Three particles, each having a charge of $10 μC$ are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is (Given $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} N - m^{2} / C^{2}$ )

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Infinite

27 J

100 J

Zero

answer is C.

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Detailed Solution

For pair of charge $U = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{1} q_{2}}{r}$

$U_{System} = \frac{1}{4 {πε}_{0}} [\frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100}]$

$= 3 \times 9 \times 10^{9} \times \frac{100 \times 10^{- 12} \times 100}{10} = 27 J$

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