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(A) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = \frac{1}{4\pi\varepsilon_0}$

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(A) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = \frac{1}{4\pi\varepsilon_0}$.At the midpoint $O$,the distance from both charges $+q$ and $-q$ is $d$.The total electric potential $V_{total}$ at the midpoint is the algebraic sum of the potentials due to both charges:$V_{total} = V_{+q} + V_{-q}$$V_{total} = \frac{k(+q)}{d} + \frac{k(-q)}{d}$$V_{total} = \frac{kq}{d} - \frac{kq}{d} = 0$.

Two particles with charges $+q$ and $-q$ are separated by a distance of $2d$. What is the electric potential at the midpoint of the line joining them?

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