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(D) The electric potential $V$ at a point due to a charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.From the fig

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(D) The electric potential $V$ at a point due to a charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.From the figure,the distance from both charges $+q$ (at $B$) and $-q$ (at $C$) to point $A$ is $r = \sqrt{a^2 + b^2}$.The total potential at $A$ is the algebraic sum of the potentials due to both charges:$V_A = V_B + V_C$$V_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{\sqrt{a^2 + b^2}} + \frac{1}{4\pi\varepsilon_0} \frac{(-q)}{\sqrt{a^2 + b^2}}$$V_A = \frac{1}{4\pi\varepsilon_0} \left( \frac{q - q}{\sqrt{a^2 + b^2}} \right) = 0$.

What is the electric potential at point $A$ due to the given charges?

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