Physics - Capacitor Question with Solution | TestHub

Let the capacitance be $C$.

The initial charge is $Q$ when the potential is $V$.

$Q=CV\Rightarrow C=\frac{Q}{V}$

When the potential is reduced by $V_1$, the new potential is $V-V_1$ and the new charge is $Q_1$.

$Q_1=C(V-V_1)$

Substituting $C=\frac{Q}{V}$:

$Q_1=\frac{Q}{V}(V-V_1)$

$Q_{1}V=Q(V-V_1)$

$Q_{1}V=QV-Q{V}_1$

$Q{V}_1=QV-Q_{1}V$

$Q{V}_1=V(Q-Q_1)$

$V=\frac{Q{V}_1}{Q-Q_1}$