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

Answer 1

Answer: (D)

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
$V =V_{1}+V_{2}$
$=\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)} $
$ \Rightarrow V =\frac{2 q a}{4 \pi \varepsilon_{0}\left(z^{2}-a^{2}\right)}$