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AP Physics C: Electricity and Magnetism Hard practice

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Question 1

A parallel-plate capacitor (area $A$ , separation $d$ , no dielectric) is connected to a constant voltage source $V$ . A dielectric slab of thickness $t < d$ and dielectric constant $κ$ is inserted so that it fills the cross-section but only part of the gap. The capacitance of the resulting configuration is:

C = \frac{ϵ _{0} A}{d - t + t / κ}

C = \frac{κ ϵ _{0} A}{d}

C = \frac{ϵ _{0} A}{d + t ( κ - 1 )}

C = \frac{κ ϵ _{0} A}{d - t}

Question 2

An RC circuit with $R = 1 k Ω$ and $C = 10 μ F$ is driven by a square-wave voltage that steps from $0$ to $V_{0} = 10 V$ at $t = 0$ (with $C$ initially uncharged). The voltage across the capacitor at $t = 2 τ$ (where $τ = R C$ ) is:

V_{C} (2 τ) = V_{0} (1 - e^{- 2}) \approx 8.65 V

V_{C} (2 τ) = V_{0} (1 - e^{- 1/2}) \approx 3.94 V

V_{C} (2 τ) = 2 V_{0} (1 - e^{- 1}) \approx 12.6 V

V_{C} (2 τ) = V_{0} e^{- 2} \approx 1.35 V

Question 3

An RC circuit has $R = 10 k Ω$ , $C = 20 μ F$ , and battery $V_{0} = 6 V$ . The switch is closed at $t = 0$ with $C$ initially uncharged. The energy dissipated in the resistor from $t = 0$ to $t \to \infty$ is:

E_{R} = \frac{1}{2} C V_{0}^{2} = 360 μ J

E_{R} = C V_{0}^{2} = 720 μ J

E_{R} = \frac{V _{0}^{2}}{2 R} = 1.8 \times 1 0^{- 3} J

E_{R} = 0

, because an ideal resistor stores no energy.

Question 4

Two large parallel conducting plates separated by distance $d$ each carry surface charge density $+ σ$ and $- σ$ respectively. A small conducting sphere of radius $a ≪ d$ with net charge $q$ is held midway between the plates. Ignoring image charges, the electric force on the sphere is:

F = q σ / ϵ_{0}

, directed from the positive to the negative plate.B.

F = q σ / (2 ϵ_{0})

, directed from the positive to the negative plate.C.

F = 0

, because the sphere is midway between the plates.D.

F = 2 q σ / ϵ_{0}

, directed from the positive to the negative plate.

Question 5

Maxwell's addition to Ampère's law introduces the displacement current $I_{d}$ . For a parallel-plate capacitor being charged, the displacement current between the plates (where there is no real conduction current) is defined as:

I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t}

I_{d} = \frac{1}{μ _{0}} \frac{d Φ _{B}}{d t}

I_{d} = μ_{0} ϵ_{0} \frac{d Φ _{E}}{d t}

I_{d} = ϵ_{0} \frac{d E}{d t}

(scalar

E

, not flux).

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Question 6

An LC circuit consists of an inductor $L = 0.10 H$ and capacitor $C = 100 μ F$ . The capacitor is initially charged to $V_{0} = 10 V$ and the switch is closed at $t = 0$ . The angular frequency of oscillation $ω$ and the maximum current $I_{m a x}$ are:

ω = 1/ L C \approx 316 rad/s

;

I_{m a x} = V_{0} C / L \approx 0.316 A

ω = 1/ L C \approx 316 rad/s

;

I_{m a x} = V_{0} / R =

undefined (no

R

)C.

ω = L C \approx 0.00316 rad/s

;

I_{m a x} = V_{0} / L C \approx 3162 A

ω = 1/ (L C) \approx 1 0^{5} rad/s

;

I_{m a x} = C V_{0} ω =

incorrect

Question 7

A thin disk of radius $R$ has uniform surface charge density $σ$ . Using integration over rings, the electric field at a point on the axis of the disk at distance $z$ from the center is:

E = \frac{σ}{2 ϵ _{0}} (1 - \frac{z}{z ^{2} + R ^{2}})

E = \frac{σ}{2 ϵ _{0}} (1 + \frac{z}{z ^{2} + R ^{2}})

E = \frac{σ R}{2 ϵ _{0} z ^{2} + R ^{2}}

E = \frac{σ z}{2 ϵ _{0} ( z ^{2} + R ^{2} )}

Question 8

In the Wheatstone bridge circuit, resistors $R_{1}$ (top-left) and $R_{2}$ (top-right) form one pair of arms; $R_{3}$ (bottom-left) and $R_{X}$ (bottom-right) form the other. A galvanometer connects the midpoints. The bridge is balanced when:

R_{1} / R_{3} = R_{2} / R_{X}

R_{1} + R_{X} = R_{2} + R_{3}

R_{1} R_{2} = R_{3} R_{X}

R_{1} / R_{2} = R_{X} / R_{3}

Question 9

Two capacitors $C_{1} = 4 μ F$ (initially charged to $V_{1} = 12 V$ ) and $C_{2} = 8 μ F$ (initially uncharged) are connected in parallel by closing a switch. The final common voltage and the energy dissipated in the connecting wires are:

V_{f} = 4 V

;

Δ U_{dissipated} = 192 μ J

V_{f} = 6 V

;

Δ U_{dissipated} = 144 μ J

V_{f} = 4 V

;

Δ U_{dissipated} = 0

V_{f} = 12 V

;

Δ U_{dissipated} = 576 μ J

Question 10

The energy density stored in the electric field of a parallel-plate capacitor filled with dielectric $κ$ and field $E$ is:

u = \frac{1}{2} κ ϵ_{0} E^{2}

u = \frac{1}{2} ϵ_{0} E^{2}

u = \frac{1}{2 κ ϵ _{0} E ^{2}}

u = κ ϵ_{0} E^{2}

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Question 11

A Hall-effect sensor: a flat conductor (width $w$ , thickness $t$ ) carries current $I$ in the $+ x$ direction in a field $B = B \overset{z}{^}$ . Charge carriers are electrons (charge $- e$ , number density $n$ ). At steady state, the magnitude of the Hall voltage across width $w$ is:

V_{H} = \frac{I B}{n e t}

V_{H} = \frac{I B}{2 n e t}

V_{H} = 0

, because the magnetic force on electrons and ions cancels.D.

V_{H} = \frac{I B}{2 n e t}

(split evenly between edges).

Question 12

A proton ( $m_{p} = 1.67 \times 1 0^{- 27} kg$ , $q = 1.60 \times 1 0^{- 19} C$ ) moving with speed $v = 2.0 \times 1 0^{6} m/s$ enters a region with both $E$ and $B$ fields. For the proton to travel in a straight line (velocity selector), with $v = v \overset{x}{^}$ , $B = B \overset{z}{^} = 0.050 T \overset{z}{^}$ , the required electric field $E$ and its magnitude are:

E = 1.0 \times 1 0^{5} V/m (- \overset{y}{^})

E = 1.0 \times 1 0^{5} V/m (+ \overset{y}{^})

E = 1.0 \times 1 0^{5} V/m (+ \overset{x}{^})

E = 4.0 \times 1 0^{- 14} V/m (+ \overset{y}{^})

Question 13

A transformer has primary coil $N_{1} = 200$ turns and secondary coil $N_{2} = 50$ turns. The primary is connected to $V_{1} = 120 V$ (rms) AC. A load $R_{L} = 10 Ω$ is connected to the secondary. Assuming an ideal transformer (no losses), the primary current $I_{1}$ (rms) is:

I_{1} = 0.75 A

I_{1} = 12 A

I_{1} = 3.0 A

I_{1} = 0.60 A

Question 14

A solid insulating cylinder of radius $R$ and infinite length has non-uniform volume charge density $ρ (r) = ρ_{0} r / R$ , where $r$ is the distance from the axis. Using Gauss's law, the electric field magnitude for $r < R$ is:

E = \frac{ρ _{0} r ^{2}}{3 ϵ _{0} R}

E = \frac{ρ _{0} r}{3 ϵ _{0}}

E = \frac{ρ _{0} r ^{2}}{2 ϵ _{0} R}

E = \frac{ρ _{0} R}{3 ϵ _{0}}

Question 15

A toroid has $N$ turns, inner radius $a$ , outer radius $b$ , and carries current $I$ . Using Ampère's law with a circular Amperean loop of radius $r$ ( $a < r < b$ ), the magnetic field inside the toroid is:

B = \frac{μ _{0} N I}{2 π r}

B = μ_{0} n I

where

n = N / (2 π r)

, same as a straight solenoid.C.

B = \frac{μ _{0} N I}{2 π a}

(uniform inside the toroid).D.

B = 0

inside the toroid.

AP Physics C: Electricity and Magnetism Hard practice — Free with Answer Explanations | Test Practice Hub