Question

Four charges each of magnitude $q$ are placed at four corners of a square of side $a$. The work done in carrying a charge $-q$ from its centre to infinity will be

• A. zero
• B. $\frac{2\sqrt{2}{q}^2}{\pi {ϵ}_{0}a}$
  
• C. $\frac{\sqrt{2}{q}^2}{\pi {ϵ}_{0}a}$
• D. $\frac{q^2}{2\pi {ϵ}_{0}a}$
  

Answer 1

The correct option is C $\frac{\sqrt{2}{q}^2}{\pi {ϵ}_{0}a}$

Here, in $△ACD$
$AD=a$
$DC=a$
$\Rightarrow AC=\sqrt{A{D}^2+D{C}^2}=\sqrt{2}a$
So, $OA=OB=OC=OD=$
$\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$

$-q$ charge is equidistant from all the $q$ charges.

The initial potential energy of the system when $-q$ charge is at $O$ is
$U_i=4\times \frac{-k{q}^2}{\frac{a}{\sqrt{2}}}$
$\Rightarrow U_i=\frac{-4\sqrt{2}k{q}^2}{a}$

Final potential energy, $U_f=0\text{J}$ as charge $-q$ is kept at infinity.

Further, work done $=$ change in potential energy
$\Rightarrow W=U_f-U_i$
$\Rightarrow W=0-(-\frac{4\sqrt{2}k{q}^2}{a})$
$\Rightarrow W=\frac{q^2}{4\pi {ϵ}_0}\times \frac{4\sqrt{2}}{a}$
$\Rightarrow W=\frac{\sqrt{2}{q}^2}{\pi {ϵ}_{0}a}$