Problem (IIT JEE 1999): Two identical metal plates are given positive charges $ q_1 $ and $ q_2 $ $( < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$, the potential difference between them is,
Solution:
Let $A$ be area of the plates and $d$ be separation between them. The electric field in the region between plates due to the left plate is
\begin{alignat}{2}
&\vec{E}_1=\frac{\sigma_1}{2\epsilon_0}\,\hat\imath=\frac{q_1}{2\epsilon_0 A}\,\hat\imath, \nonumber
\end{alignat}
and due to the right plate is
\begin{alignat}{2}
&\vec{E}_2=-\frac{\sigma_2}{2\epsilon_0}\,\hat\imath=-\frac{q_2}{2\epsilon_0 A}\,\hat\imath. \nonumber
\end{alignat}
The resultant field $\vec{E}$ and potential difference between two plates are,
\begin{align}
&\vec{E}=\vec{E}_1+\vec{E}_2=\frac{q_1-q_2}{2\epsilon_0 A}\,\hat\imath, &&V=Ed=\frac{q_1-q_2}{2(\epsilon_0 A/d)}=\frac{q_1-q_2}{2C}.\nonumber
\end{align}