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

Question

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

Dr. Abubacker · Accepted Answer

(c) For pair of change $U = \frac{1}{4\pi\varepsilon_{0}}.\frac{q_{1}q_{2}}{r}$
$U_{system} = \frac{1}{4\pi\varepsilon_{0}}\ \left\lbrack \frac{10 	imes 10^{- 6} 	imes 10 	imes 10^{- 6}}{10/100} ight.\ $
$\left. \  + \frac{10 	imes 10^{- 6} 	imes 10 	imes 10^{- 6}}{10/100} + \frac{10 	imes 10^{- 6} 	imes 10 	imes 10^{- 6}}{10/100} ightbrack$
$= 3 	imes 9 	imes 10^{9} 	imes \frac{100 	imes 10^{- 12} 	imes 100}{10} = 27J$

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

Explanation

Reviewed by Dr. Abubacker

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