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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
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Verified Answer
The correct answer is:
$\frac{2 q a}{4 \pi \varepsilon_0\left(z^2+a^2\right)}$
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
$\begin{aligned}
V & =V_1+V_2 \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{(z+a)} \\
& =\frac{q}{4 \pi \varepsilon_0} \frac{(z+a-z+a)}{(z-a)(z+a)}
\end{aligned}$
$\Rightarrow \quad V=\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}$
$V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}$
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
$\begin{aligned}
V & =V_1+V_2 \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{(z+a)} \\
& =\frac{q}{4 \pi \varepsilon_0} \frac{(z+a-z+a)}{(z-a)(z+a)}
\end{aligned}$
$\Rightarrow \quad V=\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}$
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