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Q.

Three point charges +q, + 2q and -4q where $q = 0.1 μ C,$ are placed at the vertices of an equilateral triangle of side 10cm as shown in figure. The potential energy of the system is

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a

_{$- 3 \times 10^{- 3} J$}

b

_{$- 9 \times 10^{- 3} J$}

c

_{$3 \times 10^{- 3} J$}

d

_{$9 \times 10^{- 3} J$}

answer is D.

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Detailed Solution

_{$U = \frac{1}{4 π ε_{0}} (\frac{2 q^{2}}{a} - \frac{8 q^{2}}{a} - \frac{4 q^{2}}{a})$}

_{$\Rightarrow U = \frac{1}{4 π ε_{0}} (- \frac{10 q^{2}}{a})$}

_{$\Rightarrow U = - \frac{9 \times 10^{9} \times 10 \times {(0.1 \times 10^{- 6})}^{2}}{(\frac{10}{100})}$}

_{$\Rightarrow U = - 9 \times 10^{- 3} J$}

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Three point charges +q, + 2q and -4q where q=0.1μC, are placed at the vertices of an equilateral triangle of side 10cm as shown in figure. The potential energy of the system is