Q. Two point charges $- q$ and $+ q$ are located at points $(0,0 - a)$ and $(0,\ 0a)$, respectively. The potential at a point $(0,0,z)$ where $z >$ a is
A) $\frac{qa}{4\pi\varepsilon_{0}z^{2}}$
B) $\frac{q}{4\pi\varepsilon_{0}a}$
C) $\frac{2qa}{4\pi\varepsilon_{0}{(z}^{2} - a^{2})}$
D) $\frac{2qa}{4\pi\varepsilon_{0}{(z}^{2} + a^{2})}$
Explanation
(c) Potential at $P$ due to $( + q)$ charge
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$V_{1} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{(z - a)}$
Potential at $P$ due to $( - q)$ charge
$V_{2} = \frac{1}{4\pi\varepsilon_{0}}.\frac{- q}{(z + a)}$
Total potential at $P$ due (AB) electric dipole
$V = V_{1} + V_{2}$
$= \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{(z - a)} - \frac{1}{4\pi\varepsilon_{0}}\frac{q}{(z + a)} = \frac{2qa}{4\pi\varepsilon_{0}(z^{2} - a^{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.