Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Charges of $+\frac{10}{3} \times 10^{-9} \mathrm{C}$ are placed at each of the four corners of a square of side 8 cm . The potential at the intersection of the diagonals is
Text Solution
Verified by ExpertsThe correct answer is:
B
Potential at the center O,
$V = 4 \times \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{a / \sqrt{2}}$
where $Q = \frac{10}{3} \times 10^{-9} \mathrm{C}$
and $a = 8 \, cm = 8 \times 10^{-2} \, m$
So $V = 5 \times 9 \times 10^{9} \times \frac{\frac{10}{3} \times 10^{-9}}{\frac{8 \times 10^{-2}}{\sqrt{2}}}$ $= 1500 \sqrt{2} \text{ volt}$
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