A parallel-plate capacitor has a capacitance of 112 pF, a plate area of , and a mica dielectric (). At a 55.0-V potential difference, we have to calculate (a) the electric field strength in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge.

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(a)

In a parallel-plate capacitor electric field, E, is constant and is related to the potential difference between the plates, V, and the plate separation d, by the relation

Capacitance of a parallel-plate capacitor of plate area A and with dielectric, , filling the space between its plates is

The data of the problem are

and

.

Therefore, separation between the plates is

The electric field strength in the mica will be

(b)

Let the free charge on the capacitor be q. We have

Therefore,

(c)

The induced surface charge on the plates can be calculated by applying Gauss’ law. We have

and we know

Therefore,