Charges of $ + \frac{{10}}{3} \times {10^{ - 9}}C$ are placed at each of the four corners of a square of side $8\,cm$. The potential at the intersection of the diagonals is
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(b) Potential at the centre $O$, $V = 4 \times \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{{a/\sqrt 2 }}$
where $Q = \frac{{10}}{3} \times {10^{ - 9}}C$ and $a = 8\,cm\, = \,\,8 \times {10^{ - 2}}m$
So $V = 5 \times 9 \times {10^9} \times \frac{{\frac{{10}}{3} \times {{10}^{ - 9}}}}{{\frac{{8 \times {{10}^{ - 2}}}}{{\sqrt 2 }}}}$$ = 1500\sqrt 2 \,volt$
where $Q = \frac{{10}}{3} \times {10^{ - 9}}C$ and $a = 8\,cm\, = \,\,8 \times {10^{ - 2}}m$
So $V = 5 \times 9 \times {10^9} \times \frac{{\frac{{10}}{3} \times {{10}^{ - 9}}}}{{\frac{{8 \times {{10}^{ - 2}}}}{{\sqrt 2 }}}}$$ = 1500\sqrt 2 \,volt$
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