JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Electrostatics

Consider a system of three charges

\frac{q}{3}

\frac{q}{3}

and

- \frac{2 q}{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 = 6 0^{\circ}

(A)The electric field at point O is $\frac{q}{8 π ε _{0} R ^{2}}$ directed along the negative $x$ -axis
(B)The potential energy of the system is zero
(C)The magnitude of the force between the charges at C and B is $\frac{q ^{2}}{54 π ε _{0} R ^{2}}$ ✓ Correct
(D)The potential at point O is $\frac{q}{12 π ε _{0} R}$

Answer: (C)

Solution

The force between charges at C and B can be calculated using Coulomb's law. The distance BC can be found using the geometry: since angle CAB = 60° and O is the center, BC =

R 3

. Thus:

F_{B C} = \frac{1}{4 π ε _{0}} \frac{( q /3 ) ( 2 q /3 )}{( R 3 ) ^{2}} = \frac{1}{4 π ε _{0}} \frac{2 q ^{2} /9}{3 R ^{2}} = \frac{q ^{2}}{54 π ε _{0} R ^{2}}

Option (C) is correct.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Electrostatics

Solution

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