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Charges of $+\frac{10}{3} \times 10^{-9} \mathrm{C}$ are placed at each of the four corners of a square of side $8 \mathrm{~cm}$. The potential at the intersection of the diagonals is
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$1500 \sqrt{2}$ volt
Potential at the centre 0 , $V=4 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{a / \sqrt{2}}$
Where $Q=\frac{10}{3} \times 10^{-9} \mathrm{C}$ and $a=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
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 $Q=\frac{10}{3} \times 10^{-9} \mathrm{C}$ and $a=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
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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