Question

A capacitor is charged with a battery and then removed from the battery. In this specially designed capacitor, we are able to make the plate size (area) larger without changing anything else.
If the plate area is made larger after the capacitor has been disconnected, what will happen to the charge on the plates, the voltage across the plates, and the value of capacitance for this capacitor?

Answer 1

When the battery is disconnected from charged capacitor , charge ($Q$) on the capacitor remains constant because it cannot go anywhere .

Now capacitance is given by $C=k{\epsilon }_{0}A/d$ ,

where $A=$ area of the plates ,

$d=$ distance between plates ,

when $A$ is increased , capacitance $C$ also increases .

But $CV=Q\left(constant\right)$ , to keep the $Q$ constant , with increase in $C$ , voltage across the plates $V$ decreases .