Four point-charges are placed at the vertices of a square of side $2.8 \mathrm{~m}$ as shown in the figure. Find the electric potential at the centre of the square.

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Aicharya · Accepted Answer

Given that, the point charges are placed on vertices of a square as shown in figure.

Let $r$ be the distance of $O$ from the each corner. In $ \begin{aligned} A C^2 & =A B^2+B C^2 \\ (2 r)^2 & =a^2+a^2=2 a^2 \Rightarrow r=a / \sqrt{2} \\ r & =\frac{2.8}{\sqrt{2}} \mathrm{~m} \quad(\because a=2.8 \mathrm{~m}) \end{aligned} $ Now, using expression of electric potential at a point due to point charge $ V=K q / r $ Now, total electric potential at $O$ due to all the four charges $q_1, q_2, q_3$ and $q_4$ $ \begin{aligned} & V_0=V_1+V_2+V_3+V_4 \\ & V_0=\frac{K q_1}{r}+\frac{K q_2}{r}+\frac{K q_3}{r}+\frac{K q_4}{r} \\ & V_0=\frac{K}{r}\left[q_1+q_2+q_3+q_4\right] \end{aligned} $ Substituting the values, we get $ \begin{aligned} V_0 & =\frac{9 \times 10^9 \times \sqrt{2}}{(2.8)}(20+40-34+16) \times 10^{-9} \\ & =190.89 \mathrm{~V} \end{aligned} $

Four point-charges are placed at the vertices of a square of side $2.8 \mathrm{~m}$ as shown in the figure.…

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