Q.Derive an expression for the capacitance of a parallel plate capacitor.

Parallel Plate Capacitance and Plate Separation

A parallel plate capacitor stores charge on two conducting plates of area $A$ separated by a distance $d$. Its capacitance depends only on this geometry (and the medium between the plates):

Intuition: capacitance is how much charge the plates hold per volt. Pull the plates apart (increase $d$) and $C$ decreases; the field must reach across a bigger gap for the same charge.

The key distinction — is the battery connected? Two variables ($Q$, $V$) are linked by $Q = CV$, but only one is held fixed depending on the circuit:

• Battery (cell of emf $E$) still connected (key closed): the plates stay at $V = E$. So $V$ is constant. Changing $d$ changes $C$, and therefore $Q = CV$ changes — charge flows through the circuit to keep $V$ fixed.
• Battery disconnected first (key open): the charge $Q$ on the isolated plates has nowhere to go, so $Q$ is constant. Changing $d$ changes $C$, and therefore $V = Q/C$ changes. …