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 $a$ and $b$ of the plate $A$ are

• A. $\frac{q_{1}+q_{2}}{2}, \frac{q_{1}-q_{2}}{2}$
• B. $\frac{q_{1}-q_{2}}{2}, \frac{q_{2}-q_{1}}{2}$
• 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: (A)

At the point $p$, the electric field due to charge $q_{a}$ is towards right whereas it will be towards left due to charges $q_{b}, q_{c}$ and $q_{d}$. Since, the resultant electric field at a point inside a conductor is zero,

$\frac{q_{a}}{2 \varepsilon_{0} A}-\frac{q_{b}}{2 \varepsilon_{0} A}-\frac{q_{c}}{2 \varepsilon_{0} A}-\frac{v_{d}}{2 \varepsilon_{0} A}=0$
or $q_{a} =q_{b}+q_{c}+q_{d}$
$=\left(q_{1}-q_{a}\right)+\left(q_{2}-q_{d}\right)+q_{d}=q_{1}-q_{a}+q_{2}$
where, $q_{a}=\frac{q_{1}+q_{2}}{2}$
and $q_{b}=q_{1}-q_{a}=q_{1}-\frac{q_{1}+q_{2}}{2}=\frac{q_{1}-q_{2}}{2}$