Question

Consider a system of three charges $\frac{q}{3},\frac{q}{3}$ and $\frac{-2q}{3}$ placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius $R$ and angle $CAB={60}^{\circ }$. Then, which of the following option(s) is/are correct?

• A. The electric field at point O is $\frac{q}{8\pi {\in }_0{R}^2}\hat{i}$.
• B. The magnitude of the force between the charges at C and B is $\frac{q^2}{54\pi {\in }_0{R}^2}$.
• C. The potential energy of the system is zero.
• D. The potential at point O is $\frac{q}{12\pi {\in }_{0}R}$.

Answer 1

The correct option is B The magnitude of the force between the charges at C and B is $\frac{q^2}{54\pi {\in }_0{R}^2}$.

The electric field at O is due to the negative charge at C only, since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel).

The field at O is therefore along negative x-axis and option (A) is obviously wrong.

Potential energy of the system is given by
$U=\frac{k\times \frac{q}{3}\times \frac{q}{3}}{2R}+\frac{k\times \frac{q}{3}\times \frac{-2q}{3}}{BC}+\frac{k\times \frac{q}{3}\times \frac{-2q}{3}}{AC}$
From the figure, $BC=2R\sin {60}^{\circ }=R\sqrt{3};AC=2R\cos {60}^{\circ }=R$

Hence, $U\ne 0$
Option (C) also is wrong.

Potential at point O, $U_O=\frac{k\times \frac{q}{3}}{R}+\frac{k\times \frac{q}{3}}{R}+\frac{k\times \frac{-2q}{3}}{R}=0$
Option (D) is wrong.

The magnitude of the force $\left(F\right)$ between the charges at C and B is given by
$F=\frac{1}{4\pi {\in }_0}\frac{\left(\frac{2q}{3}\right)\left(\frac{q}{3}\right)}{(2R\sin {60}^{\circ }{)}^2}$
Or $F=\frac{q^2}{54\pi {\in }_0{R}^2}$

Hence, only option (B) is correct.