Question

Three point charges q, - 2q and - 2q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to in crease their separation to 2a will be

• A. $\frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ 2q^2 }{ a}$
• B. $\frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ q^2}{ 2a}$
• C. $\frac{ 1}{ 4 \pi \varepsilon_0 } \frac{ 8 q}{ a^2 }$
• D. zero

Answer 1

Answer: (D)

zero

Answer 2

According to the question work done in increasing the separation from a to 2a is W = $U_f - U_i$ Here, $U_i = \frac{ 1}{ 4 \pi \varepsilon_0 } \bigg [ \frac{ q ( - 2q) }{ a} + \frac{ q ( - 2q) }{ a} + \frac{ ( - 2q) ( - 2q) }{ a} \bigg ]$ = $\frac{ 1}{ 4 \pi \varepsilon_0 } [ - 2q^2 - 2q^2 + 4q^2 ] = 0$ Similariy, $U_f$ is aiso zero Hence, W = 0