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Eight charges having the values as shown in fig are arranged symmetrically on a circle of radium 0.1 m in air. Potential at center ‘O’ will be

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a

252×10⁶V

b

zero

c

252×10¹⁰V

d

252×10⁴V

answer is B.

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Detailed Solution

$\begin{array}{l} V = V_{1} + V_{2} + V_{3} + - - - - + V_{8} \\ \frac{1}{4 {πε}_{0}} \times \frac{1}{r} (q_{1} + q_{2} + - - - - + q_{8}) \\ = \frac{9 \times 10^{9}}{0 .1} [11 - 5 + 7 - 5 + 5 - 7 + 15 + 7] \times 10^{- 6} \\ = 9 \times 10^{10} \times 10^{- 6} \times 28 = 252 \times 10^{4} V \end{array}$

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