phy 112 midterm practice problems

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

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?

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.

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.

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?

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

A component in a circuit that hinders the flow of charge through the circuit

What is an electric resistor?

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.

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?

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?

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.?

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 = 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?

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?

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?

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.

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?

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?

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.

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?

· 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.

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?

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.

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.

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.

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.

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.

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.

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.

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?

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?

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

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?

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

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?

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.

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.

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?

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.

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?

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?

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?

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.

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.

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.

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?

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.

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?

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?

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

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?

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 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.