Question

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 $\dfrac{L{2}$ is}

• A. $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q^2}{L}$
• B. $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{2q^2}{L}$
• C. $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{3q^2}{L}$
• D. $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{3q^2}{2L}$

Hint:

Work done = change in potential energy. Count all pairs in the triangle (3).

Answer 1

The Correct Option is C

Initial potential energy: $U_1 = 3 \cdot \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q^2}{L}$
Final potential energy (at $L/2$): $U_2 = 3 \cdot \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q^2}{L/2} = 6 \cdot \dfrac{q^2}{4\pi\varepsilon_0 L}$
Work done $= U_2 - U_1 = \dfrac{6 - 3}{4\pi\varepsilon_0} \cdot \dfrac{q^2}{L} = \dfrac{3q^2}{4\pi\varepsilon_0 L}$