Energy Stored in an Electric Field — Question 2

Question 2

An isolated plate capacitor resists separation with a constant ideal force

A vacuum parallel-plate capacitor has area A, gap d and fixed charges $\pm Q$. Neglect fringing and allow the gap to increase quasistatically.

1. Express U(d) and find the electric force conjugate to increasing d.

2. Find external work to double d and compare the force with the field of the opposite plate acting on Q.

Question 2 - Solution

An isolated plate capacitor resists separation with a constant ideal force

(a) Method and calculation

$$C=\epsilon_0A/d,\quad U=\frac{Q^2}{2C}=\frac{Q^2d}{2\epsilon_0A}.$$$$F_d=-\left(\frac{\partial U}{\partial d}\right)_Q=-\frac{Q^2}{2\epsilon_0A}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}=U(2d)-U(d)=\frac{Q^2d}{2\epsilon_0A}.$$$$E_{\mathrm{other}}=\frac{Q}{2\epsilon_0A}\Rightarrow |F|=QE_{\mathrm{other}}=\frac{Q^2}{2\epsilon_0A}.$$

Physical check

A plate does not exert a net force on itself; using the full gap field in QE would double the correct force.