A capacitor is charged up with a battery and then removed from the battery. What will happen to the stored potential energy when the plates are pushed closer together?
My attempt:
Let the subscript $1$ denote the initial values of the capacitor, while the subscript $2$ denotes the final values of the capacitor after the plates are pushed closer together. Let the charge stored in the capacitor be $Q$. We note that the charge stored in the capacitor does not change when the plates are pushed closer together.
$Q=C_1V_1=C_2V_2$
$\displaystyle C_1=\epsilon_0\frac{A}{d_1}$ and $\displaystyle C_2=\epsilon_0\frac{A}{d_2}$
$d_2<d_1$
$\implies C_2>C_1$
$\displaystyle U_1=\frac{Q^2}{2C_1}$ and $\displaystyle U_2=\frac{Q^2}{2C_2}$
$\implies U_2<U_1$
So, the stored potential energy decreases when the plates are pushed closer together. Is this correct?
Best Answer
Correct. But the plates don't get "pushed" together, they get pulled together by the attractive force between them. In order that they can move together it would have to be an air gap capacitor. An air gap capacitor needs some means to hold the plates apart. Without getting into the practical aspects of air gap capacitors, let's imagine the following:
On end of a relaxed coil spring is connected to one plate of the capacitor and the other end of the spring connected to some fixed object. The mechanism for keeping the plates apart is then removed. The force of attraction between the plates pulls on the plate connected to the spring stretching the spring. This causes the spring to acquired elastic potential energy, which comes from the decrease in the electrical potential energy of the capacitor.
In effect the capacitor loses potential energy because negative work must be done by an external agent when the plates to move together.
Hope this helps.