# Let there be a PP capacitor where d increases, what happens to the charge?

If a fully charged parallel plate capacitor is disconnected from a battery and its plates are pulled apart to increase the distance between the plates, what is the value of charge on the capacitor?

I know the charge and the electric field will both remain the same, but I don't know why or how?
Since d increases, the capacitance C will decrease. Also since positive work is done on the system to seperate the attracted plates, the potential difference also increases.

cepheid
Staff Emeritus
Gold Member
If a fully charged parallel plate capacitor is disconnected from a battery and its plates are pulled apart to increase the distance between the plates, what is the value of charge on the capacitor?

I know the charge and the electric field will both remain the same, but I don't know why or how?
Since d increases, the capacitance C will decrease. Also since positive work is done on the system to seperate the attracted plates, the potential difference also increases.

Welcome to PF Ling2,

If the capacitor is disconnected from the circuit, then the charge stored on the capacitor must remain the same, simply because that charge has nowhere to go.

The electric field remains the same, because it depends only on the surface charge density of the plates.

Another way to understand why the voltage increases: Q = CV. If Q is constant, and C goes down, then V goes up.

A third (more physical) way to understand why the voltage increases. The electric field between the plates remains the same, but the distance between them increases. Recall that for a uniform electric field it's simply true that V = E*d.

So $V \propto d$, which is why $C \propto 1/d$.

Thank you so much! Just the clarification I needed :)