AP Physics C E and M

Question

A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected.

The plate separation is now increased to a distance of 2d. Which of the following correctly describes the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?

1) Voltage does not change

Electric field does not change

Energy does not change

2) Voltage does not change

Electric field doubles

Energy does not change

3) Voltage doubles

Electric field doubles

Energy does not change

4) Voltage doubles

Electric field does not change

Energy doubles

5) Voltage doubles

Electric field doubles

Energy doubles

Answer 1

Initially, the battery supplies a voltage V

that leaves a charge Q for the capacitor

of capacitance C and spacing d.

Capacitance determines the amount of charge on the plates

at a voltage V: Q = C V.

Disconnecting the battery means no current flows thereafter, so the resulting

charge Q' = Q.

Capacitance is proportional to the plate area A divided by the separation d.

Doubling d to d' = 2d

makes the capacitance become C' = C / 2.

The voltage was V = Q / C and becomes V' = Q / C' = 2V, thus doubling.

The electric field is E = V / d, and it becomes E' = V' / d' = 2V / 2d = E, unchanged.

The energy is proportional to the work done to move a charge the distance d in the E field:

U = E d and thus U' = E' d' = E (2d) = 2U, thus doubling.

The formula for the energy of a parallel-plate capacitor is

U = (1/2) C V^2

Doubling d halves the capacitance and doubles the voltage, increasing U by the factor (1/2)(2)^2 = 2.

Thus, the answer is choice (4).