Question

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

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.
$\frac{q_a}{2{\epsilon }_{0}A}+\frac{q_b}{2{\epsilon }_{0}A}+\frac{q_c}{2{\epsilon }_{0}A}-\frac{q_d}{2{\epsilon }_{0}A}=0$
where, $q_d=\frac{q_1+q_2}{2}$ and $q_c=q_2-q_d$
$\therefore q_c=\frac{q_2-q_1}{2}$