Question

A parallel plate vacuum capacitor with plate area A and separation x has charges `+Q` and `-Q` on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed.
(a) What is the total energy stored in the capacitor?
(b) The plates are pulled apart an additional distance dx. What is the change in the stored energy?
(c) If `F` is the force with which the plates attract each other, then the change in the stored energy must equal the work `dW = Fdx` done in pulling the plates apart. Find an expression for `F`.
(d) Explain why `F` is not equal to `QE`, where `E` is the electric field between the plates.

Answer 1

Correct Answer - A::B
a. `C=(epsilon_0A)/x, U=q^2/(2C)=(Q^2x)/(2epsilon_0A)`
b. `(dU)/(dx)=Q^2/(2epsilon_0A)`
`:. dU=(Q^2/(2epsilon_0A))dx`
c. `(Q^2/(2epsilon_0A))dx=dW=Fdx`
`:. F=Q^2/(2epsilon_0A)`
b. Because `E` between the plates is due to both the plates.
While `F=(Q)` (field due to other plate)