive net charge)

then the objects will repel each other.

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?

v^2=2qEd/m v^2= (2)(0.05)(2000)(0.5)/0.01 v=sqt 10,000 v=100

Current is caused by

voltage differences.

A helium nucleus with a charge of +3.2x10-19 C and a mass of 6.64x10-27 kg moving at a speed of 4x105 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?

ΣF = ma qvBsinθ = mv2/r qB = mv/r (3.2x10-19 C)(4.2 T)(sin90) = (6.64x10-27 kg)(4x105 m/s) / r r = 0.00198 m or 1.98 mm

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?

ΣF = ma qvBsinθ = mv^2/r (2E-3 C)(3.0 T) = (m)(20 m/s) / (0.06 m) m=1.8 x 10-5 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?

ΣF = ma qvBsinθ = mv^2/r (4E-3 C)(3.5 T) = (m)(200 m/s) / (0.1 m) m = 7E-6 kg

To summarize, inserting a dielectric with no battery connected will cause the following changes:

εo will increase by a factor of k. C will increase by a factor of k. V will decrease by a factor of k. E will decrease by a factor of k. Q will not be affected.

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.

τ = NIABsinθ τ = (20)(1.0 A)(0.02 m)^2(2 T) τ = 0.016 N-m This is a very small torque.

torque on a current-bearing loop in a magnetic field

τ=NIABsinθ Where τ is the torque, N is the number of loops, I is the current through the loops, A is the area of the loop, and B is the magnetic field strength. You will study the properties of motors and generators in much more detail in the next lesson.