Two point charges $-q$ and $+q$ are located at points $(0,0,-a)$ and $(0,0, a)$ respectively. The electric potential at a point $(0,0, z)$, where $z>a$ is

• A

  $\frac{q a}{4 \pi \varepsilon_0 z^2}$
• B

  $\frac{q}{4 \pi \varepsilon_0 a}$
• C

  $\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}$
• Correct answer
  D

  $\frac{2 q a}{4 \pi \varepsilon_0\left(z^2+a^2\right)}$

Solution

Potential at $P$ due to $(+q)$ charge $V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}$
Potential at $P$ due to $(-q)$ charge $V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z+a)}$ Total potential at $P$ due to $(A B)$ electric dipole $\begin{aligned} V & =V_1+V_2 \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{(z+a)} \\ & =\frac{q}{4 \pi \varepsilon_0} \frac{(z+a-z+a)}{(z-a)(z+a)} \end{aligned}$ $\Rightarrow \quad V=\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}$

Hence, the correct answer is option: 'D'.