Capacitors AP Physics 2 FRQ – Exam Style Questions etc.
Capacitors AP Physics 2 FRQ
Unit 10: Electric Force, Field, and Potential
Weightage : 15–18%
Exam Style Practice Questions , Capacitors AP Physics 2 FRQ
Question
An uncharged air-filled capacitor has square plates of side L whose separation can be varied. The initial plate separation is \(d_{1}\), which is comparable to L. The capacitor is connected to a battery with potential difference \(V_{0}\) and a switch S, as shown above. Assume the dielectric constant of the air is 1.0.
(a) The switch is closed and remains closed for a long time.
i. On the diagram below, draw a vector at each dot to represent the electric field created by the charge on the capacitor plates.
ii. The plates are now moved to a new separation \(d_{2}\) << \(d_{1}\) and allowed to reach equilibrium. Calculate the magnitude of the charge on each plate of the capacitor in terms of \(V_{0}\), L, \(d_{2}\), and physical constants, as appropriate.
(b) The plates are pulled apart to a separation of 2\(d_{2}\), where 2\(d_{2}\) << \(d_{1}\).
i. Suppose the switch remains closed while the plates are pulled apart. Compare the magnitude of the new electric field between the plates when their separation is 2\(d_{2}\) to the magnitude of the field for plate separation \(d_{2}\). Support your comparison using appropriate physics principles.
ii. Suppose the switch is opened before the plates are pulled apart. Compare the magnitude of the new electric field between the plates when their separation is 2\(d_{2}\) to the magnitude of the field for plate separation \(d_{2}\). Support your claim using appropriate physics principles.
(c) The new circuit shown below is now assembled. The plates of the original capacitor, \(C_{1}\), are at separation \(d_{2}\). A second capacitor, \(C_{2}\), with square plates of side 2L and plate separation \(d_{2}\), is connected in parallel with the original capacitor. Both capacitors are initially uncharged. The switch is closed and the circuit reaches equilibrium. Let \(Q_{0}\) be the magnitude of the charge on each plate of capacitor \(C_{1}\) in the previous circuit, as determined in part (a)(ii). In terms of \(Q_{0}\), calculate the magnitude of the charge on each plate of the two capacitors in the new circuit.
Answer/Explanation
Ans:
(a) i) 
ii) \(Q=C\Delta V=(\varepsilon _{0}L^{2}/d_{2})V_{0}\)
\( Q=\varepsilon _{0}L^{2}V_{0}/d_{2}\)
(b) i) Since the battery remains connected, the potential difference across the capacitor remains the same. The electric field must decrease since \(E=\Delta V/d\) and d has increased
ii) The charge on the plates does not change, so E remains the same. \(E=\Delta V/d\) = Q/Cd. Since C \(\alpha\) 1/d , E is independent of the plate spacing which is the only thing that changes.
(c) The conditions for \(C_{1}\) are the same as in part (a)(ii), so it has a charge \(Q_{0}\). The area of \(C_{2}\) is four times that of \(C_{1}\), so it has four times the capacitance. In order to have the same potential difference as \(C_{1}\), it must therefore have four times the charge.