A point charge $q$ is placed at $O$ as shown in the figure.
Is $V_{P} - V_{Q}$ positive or negative when a) $q > 0$ b) $q < 0$ ?
Justify your answer.

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Solution

$V = \frac{q}{4 π ϵ_{o} r} = \frac{k q}{r}$ where $k = \frac{1}{4 π ϵ_{o}}$

$V_{P} = \frac{k q}{O P}$

And $V_{Q} = \frac{k q}{O Q}$

$⟹ V_{P} - V_{Q} = k q \frac{O Q - O P}{O P . O Q}$

Since, $O Q > O P$

Hence, $\frac{O Q - O P}{O Q . O P} > 0$

Hence $V_{P} - V_{Q} > 0$ if $q > 0$

And, $V_{P} - V_{Q} < 0$ if $q < 0$

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