Three particles of each charge $q$ are placed at the vertices of an equilateral triangle of side $L$. The work to be done to decrease the side of the triangle to $\frac{L}{2}$ is

• A

  $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{~L}}$
• B

  $\frac{1}{4 \pi \varepsilon_0} \frac{2 q^2}{L}$
• Correct answer
  C

  $\frac{1}{4 \pi \varepsilon_0} \frac{3 q^2}{L}$
• D

  $\frac{1}{4 \pi \varepsilon_0} \frac{3 q^2}{2 L}$

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