Question

The electrostatic potential due to the charge configuration ot point P as shown in figure for $b<<a$ is

• A. $\frac{2q}{4\pi {\epsilon }_{0}a}$
• B. $\frac{2q{b}^2}{4\pi {\epsilon }_0{a}^3}$
• C. $\frac{q{b}^2}{4\pi {\epsilon }_0{a}^3}$
• D. zero

Answer 1

The correct option is D zero

Here, $PA=PC=a,AB=CD=b,PB=PD=\sqrt{a^2+b^2}$

Potential at P is $V=\frac{1}{4\pi {ϵ}_0}[\frac{q}{PA}+\frac{(-q)}{PB}+\frac{q}{PC}+\frac{(-q)}{PD}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{q}{a}+\frac{(-q)}{\sqrt{a^2+b^2}}+\frac{q}{a}+\frac{(-q)}{\sqrt{a^2+b^2}}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{2q}{a}-\frac{2q}{\sqrt{a^2+b^2}}]$

since, $b<<a,$ so, $\sqrt{a^2+b^2}\sim a$

$V=\frac{1}{4\pi {ϵ}_0}[\frac{2q}{a}-\frac{2q}{a}]=0$