Question

The electrostatic potential due to the charge configuration at point $\text{P}$ as shown in figure for $b<<a$ is:

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

Answer 1

The correct option is D $\frac{q{b}^2}{4\pi {ϵ}_0{a}^3}$

The distance of point $\text{P}$ from both $-q$ charges is,

$r=\sqrt{b^2+a^2}$

There for applying the sum of potential due to dipoles constitute charges is given as

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

$\Rightarrow V_P=\frac{2q}{4\pi {ϵ}_{0}a}[1-{(1+\frac{b^2}{a^2})}^{-1/2}]$

Given that $b<<a$, therefore we can use binomial expansion for simplification

${(1+\frac{b^2}{a^2})}^{-1/2}=[1-\frac{1}{2}\left(\frac{b^2}{a^2}\right)]$


Substituting this in $V_P$, we get

$\Rightarrow V_P=\frac{2q}{4\pi {ϵ}_{0}a}[1-(1-\frac{b^2}{2a^2})]$

$\Rightarrow V_P=\frac{2q}{4\pi {ϵ}_{0}a}\left[\frac{b^2}{2a^2}\right]$

$\therefore V_P=\frac{q{b}^2}{4\pi {ϵ}_0{a}^3}$

Therefore, $\left(d\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip:The dipole is just an arrangement}\\ \text{of two equal and opposite point charges}\\ \text{hence instead of using formula of potential}\\ \text{"We can use summation of potential due to}\\ \text{each charge", if we find geometry to be}\\ \text{simple and useful}\end{array}$