Question

The electric field due to a plane sheet of charge having charge $q$ and area $A$ is given by $E=\frac{q}{2 \varepsilon_{0} A}$. Further, with usual notation, in case of a parallel-plate capacitor,
$E=\frac{q}{\varepsilon_{0} A}, V=\frac{q d}{\varepsilon_{0} A}$ and $C=\frac{\varepsilon_{0} A}{d}$

Let us now consider that the charges given to the two plates $A$ and $B$ of a parallel-plate capacitor are different, i.e., $q_{1}$ and $q_{2}$ as shown in figure.
With the help of the paragraph given above, choose the most appropriate alternative for each of the following questions.
The charges on the surfaces $c$ and $d$ of the plate $B$, are

• A. $\frac{q_{1}-q_{2}}{2}, \frac{q_{2}-q_{1}}{2}$
• B. $\frac{q_{2}-q_{1}}{2}, \frac{q_{1}+q_{2}}{2}$
• C. $\frac{q_{1}+q_{2}}{2}, \frac{q_{1}+q_{2}}{2}$
• D. $\frac{q_{1}+q_{2}}{2}, \frac{q_{2}-q_{1}}{2}$

Answer 1

Answer: (B)

Since, the electric field at $Q$ is zero.
$\frac{q_{a}}{2 \varepsilon_{0} A}+\frac{q_{b}}{2 \varepsilon_{0} A}+\frac{q_{c}}{2 \varepsilon_{0} A}-\frac{q_{d}}{2 \varepsilon_{0} A}=0$
where, $q_{d}=\frac{q_{1}+q_{2}}{2}$ and $q_{c}=q_{2}-q_{d}$
$\therefore q_{c}=\frac{q_{2}-q_{1}}{2}$