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

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(C) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.Potential at poi

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$\frac{2 q a}{4 \pi \varepsilon_0(z^2 - a^2)}$

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(C) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.Potential at point $P(0, 0, z)$ due to charge $+q$ at $(0, 0, a)$ is:$V_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z - a)}$Potential at point $P(0, 0, z)$ due to charge $-q$ at $(0, 0, -a)$ is:$V_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z - (-a))} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z + a)}$Total potential $V$ at point $P$ is:$V = V_1 + V_2$$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{1}{z - a} - \frac{1}{z + a} \right)$$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{(z + a) - (z - a)}{(z - a)(z + a)} \right)$$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{2a}{z^2 - a^2} \right)$$V = \frac{2qa}{4 \pi \varepsilon_0(z^2 - a^2)}$

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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