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1\vo charges of 4μC each are placed at the comers A and B of an equilateral triangle of side length 0.2 m in air. The electric potential at C is $[\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{C^{2}}]$

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9 X 10⁴V

18 x 10⁴ V

36 X 10⁴ V

72 X 10⁴ V

answer is C.

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Detailed Solution

$Potential at C = (9 \times 10^{9} \times \frac{4 \times 10^{- 6}}{0.2}) \times 2 = 36 \times 10^{4} V$

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