AP Physics C · Electricity and Magnetism

A solid conducting sphere of radius R carries a total charge Q distributed on its surface. What is the magnitude of the electric field at a distance r from the centre, where r < R?

1. Azero
2. BQr/(4πε₀R³)
3. CQ/(4πε₀r²)
4. DQ/(4πε₀R²)

A parallel-plate capacitor is charged by a battery and the battery is then disconnected. A dielectric slab of dielectric constant κ > 1 is inserted, filling the space between the plates. What happens to the charge on the plates and to the potential difference across them?

1. Acharge unchanged, potential difference decreases
2. Bcharge unchanged, potential difference increases
3. Ccharge increases, potential difference unchanged
4. Dcharge decreases, potential difference unchanged

In a series RC circuit connected to a battery of e.m.f. ε at t = 0 with the capacitor initially uncharged, what are the current immediately after the switch is closed and a long time later?

1. Aε/R immediately, and zero a long time later
2. Bzero immediately, and ε/R a long time later
3. Cε/R at both times
4. Dzero at both times

A long solenoid has n turns per unit length and carries a current I. Using Ampère's law, what is the magnitude of the magnetic field well inside the solenoid, far from its ends?

1. Aμ₀I/(2πr)
2. Bμ₀nI
3. Cμ₀nI/2
4. Dμ₀I/(2R)

A circular loop of wire lies in the plane of the page in a uniform magnetic field directed into the page. The magnitude of the field is increasing with time. What is the direction of the induced current in the loop?

1. Aclockwise, because the induced current opposes the field
2. Bcounterclockwise, because the induced current opposes the increase in flux into the page
3. Cclockwise, because the induced current reinforces the increase in flux
4. Dthere is no induced current, because the loop is not moving

In a region of space the electric potential is given by V(x) = 3x² − 2x, where V is in volts and x in metres. What is the x-component of the electric field at x = 2.0 m?

1. A−10 V/m
2. B−8.0 V/m
3. C+8.0 V/m
4. D+10 V/m

An insulating sphere of radius R carries a volume charge density that varies with distance from the centre as ρ(r) = ρ₀ r/R for r ≤ R, where ρ₀ is a positive constant. There is no charge outside the sphere.

1. (a)

   Derive Derive an expression for the total charge Q on the sphere.

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   1. 1

      1 point: uses a spherical shell of volume dV = 4πr²dr as the element of integration

   2. 2

      1 point: Q = ∫₀^R ρ₀(r/R)4πr² dr = (4πρ₀/R)∫₀^R r³ dr

   3. 3

      1 point: Q = πρ₀R³

2. (b)

   Derive Using Gauss's law, derive an expression for the magnitude of the electric field at a distance r from the centre, for r < R.

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   1. 1

      1 point: chooses a concentric spherical Gaussian surface of radius r and states that E is radial and constant on it, so ∮E·dA = E(4πr²)

   2. 2

      1 point: computes the enclosed charge by integrating, q_enc = (4πρ₀/R)(r⁴/4) = πρ₀r⁴/R

   3. 3

      1 point: sets E(4πr²) = q_enc/ε₀

   4. 4

      1 point: E = ρ₀r²/(4ε₀R)

3. (c)

   Derive Derive an expression for the magnitude of the electric field at a distance r from the centre, for r > R, and verify that your two expressions agree at r = R.

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   1. 1

      1 point: outside the sphere the whole charge is enclosed, so E = Q/(4πε₀r²) = ρ₀R³/(4ε₀r²)

   2. 2

      1 point: substitutes r = R into the inside expression to get ρ₀R/(4ε₀)

   3. 3

      1 point: substitutes r = R into the outside expression to get the same value, confirming the field is continuous at the surface

4. (d)

   Derive Derive an expression for the electric potential at the centre of the sphere, taking the potential to be zero at infinity.

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   1. 1

      1 point: V(0) = ∫₀^∞ E dr, splitting the integral at r = R

   2. 2

      1 point: outside contribution ∫_R^∞ ρ₀R³/(4ε₀r²) dr = ρ₀R²/(4ε₀)

   3. 3

      1 point: inside contribution ∫₀^R ρ₀r²/(4ε₀R) dr = ρ₀R²/(12ε₀)

   4. 4

      1 point: V(0) = ρ₀R²/(4ε₀) + ρ₀R²/(12ε₀) = ρ₀R²/(3ε₀)

A capacitor of capacitance C, initially uncharged, is connected in series with a resistor of resistance R and a battery of e.m.f. ε. The switch is closed at t = 0.

1. (a)

   Derive Write the loop equation for the circuit and derive an expression for the charge on the capacitor as a function of time.

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   1. 1

      1 point: ε − IR − q/C = 0 with I = dq/dt

   2. 2

      1 point: separates variables, dq/(εC − q) = dt/(RC)

   3. 3

      1 point: integrates with q = 0 at t = 0

   4. 4

      1 point: q(t) = εC(1 − e^(−t/RC))

2. (b)

   Sketch On separate axes, sketch the charge on the capacitor and the current in the circuit as functions of time. Label the initial value and the asymptotic value on each.

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   1. 1

      1 point: charge rises from zero, concave down, approaching the asymptote q = εC

   2. 2

      1 point: εC labelled on the charge graph

   3. 3

      1 point: current starts at ε/R and decays, concave up, approaching zero

   4. 4

      1 point: ε/R labelled on the current graph

3. (c)

   Determine Determine the time, as a multiple of RC, at which the capacitor holds half of its final charge.

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   1. 1

      1 point: sets εC(1 − e^(−t/RC)) = ½εC

   2. 2

      1 point: e^(−t/RC) = ½, so t/RC = ln 2

   3. 3

      1 point: t = 0.69RC

4. (d)

   Explain The total energy delivered by the battery over the whole charging process is Cε², but only ½Cε² is stored in the capacitor. Explain where the remaining energy goes, and explain why this fraction does not depend on the value of R.

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   1. 1

      1 point: the difference, ½Cε², is dissipated as thermal energy in the resistor

   2. 2

      1 point: the battery moves the full charge εC through the full potential difference ε, while the capacitor's stored energy is the integral of q/C dq, which averages to half that

   3. 3

      1 point: a larger R makes the current smaller but the process correspondingly longer

   4. 4

      1 point: the two effects cancel exactly in ∫I²R dt, so the dissipated energy is ½Cε² whatever the resistance

A student is given a capacitor of unknown capacitance, a resistor of known resistance R = 47 kΩ, a battery, a switch, a voltmeter of very high resistance and a stopwatch. The student wants to determine the capacitance.

1. (a)

   Describe Describe a procedure the student could use, and state which quantities should be recorded.

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   1. 1

      1 point: charge the capacitor fully through the battery, then disconnect the battery and discharge it through the known resistor

   2. 2

      1 point: record the voltmeter reading V at measured time intervals t from the start of the discharge

   3. 3

      1 point: repeat the whole discharge several times and average the readings at each time

   4. 4

      1 point: states why the voltmeter must have a very high resistance — otherwise it provides a parallel discharge path and shortens the time constant

2. (b)

   Describe Describe how the student should linearise the data, and state what the slope of the resulting graph represents.

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   1. 1

      1 point: V = V₀e^(−t/RC), so ln V = ln V₀ − t/(RC)

   2. 2

      1 point: plot ln V on the vertical axis against t on the horizontal axis to obtain a straight line

   3. 3

      1 point: the slope is −1/(RC), so C = −1/(R × slope)

3. (c)

   Determine The student's graph of ln(V/V) against t has a slope of −0.43 s⁻¹. Determine the capacitance.

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   1. 1

      1 point: RC = 1/0.43 = 2.33 s

   2. 2

      1 point: C = 2.33 / 47 × 10³

   3. 3

      1 point: C = 5.0 × 10⁻⁵ F (50 μF)

4. (d)

   Explain Explain why plotting ln V against t is preferable to determining the time constant by reading off the time at which V has fallen to 37% of its initial value.

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   1. 1

      1 point: the single-point method relies on two readings only, V₀ and one other, so an error in either propagates directly into the answer

   2. 2

      1 point: the graph uses every data point, so random errors average out and the slope is better determined

   3. 3

      1 point: the straightness of the line is itself a test that the discharge really is exponential, which the single-point method assumes without checking

A conducting rod of mass m and length L slides without friction on two long horizontal parallel rails separated by L. The rails are joined at one end by a resistor of resistance R, and the whole apparatus sits in a uniform magnetic field of magnitude B directed vertically, perpendicular to the plane of the rails. The rod is given an initial speed v₀ to the right at t = 0 and then released. The rod and rails have negligible resistance.

1. (a)

   Derive Derive expressions for the e.m.f. induced in the circuit and the current in the resistor when the rod moves with speed v.

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   1. 1

      1 point: the flux through the circuit changes as the enclosed area changes, ε = −dΦ/dt = −BL(dx/dt)

   2. 2

      1 point: magnitude of the e.m.f. is BLv

   3. 3

      1 point: I = BLv/R

2. (b)

   Derive Derive an expression for the speed of the rod as a function of time.

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   1. 1

      1 point: the force on the current-carrying rod is F = BIL = B²L²v/R, directed opposite to the motion by Lenz's law

   2. 2

      1 point: m dv/dt = −B²L²v/R

   3. 3

      1 point: separates and integrates from v₀ at t = 0

   4. 4

      1 point: v(t) = v₀e^(−B²L²t/(mR))

3. (c)

   Determine Determine the total energy dissipated in the resistor from t = 0 until the rod has effectively stopped, and determine the total distance the rod travels.

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   1. 1

      1 point: the only energy source is the rod's kinetic energy, and it all ends up in the resistor, so the total dissipated energy is ½mv₀²

   2. 2

      1 point: distance x = ∫₀^∞ v dt = v₀ mR/(B²L²)

   3. 3

      1 point: states both results with correct symbols

4. (d)

   Explain Explain, without deriving anything further, why the rod slows down, and explain what would be observed instead if the resistor were replaced by a superconducting wire of zero resistance.

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   1. 1

      1 point: the motion changes the flux through the circuit, which drives a current

   2. 2

      1 point: by Lenz's law that current is in the direction whose magnetic force opposes the motion, so the rod decelerates — the kinetic energy is transferred to the resistor as thermal energy

   3. 3

      1 point: with zero resistance the current, and hence the retarding force, would become arbitrarily large for any non-zero speed

   4. 4

      1 point: so the rod would stop almost immediately, and in the idealised limit the flux through the circuit cannot change at all — which is the behaviour that makes a superconducting loop hold its flux fixed