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 $q=10 μC$) is:

-27 J

The correct answer is Option (3) → -27 J

Given,

$q$ = charge = $10 μC=10×10^{-6}C$

Now,

$q_1=+q=10×10^{-6}C$

$q_2=-2q=-20×10^{-6}C$

$q_3=+4q=40×10^{-6}C$

Potential energy, $U=\frac{1}{4πε_0}\left(\frac{q_1q_2}{d}+\frac{q_2q_3}{d}+\frac{q_3q_1}{d}\right)$

$=\frac{1}{4πε_0}\left(\frac{10×10^{-6}×10×10^{-6}}{0.2}+\frac{-20×10^{-6}×40×10^{-6}}{0.2}+\frac{40×10^{-6}×10×10^{-6}}{0.2}\right)$

$=9×10^9[-10^{-6}-4×10^{-6}+2×10^{-6}]$

$=9×10^9×(-3×10^{-6})$

$=-27J$