Two point charges -q and +q are located at po | vedclass

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Question

(C) The potential at point $P(0, 0, z)$ due to a point charge $Q$ at distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}$.$1$. Potenti

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$\frac{2qa}{4\pi \epsilon_0 (z^2 - a^2)}$

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(C) The potential at point $P(0, 0, z)$ due to a point charge $Q$ at distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}$.$1$. Potential at $P$ due to charge $+q$ located at $(0, 0, a)$:The distance of $P$ from $(0, 0, a)$ is $r_1 = z - a$.$V_1 = \frac{1}{4\pi \epsilon_0} \frac{q}{z - a}$.$2$. Potential at $P$ due to charge $-q$ located at $(0, 0, -a)$:The distance of $P$ from $(0, 0, -a)$ is $r_2 = z - (-a) = z + a$.$V_2 = \frac{1}{4\pi \epsilon_0} \frac{-q}{z + a}$.$3$. Total potential at $P$:$V = V_1 + V_2 = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{z - a} - \frac{1}{z + a} \right)$.$V = \frac{q}{4\pi \epsilon_0} \left( \frac{(z + a) - (z - a)}{(z - a)(z + a)} \right)$.$V = \frac{q}{4\pi \epsilon_0} \left( \frac{2a}{z^2 - a^2} \right) = \frac{2qa}{4\pi \epsilon_0 (z^2 - a^2)}$.

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

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