Q131 / 150 · MHT CET 2011 · Physics

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

  • A
    Zero
  • B
    $\infty$
  • Correct answer
    C
    $27 \mathrm{~J}$
  • D
    $100 \mathrm{~J}$

Solution

For a pair of charges $$ \begin{array}{c} U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \\ U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \\ +\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \\ \left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \\ =3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \\ =27 \mathrm{~J} \end{array} $$

Hence, the correct answer is option: 'C'.

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