Question

The plates of a parallel plate capacitor are given charges $+4Q$ and $-2Q$. The capacitor is then connected across an uncharged capacitor of same capacitance as first one $(=C)$. Find the final potential difference between the plates of the first capacitor.

Answer 1

Based on the symmetry of charges at equilibrium condition, $+4Q$ will divide into $-2Q$ each. Similarly$-2Q$ will divide into $-Q$ each.

Effictively capacitance$=C+C=2C$

So, the final potential difference $=\frac{2Q-\left(Q\right)}{2C}=\frac{3Q}{2C}$