Physics - Capacitor Question with Solution | TestHub

When two identical metal plates are given charges $q_1$ and $q_2$, the charge on the outer surfaces of both plates is $\frac{q_1+q_2}{2}$.

The charge on the inner surface of the first plate is $q_1-\frac{q_1+q_2}{2}=\frac{q_1-q_2}{2}$.

The charge on the inner surface of the second plate is $q_2-\frac{q_1+q_2}{2}=-\left(\frac{q_1-q_2}{2}\right)$.

The charge on the capacitor is the magnitude of the charge on the inner surfaces, which is $q=\frac{q_1-q_2}{2}$.

The potential difference $V$ between the plates is given by $V=\frac{q}{C}$.

Substituting the value of $q$, we get $V=\frac{q_1-q_2}{2C}$.

Answer: $\frac{q_1-q_2}{2C}$