Question

A parallel plate vacuum capacitor eith 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 each case.
c. If F is the force with which the plates attract each other, the change in the stored energy must equal the work `dW =F dx` done in pulling the plates apart. Find an expression for F
d. Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series for parallel combinations?

Answer 1

a. `U_(0)=Q^(2)/(2C)=(xQ^(2))/(2epsilon_(0)A)`
b. Increase the separation by `dx`.
`U=((x+dx)Q^(2))/(2epsilon_()A)=U_(0)(1+dx//x)`
The charge is then `(Q^(2)//2epsilon_(0)A)dx`.
c. The work done in increasing the separation is given by
`dW=U-U_(0)=(dxQ^(2))/(2epsilonA)=Fdx` or `F=Q^(2)/(2epsilon_(0)A)`
d. The reason for the differnce is that `E` is the field due to both plates. The force is `QE` if `E` is the field due to one plate and `Q` is the charge on the other plate.