Question

Charges $2q$ and $-3q$ are given to two identical metal plates of cross-sectional area $A$. The distance between the plates is $d$. The capacitance of system and potential difference between plates respectively will be:

• A. $\frac{2A{ϵ}_0}{d},\frac{qd}{2{ϵ}_{0}A}$
• B. $\frac{A{ϵ}_0}{d},\frac{2.5qd}{{ϵ}_{0}A}$
• C. $\frac{A{ϵ}_0}{2d},\frac{2.5qd}{{ϵ}_{0}A}$
• D. $\frac{A{ϵ}_0}{d},\frac{5qd}{{ϵ}_{0}A}$

Answer 1

The correct option is B $\frac{A{ϵ}_0}{d},\frac{2.5qd}{{ϵ}_{0}A}$

The charge appearing on outermost faces is given by

$q_{outer}=\frac{q_{total}}{2}$

$\Rightarrow q_{outer}=\frac{2q-3q}{2}=\frac{-q}{2}$

Thus, $\frac{-q}{2}$ will reside at the outermost surfaces.


The charge distribution is shown in the above figure.

Here we can see that the charge on plates facing each other is $2.5q$ & $-2.5q$ respectively. Thus, they form a parallel plate capacitor.

using formula, $Q=CV$

$\Rightarrow 2.5q=CV........\left(1\right)$

Since the area of plates is $A$ and separation between them is $d$. The capacitance will be,

$C=\frac{A{ϵ}_0}{d}........\left(2\right)$

From eqs. $\left(1\right)\&\left(2\right)$

$\Rightarrow 2.5q=\frac{A{ϵ}_0}{d}V$

$\therefore V=\frac{2.5qd}{A{ϵ}_0}$

Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip : The charged appearing on outer faces}\\ \text{of the outermost plate is}\\ q_{outer}=\frac{\sum q}{2}=\frac{q_{total}}{2}\\ \text{This trick is widely used for such problems}\\ \text{and it comes as a result of application of}\\ \text{Gauss law and charge conservation.}\end{array}$