Question

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to
A)4Q
B)2Q
C)Q
D)Q/2
E)Q/4

Answer 1

Answer:

D) Q/2

Explanation:

The amount of charge on a capacitor is given by:

Q = CV

where

C is the capacitance

V is the voltage

The capacitance of a parallel-plate capacitor is given by

$C=\epsilon_0 \frac{A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

Substituting the last formula into the first one, we can write

$Q=\epsilon_0 \frac{AV}{d}$

In this problem, the two plates are pulled apart to twice their original separation, so

d' = 2d

While the voltage V is kept constant. Therefore, the new charge stored in the capacitor will be

$Q'=\epsilon_0 \frac{AV}{2d}=\frac{1}{2} \epsilon_0 \frac{AV}{d}=\frac{Q}{2}$