Question

In the figure shown, the plates of a parallel plate capacitor have unequal charges. Its capacitance is $C$. The distance between the plates of cross-sectional area $A$ is $d$.


Consider the following statements:

$1).$ The energy stored in the electric field region between plates is $\frac{9Q^2}{8C}$
$2).$ Force exerted by one plate on the another one is $\frac{Q^2}{4A{ϵ}_0}$
$3).$ The potential difference between the plates will be $\frac{3Q}{2C}$
$4).$ Force exerted by one plate on the another one is $\frac{Q^2}{A{ϵ}_0}$
Choose the correct option, which represents a set of true statements:

• A. $\left(1\right),\left(2\right),\left(3\right)$ only
• B. $\left(1\right),\left(3\right),\left(4\right)$ only
• C. $\left(2\right)\&\left(3\right)$ only
• D. $\left(3\right)$ only

Answer 1

The correct option is B $\left(1\right),\left(3\right),\left(4\right)$ only

The charge distribution on given plates is shown below


The charge distribution follows the law of charge conservation and

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

Therefore, charge on capacitor is $\frac{3Q}{2}$ as shown in the above figure.

(charges on opposite plates facing each other)

Inside the capacitor, the net electric field is:

$E_{net}=E_1+E_2$

$\Rightarrow E=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$

$[\text{Since,}\sigma =\frac{\left(3Q/2\right)}{A}=\frac{3Q}{2A}]$

$\Rightarrow E=\frac{3Q}{2A{ϵ}_0}$

Since the $\overrightarrow{E}$ is uniform inside capacitor, the potential difference between plates will be

$V=Ed\Rightarrow V=\frac{3Qd}{2A{ϵ}_0}$

Substituting, $C=\frac{A{ϵ}_0}{d}$

$\therefore V=\frac{3Q}{2C}$


The charge $2Q$ on left plate is unequally distributed on both sides of the plate, hence it can be visualized as combination of two large plates with the charges shown.
Thus, the net electric field at the position of second plate will be;

$E^{\prime }=\frac{{\sigma }_1}{2{ϵ}_0}+\frac{{\sigma }_2}{2{ϵ}_0}$

$\Rightarrow E^{\prime }=\frac{\left(\frac{3Q/2}{A}\right)}{2{ϵ}_0}+\frac{\left(\frac{Q/2}{A}\right)}{2{ϵ}_0}=\frac{Q}{A{ϵ}_0}$

Hence, force experienced by the second plate (right plate) due to the left one will be,

$F=(-Q)E^{\prime }=-\frac{Q^2}{A{ϵ}_0}$
(-ve sign represents attractive force)

Thus, both the plates will attract each other with the above force as per Newton's third law.

Again, energy stored in capacitor:
$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times C\times {\left(\frac{3Q}{2C}\right)}^2$

$\therefore U=\frac{9Q^2}{8C}$

Hence, statement $\left(1\right),\left(3\right)$ and $\left(4\right)$ are correct.

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: In problems involving plates with}\\ \text{unequal charges, first show the charge}\\ \text{distribution. Then equal and opposite}\\ \text{charges on facing plates will be charge}\\ \text{on capacitor.}\end{array}$$