Question

Two point charges $-q$ and $+ q$ are located at point's $(0, 0, - a)$ and, $(0, 0, a)$ respectively. The electric potential at a point $(0, 9, z)$, where $z > a$ is

• A. $\frac{q_a}{4 \pi \varepsilon_0 z^2}$
• B. $\frac{q}{4 \pi \varepsilon_0 a}$
• C. $\frac{2q a}{4 \pi \varepsilon_0 z^2 - a^2}$
• D. $\frac{2qa}{4 \pi \varepsilon_0 z^2 + a^2}$

Answer 1

The Correct Option is C

Potential at $P$ due to $(+q)$ charge
$V_{1} = \frac{1}{4 \pi \varepsilon_{0} . \frac{q}{z - a}}$
Potential at $P$ due to $(-q)$ charge
$V_2 = \frac{1}{4 \pi \varepsilon_0} . \frac{-q}{z + a}$
Total potential at P due to (AB) electric dipole
$V = V_1 + V_2$
$ = \frac{1}{4 \pi \varepsilon_0} . \frac{q}{z -z} - \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}$
$\Rightarrow \, \, V = \frac{2qa}{4 \pi \varepsilon_0 \, \, z^2 - a^2}$