# What happens to the potential energy stored in a capacitor when the plates are pushed closer together?

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

$$\implies C_2>C_1$$

$$\displaystyle U_1=\frac{Q^2}{2C_1}$$ and $$\displaystyle U_2=\frac{Q^2}{2C_2}$$

$$\implies U_2

So, the stored potential energy decreases when the plates are pushed closer together. Is this correct?

• You don't have to push them: they attract each other. Commented Nov 25, 2023 at 12:15
• Presumably you’re talking about an air gap capacitor? Commented Nov 25, 2023 at 13:50