Question

If 3 charges are placed at the vertices of equilateral triangle of charge ${}^{\prime }{q}^{\prime }$ each. What is the net potential energy, if the side of equilateral triangle is 1$cm$.

• A. $\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l}$
• B. $\frac{1}{4\pi {ϵ}_0}\frac{2q^2}{l}$
• C. $\frac{1}{4\pi {ϵ}_0}\frac{3q^2}{l}$
• D. $\frac{1}{4\pi {ϵ}_0}\frac{4q^2}{l}$

Answer 1

The correct option is C $\frac{1}{4\pi {ϵ}_0}\frac{3q^2}{l}$

Potential energy,$U=\frac{1}{4\pi {ϵ}_0}{\sum }_{i<j}\frac{q_i{q}_j}{r_{ij}}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_{12}}+\frac{1}{4\pi {ϵ}_0}\frac{q_2{q}_3}{r_{23}}+\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_3}{r_{13}}$

Here, $q_1=q_2=q_3=q,r_{12}=r_{23}=r_{13}=l$

so, $U=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l}+\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l}+\frac{1}{4\pi {ϵ}_0}\frac{q^2}{l}=\frac{1}{4\pi {ϵ}_0}\frac{3q^2}{l}$

Ans: $\left(C\right)$