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

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(C) The potential at point $P(0,0,z)$ due to the charge $+q$ located at $(0,0,a)$ is:$V_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}$The po

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

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