Four point charges each $+q$ is placed on the circumference of a circle of diameter 2 d in such a way that…
Four point charges each $+q$ is placed on the circumference of a circle of diameter 2 d in such a way that they form a square. The potential at the centre is proportional to
$\frac{\mathrm{q}^2}{\mathrm{~d}^2}$
$\frac{q}{d}$
$\frac{d}{q}$
$\frac{\mathrm{d}^2}{\mathrm{q}^2}$
Solution
As charges are placed on the circumference of circle, each charge is at a distance,
$\mathrm{r}=\frac{\text { Diameter }}{2}=\frac{2 \mathrm{~d}}{2}=\mathrm{d}$
$\therefore \quad$ Potential at centre due to four point charges
$\begin{aligned}
& =\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{~d}}+\frac{\mathrm{q}}{\mathrm{~d}}+\frac{\mathrm{q}}{\mathrm{~d}}+\frac{\mathrm{q}}{\mathrm{~d}}\right] \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{4 \mathrm{q}}{\mathrm{~d}}=4 \dot{K}\left[\frac{\mathrm{q}}{\mathrm{~d}}\right]
\end{aligned}$
Potential $\propto \frac{q}{d}$