Question

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. $\frac{qa}{4\pi {\epsilon }_0{z}^2}$
• B. $\frac{q}{4\pi {\epsilon }_{0}a}$
• C. $\frac{2qa}{4\pi {\epsilon }_0\left(z^2-a^2\right)}$
• D. $\frac{2qa}{4\pi {\epsilon }_0\left(z^2+a^2\right)}$

Answer 1

Answer: (D)

Potential at $P$ due to $(+q)$ charge
$V_1=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{(z-a)}$

Potential at $P$ due to $(-q)$ charge
$V_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{-q}{(z+a)}$
Total potential at $P$ due to $(AB)$ electric dipole
$V=V_1+V_2$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{(z-a)}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{(z+a)}$
$=\frac{q}{4\pi {\epsilon }_0}\frac{(z+a-z+a)}{(z-a)(z+a)}$
$\Rightarrow V=\frac{2qa}{4\pi {\epsilon }_0\left(z^2-a^2\right)}$