Question

Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and the negative plate has a charge −q.The plate area is doubled, and the plate separation is reduced to half its initial separation.
a) What is the new charge on the negative plate?

Answer 1

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

$c=\epsilon_0 \cdot \frac{A}{d}$

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q