2024 AP AP Physics C: Electricity and Magnetism Practice Paper with Answers

(a) Using Gauss's Law:
$$\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}$$
$$\Phi_E = \frac{3.0 \times 10^{-9}\text{ C}}{8.85 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2)} \approx 339\text{ N}\cdot\text{m}^2/\text{C}$$

(b)
i. The work done by an external force is \(W_{\text{ext}} = q_0 \Delta V\).
- From \(K\) to \(M\): \(\Delta V = V_M - V_K = 9.0\text{ V} - 9.0\text{ V} = 0\text{ V}\).
$$W_{KM} = q_0(0\text{ V}) = 0\text{ J}$$
- From \(J\) to \(K\): \(\Delta V = V_K - V_J = 9.0\text{ V} - 15.0\text{ V} = -6.0\text{ V}\).
$$W_{JK} = (2.0 \times 10^{-9}\text{ C})(-6.0\text{ V}) = -1.2 \times 10^{-8}\text{ J}$$

ii. The approximate electric field is:
$$|E_y| = \left| -\frac{\Delta V}{\Delta y} \right| = \left| -\frac{V_K - V_J}{y_K - y_J} \right| = \left| -\frac{9.0\text{ V} - 15.0\text{ V}}{3(0.20\text{ m}) - 2(0.20\text{ m})} \right| = \frac{6.0\text{ V}}{0.20\text{ m}} = 30\text{ V/m}$$

(c) Direction: \(+y\)
Justification: The positive point charge at \((0, d)\) repels the positive test charge placed at \((0, 2d)\) along the line connecting them, exerting an electric force in the \(+y\)-direction. Due to the symmetrical placement of the charged rod from \(x = -d\) to \(x = +d\) about the \(y\)-axis, the horizontal (\(x\)) components of the repulsive electric force exerted by symmetric charge elements cancel out completely. The vertical components of the repulsive forces from the rod all point in the \(+y\)-direction (away from the rod since \(y = 2d > 0\)). Thus, the net electric force points strictly in the \(+y\)-direction.

(d)
i. An infinitesimal charge element along the rod is \(dq = \lambda\, dx'\), located at \(x'\) where \(0 \le x' \le L\).
The distance from \(dq\) to Point \(P\) at \(x_P\) is \(r = x_P - x'\).
Using the integral for electric potential with \(V(\infty) = 0\):
$$V_P = \int \frac{k\, dq}{r} = \int_0^L \frac{k \lambda\, dx'}{x_P - x'}$$
Evaluating the integral:
$$V_P = k\lambda \left[ -\ln(x_P - x') \right]_0^L = -k\lambda \left( \ln(x_P - L) - \ln(x_P) \right) = k\lambda \ln\left(\frac{x_P}{x_P - L}\right)$$

ii. The electric field component is \(E_x = -\frac{dV_P}{dx_P} = \frac{k\lambda L}{x_P(x_P - L)}\).
For \(x > L\):
- \(E_x\) is positive (directed in the \(+x\)-direction away from the positive charge distribution).
- \(E_x\) approaches \(+\infty\) asymptotically as \(x \to L^+\).
- \(E_x\) decreases continuously and approaches \(0\) asymptotically as \(x \to \infty\).
- The graph is concave up throughout \(x > L\).

Part (a): 2 points
- 1 point: For using Gauss's law with the correct enclosed charge (\(\Phi_E = Q_{\text{enc}}/\varepsilon_0\)).
- 1 point: For the correct numerical value with or without units (\(\approx 339\text{ N}\cdot\text{m}^2/\text{C}\) or \(3.4 \times 10^2\text{ V}\cdot\text{m}\)).

Part (b)(i): 2 points
- 1 point: For correctly determining that \(W_{KM} = 0\text{ J}\) because points \(K\) and \(M\) are at the same electric potential.
- 1 point: For correctly calculating \(W_{JK} = -1.2 \times 10^{-8}\text{ J}\) (or stating magnitude \(1.2 \times 10^{-8}\text{ J}\)) using \(W = q\Delta V\).

Part (b)(ii): 2 points
- 1 point: For using a correct relationship between electric field and potential gradient (e.g., \(|E_y| = |\Delta V / \Delta y|\)).
- 1 point: For substituting correct potential values and distance \(\Delta y = 0.20\text{ m}\) to obtain \(30\text{ V/m}\).

Part (c): 3 points
- 1 point: For selecting only \(+y\) with an attempt at a relevant justification.
- 1 point: For explaining that horizontal components of force from the rod cancel by symmetry.
- 1 point: For explaining that repulsive forces from both the point charge and the rod act in the \(+y\)-direction on the positive test charge.

Part (d)(i): 4 points
- 1 point: For writing a correct integral expression for potential \(V = \int \frac{k\, dq}{r}\).
- 1 point: For correctly expressing \(dq = \lambda\, dx'\) and the distance \(r = x_P - x'\).
- 1 point: For setting the correct limits of integration from \(0\) to \(L\).
- 1 point: For integrating correctly to obtain \(V_P = k\lambda \ln\left(\frac{x_P}{x_P - L}\right)\).

Part (d)(ii): 2 points
- 1 point: For indicating a curve that is strictly positive and approaches zero as \(x \to \infty\).
- 1 point: For indicating a concave-up curve that approaches \(+\infty\) as \(x \to L^+\).

(a) i.
Immediately after the switch is closed, the uncharged capacitor has \(\Delta V_C = 0\), and the entire emf \(\mathcal{E}\) falls across the resistor combination. As charge accumulates on the capacitor, the current in the circuit decreases exponentially toward zero, causing the potential difference across the resistors to decrease from \(\mathcal{E}\) to \(0\). Therefore, the voltmeter must be connected in parallel across either one of the resistors or across the parallel pair of resistors.

ii.
1. Construct the circuit with the battery of emf \(\mathcal{E}\), open switch, uncharged capacitor \(C\), and the two identical resistors of resistance \(R\) connected in parallel with each other.
2. Connect the voltmeter in parallel across the parallel combination of resistors.
3. Close the switch to begin charging the capacitor and simultaneously start recording the potential difference \(\Delta V_R(t)\) as a function of time \(t\).
4. Record data continuously at regular time intervals until the potential difference approaches zero (steady state is reached).

(b) i.
- Vertical axis: Potential difference \(\Delta V\) (or \(\Delta V_R\))
- Horizontal axis: Time \(t\)
- Curve shape: Concave up, exponentially decaying from a vertical intercept at \(\Delta V = \mathcal{E}\) at \(t = 0\) and asymptotically approaching \(\Delta V = 0\) as \(t \to \infty\).

(b) ii.
The potential difference across the parallel resistor combination is:
\[ \Delta V_R(t) = \mathcal{E} e^{-t/\tau} \implies \ln(\Delta V_R) = \ln(\mathcal{E}) - \frac{t}{\tau} \]
where the equivalent resistance is \(R_{\text{eq}} = \frac{R \cdot R}{R + R} = \frac{R}{2}\), so the time constant is \(\tau = R_{\text{eq}}C = \frac{RC}{2}\).
- Plot \(\ln(\Delta V_R)\) on the vertical axis versus \(t\) on the horizontal axis.
- Fit a straight line to the data. The slope \(m\) of the best-fit line is equal to \(-\frac{1}{\tau} = -\frac{2}{RC}\).
- The experimental capacitance is calculated as:
\[ C = -\frac{2}{R \cdot m} = \frac{2}{R |m|} \]

(c)
Applying Kirchhoff's loop rule around the single loop:
\[ \sum \Delta V = 0 \implies \mathcal{E} - \Delta V_R - \Delta V_C = 0 \]
The two resistors in parallel have an equivalent resistance of \(R_{\text{eq}} = \frac{R}{2}\). The potential difference across the combination carrying total current \(I = \frac{dq}{dt}\) is \(\Delta V_R = I R_{\text{eq}} = \frac{R}{2}\frac{dq}{dt}\).

The potential difference across the capacitor with charge \(q\) is \(\Delta V_C = \frac{q}{C}\).

Substituting these expressions gives the differential equation:
\[ \mathcal{E} - \frac{R}{2}\frac{dq}{dt} - \frac{q}{C} = 0 \quad \text{or} \quad \frac{dq}{dt} = \frac{2}{R}\left(\mathcal{E} - \frac{q}{C}\right) \]

(d)
Correct selection: \(t_2 > t_1\)

Justification:
Inserting a dielectric of dielectric constant \(\kappa > 1\) increases the capacitance of the capacitor according to \(C' = \kappa C > C\). Because the equivalent resistance \(R_{\text{eq}} = R/2\) of the circuit remains constant, the capacitive time constant of the circuit \(\tau' = R_{\text{eq}}C' = \frac{\kappa R C}{2}\) increases. Since the potential difference across the resistors decays according to \(\Delta V_R(t) = \mathcal{E}e^{-t/\tau}\), the half-life time required to reach \(\Delta V_R = 0.5\mathcal{E}\) is \(t_{1/2} = \tau \ln(2)\). Because \(\tau' > \tau\), the time \(t_2\) required to drop to half of the initial potential difference is greater than \(t_1\).

Part (a)(i): 1 point
- 1 pt: For correctly indicating that the voltmeter must be connected in parallel across one of the resistors or across the parallel resistor pair.

Part (a)(ii): 2 points
- 1 pt: For describing a procedure that involves measuring potential difference across the resistor(s) over time starting immediately upon closing the switch.
- 1 pt: For indicating that data is collected continuously until steady-state conditions are reached (or for a sufficient time interval to establish the decay curve/time constant).

Part (b)(i): 4 points
- 1 pt: For correctly labeling the vertical axis as potential difference and the horizontal axis as time.
- 1 pt: For sketching a concave-up, decreasing exponential curve.
- 1 pt: For indicating a vertical intercept at \(\Delta V = \mathcal{E}\).
- 1 pt: For showing and labeling a horizontal asymptote at \(\Delta V = 0\).

Part (b)(ii): 2 points
- 1 pt: For describing a valid linearization method (e.g., plotting \(\ln(\Delta V)\) vs. \(t\)) or identifying the time constant \(\tau\) from the graph (such as the time when \(\Delta V = \mathcal{E}/e \approx 0.37\mathcal{E}\)).
- 1 pt: For correctly relating the slope/time constant to the equivalent resistance \(R_{\text{eq}} = R/2\) to solve for \(C\) (e.g., \(\text{slope} = -2/(RC)\implies C = -2/(R \cdot \text{slope})\)).

Part (c): 3 points
- 1 pt: For starting with a valid statement of Kirchhoff's loop rule (e.g., \(\mathcal{E} - \Delta V_R - \Delta V_C = 0\)).
- 1 pt: For correctly expressing \(\Delta V_R\) using the parallel equivalent resistance \(R_{\text{eq}} = R/2\) and \(I = dq/dt\) as \(\frac{R}{2}\frac{dq}{dt}\).
- 1 pt: For expressing \(\Delta V_C\) as \(q/C\) and presenting a correct unintegrated differential equation in terms of the specified variables.

Part (d): 3 points
- 1 pt: For selecting '\(t_2 > t_1\)' with an attempt at a relevant justification.
- 1 pt: For stating that the capacitance increases due to the dielectric (\(C' = \kappa C\)).
- 1 pt: For explaining that a larger capacitance increases the time constant (\(\tau = R_{\text{eq}}C\)), thereby increasing the time required for the potential difference to drop to half of its initial value.

(a) i. The magnetic flux through the closed loop formed by the rails, resistor, and rod is given by:
\[ \Phi_B = \int \vec{B} \cdot d\vec{A} = B_0 A = B_0 L x \]
Applying Faraday's law of induction:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(B_0 L x) = -B_0 L \frac{dx}{dt} = -B_0 L v \]
Taking the magnitude:
\[ \mathcal{E} = B_0 L v \]

(a) ii. Upward (toward the top rail)
Justification: As the rod moves to the right, the area of the loop increases, which increases the magnetic flux directed into the page (\(-z\)-direction). According to Lenz's law, the induced current must produce an opposing magnetic field directed out of the page (\(+z\)-direction). By the right-hand rule, this requires a counterclockwise induced current around the closed loop. Therefore, the induced current flows upward through the resistor \(R\) (from the bottom rail to the top rail).

(b) By Ohm's law, the magnitude of the induced current in the loop is:
\[ I = \frac{\mathcal{E}}{R} = \frac{B_0 L v}{R} \]
The magnetic force on the conducting rod carrying current \(I\) downward through the rod is:
\[ \vec{F}_B = I (\vec{L} \times \vec{B}) \]
By the right-hand rule, this force points to the left (\(-x\)-direction), with magnitude:
\[ F_B = I L B_0 = \left(\frac{B_0 L v}{R}\right) L B_0 = \frac{B_0^2 L^2 v}{R} \]
Applying Newton's second law to the rod:
\[ \Sigma F_x = m a_x \implies F_0 - F_B = m \frac{dv}{dt} \]
\[ F_0 - \frac{B_0^2 L^2}{R}v = m\frac{dv}{dt} \]

(c) At terminal velocity \(v_{\text{term}}\), the acceleration of the rod becomes zero (\(\frac{dv}{dt} = 0\)):
\[ F_0 - \frac{B_0^2 L^2}{R}v_{\text{term}} = 0 \implies v_{\text{term}} = \frac{F_0 R}{B_0^2 L^2} \]

(d) i.
- The graph of \(v(t)\) starts at \((0,0)\).
- It is concave-down and increases monotonically.
- It asymptotically approaches the horizontal asymptote at \(v = \frac{F_0 R}{B_0^2 L^2}\).

(d) ii.
The power dissipated in the resistor is given by:
\[ P(t) = I(t)^2 R = \frac{\mathcal{E}(t)^2}{R} = \frac{B_0^2 L^2 v(t)^2}{R} \]
At \(t = 0\), \(v(0) = 0\), so \(P(0) = 0\).
As \(t \to \infty\), \(v \to v_{\text{term}}\), so the maximum power dissipated is:
\[ P_{\text{max}} = \frac{B_0^2 L^2}{R} \left(\frac{F_0 R}{B_0^2 L^2}\right)^2 = \frac{F_0^2 R}{B_0^2 L^2} \]
The graph starts at \((0,0)\), increases continuously, and asymptotically approaches \(P_{\text{max}} = \frac{F_0^2 R}{B_0^2 L^2}\).

(e) \(v_{\text{term, new}} > v_{\text{term}}\)
Justification: Doubling the resistance from \(R\) to \(2R\) reduces the induced current for any given velocity by half (\(I = B_0 L v / (2R)\)), which in turn reduces the opposing magnetic drag force \(F_B\) at that speed. Because a higher velocity is now required for the magnetic braking force to balance the constant applied external force \(F_0\), the new terminal velocity \(v_{\text{term, new}} = \frac{F_0 (2R)}{B_0^2 L^2} = 2 v_{\text{term}}\) is greater than the original.

Part (a)(i): 2 points
- 1 point: For setting up Faraday's law \(\mathcal{E} = -\frac{d\Phi_B}{dt}\) with correct flux definition \(\Phi_B = B_0 L x\).
- 1 point: For substituting \(\frac{dx}{dt} = v\) to derive \(\mathcal{E} = B_0 L v\).

Part (a)(ii): 2 points
- 1 point: For correctly selecting 'Upward (toward the top rail)'.
- 1 point: For a valid justification referencing Lenz's law (increasing downward flux produces an upward/out-of-page induced field, requiring counterclockwise current).

Part (b): 3 points
- 1 point: For a valid expression for the magnetic braking force \(F_B = I L B_0 = \frac{B_0^2 L^2 v}{R}\).
- 1 point: For applying Newton's second law \(\Sigma F = m\frac{dv}{dt}\) with opposing forces \(F_0 - F_B\).
- 1 point: For the correct single differential equation in terms of the specified variables: \(F_0 - \frac{B_0^2 L^2}{R}v = m\frac{dv}{dt}\).

Part (c): 2 points
- 1 point: For setting the acceleration \(\frac{dv}{dt} = 0\) or setting \(F_B = F_0\).
- 1 point: For correctly solving for \(v_{\text{term}} = \frac{F_0 R}{B_0^2 L^2}\).

Part (d)(i): 2 points
- 1 point: For sketching a curve starting at the origin that is concave down and strictly increasing.
- 1 point: For clearly indicating and labeling the horizontal asymptote at \(v = \frac{F_0 R}{B_0^2 L^2}\).

Part (d)(ii): 2 points
- 1 point: For sketching an increasing curve starting at the origin and asymptotically leveling off.
- 1 point: For correctly identifying and labeling the asymptote at \(P = \frac{F_0^2 R}{B_0^2 L^2}\).

Part (e): 2 points
- 1 point: For selecting '\(v_{\text{term, new}} > v_{\text{term}}\)' with an attempt at a relevant justification.
- 1 point: For correctly justifying that a larger resistance reduces current and magnetic braking force at a given speed, thus requiring a higher speed to reach force equilibrium (or referencing \(v_{\text{term}} \propto R\)).