Plates pulled apart with the battery disconnected

A charged capacitor is disconnected from the battery and then its plates are pulled apart. What happens to the potential difference between the plates? And to the electric field between them? And to the stored energy? Where does any extra energy come from?

Solution

Key constraint. Disconnected from the battery, the capacitor is isolated: the charge $Q$ stays constant (it has nowhere to go). This is the constraint governing all the answers.

Electric field. For a parallel-plate capacitor the field depends only on the surface charge density: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$ Since $Q$ and $A$ don’t change, the field $E$ stays unchanged as you pull the plates apart.

Potential difference. In a uniform field $V = E\,d$. With $E$ constant and $d$ increasing, $V = E\,d \ \uparrow$ so the p.d. increases proportionally to the distance.

Stored energy. It’s convenient to use the constant-charge form $U = \frac{Q^2}{2C}, \qquad C = \varepsilon_0\frac{A}{d}$ Pulling the plates apart, $C$ decreases, so $\ev{Q = \text{cost}, \quad E = \text{cost}, \quad V \uparrow, \quad U \uparrow}$

Where the extra energy comes from. The two plates carry opposite charges and attract each other. To pull them apart you must do work against this attractive force. That mechanical work is exactly the extra energy you find stored in the field: the energy balance checks out, no energy appears from nowhere.

Links

Topics: Electric field and potential Concepts: Capacitance and capacitor · Field energy density Skills: Conservation of energy Methods: Gradient method Objects: Parallel-plate capacitor