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?

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

[tex]c=\epsilon_0 \cdot \frac{A}{d}[/tex]

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

adioabiola adioabiola · Answer 2 · 2021-10-29T06:39:52+02:00

The new charge on the negative plate of the capacitor as described in the question remains constant (remains q-).

According to the question:

The capacitor is fully charged.

The capacitor is then disconnected from the battery.

When the plate Area is doubled, and the plate separation is reduced to half. We must remember that, the capacitance of a capacitor is given by;

[tex]c = e \frac{a}{d} [/tex]

where:

c = capacitance
a = area of plates
d = distance between the two plates

As such, the capacitance increases by a factor of 4.

However, this increase in capacitance doesn't lead to an increase in the quantity of charge on the negative plate.

This is so because, the battery has been disconnected and therefore, the potential across the plates is adjusted so that the quantity of charge on the negative plate remains constant.