Three charges +q, -2q and 4q are placed on the vertices of an equilateral triangle of side 0.2 m. The electrostatic potential energy of the system (Take \( q = 10\,\mu\mathrm{C} \)) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -27 J
The electrostatic potential energy of a system of point charges is the sum of \( \frac{1}{4\pi\varepsilon_0}\frac{q_iq_j}{r_{ij}} \) over every pair. Three charges make three pairs, and in an equilateral triangle each pair is the same distance \( d = 0.2\,\mathrm{m} \) apart. Take \( q_1 = +q \), \( q_2 = -2q \), \( q_3 = +4q \).
\( \displaystyle U = \frac{1}{4\pi\varepsilon_0 d}\left(q_1q_2 + q_2q_3 + q_3q_1\right) \)
\( \displaystyle U = \frac{q^{2}}{4\pi\varepsilon_0 d}\left(-2 - 8 + 4\right) = -\frac{6q^{2}}{4\pi\varepsilon_0 d} \)
Put \( q = 10\,\mu\mathrm{C} = 10^{-5}\,\mathrm{C} \) and \( \frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \):
\( \displaystyle U = -6\times\frac{9\times10^{9}\times\left(10^{-5}\right)^{2}}{0.2}\,\mathrm{J} = -\frac{6\times0.9}{0.2}\,\mathrm{J} = -27\,\mathrm{J} \)