Question

Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an identical charge from infinity to the third vertex is :

• A. $U$
• B. $2U$
• C. $3U$
• D. $4U$

Answer 1

Answer: (B)

$2U$

Let $a$ be the side of equilateral triangle.

Initial the potential energy of the system is $U_i=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_{12}}=\frac{1}{4\pi {ϵ}_0}\frac{qq}{a}=U$

When another charge charge q bring at the third vertex of triangle, the potential energy of the system is

$U_f=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_{12}}+\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_3}{r_{13}}+\frac{1}{4\pi {ϵ}_0}\frac{q_2{q}_3}{r_{23}}$

$=\frac{1}{4\pi {ϵ}_0}\frac{qq}{a}+\frac{1}{4\pi {ϵ}_0}\frac{qq}{a}+\frac{1}{4\pi {ϵ}_0}\frac{qq}{a}$

$=\frac{1}{4\pi {ϵ}_0}\frac{3qq}{a}=3U$

The work done $=$ change in potential energy $=U_f-U_i=3U-U=2U$