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Electrostatics

Section: Physics

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Question : 60 of 61

Marks: +1, -0

Consider a system of three charges q/3, q/3 and –2q/3 at points A, B and C respectively, as shown in the figure below. Take O to be the centre of the circle of radius R and angle CAB = 60°. Which among the following is correct?

[JEE Adv 2008 P2]

The electric field at point O is $\frac{q}{8\pi \varepsilon_0 R^2}$ directed along the negative x-axis.
The potential energy of the system is zero.
The magnitude of the force between the charges at C and B is $\frac{q^2}{54\pi \varepsilon_0 R^2}$ .
The potential at point O is $\frac{q}{12\pi \varepsilon_0 R^2}$ .

Solution:

At the centre O, we have

\overset{\rightarrow}{E_A}

=

-\overset{\rightarrow}{E_B}

E_C

=

\frac{k[(2q)/3]}{R^2}

(directed along x-axis)

Therefore,

U_{sys}

=

\frac{k(q/3)^2}{2R} - \frac{k(q/3)(2q/3)}{2R\sin 60^{\circ}} - \frac{K(q/3)(2q/3)}{2R\cos 60^{\circ}}

≠ 0

F_{BC}

=

\frac{k(q/3)(2q/3)}{(2R\sin 60^{\circ})^2}

=

\frac{q^2}{18\pi t_0 (\sqrt{3}R)^2}

=

\frac{q^2}{54\pi t_0 R^2}

⇒

V_0

= 0

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