Three point charges 3nC, 6nC and 9nC are placed at the corners of an equilateral triangle of side 0.1 m. The potential energy of the system is:

9910 J

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8910 J

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99100 J

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None of these

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Solution

The correct option is D
None of these

$P E = \frac{1}{4 π ϵ_{0}} \frac{Q_{1} Q_{2}}{d} P E = \frac{9 \times 10^{9}}{0.1} [18 + 54 + 27] \times 10^{- 18} = 90 \times 99 \times 10^{- 9} = 8910 \times 10^{- 9} J$

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Three point charges 3nC, 6nC and 9nC are placed at the corners of an equilateral triangle of side 0.1 m. The potential energy of the system is:

The correct option is D None of these PE=14πϵ0Q1Q2dPE=9×1090.1[18+54+27]×10−18=90×99×10−9=8910×10−9J

The correct option is D
None of these

$P E = \frac{1}{4 π ϵ_{0}} \frac{Q_{1} Q_{2}}{d} P E = \frac{9 \times 10^{9}}{0.1} [18 + 54 + 27] \times 10^{- 18} = 90 \times 99 \times 10^{- 9} = 8910 \times 10^{- 9} J$