IIT-JEE 2008 · Paper 2 · Physics Q23

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PhysicsSingle correct+3 / -1Question 23 of 66

Consider a system of three charges $\frac q3$ , $\frac q3$ 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°.

A circle of radius R centred at O with x (horizontal) and y (vertical) axes dashed. C is at the left end of the horizontal diameter; A and B are ends of a diameter through O, with A in the lower left and B in the upper right; angle CAB = 60°. Charges q/3 at A, q/3 at B and −2q/3 at C.

The electric field at point O is

\dfrac{q}{8\pi\varepsilon_0R^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

\dfrac{q^2}{54\pi\varepsilon_0R^2}

The potential at point O is

\dfrac{q}{12\pi\varepsilon_0R}

Answer and worked solution

Official answer: C

Geometry. AB is a diameter, so the angle at C is 90°. With $\angle CAB=60°$ : $AC=2R\cos60°=R$ and $BC=2R\sin60°=\sqrt3R$ . Let $k=\frac1{4\pi\varepsilon_0}$ .

(A) The equal charges at A and B (ends of a diameter) cancel at O. The $-\frac{2q}3$ at C, distance R, gives $\dfrac{k(2q/3)}{R^2}=\dfrac q{6\pi\varepsilon_0R^2}$ towards C (along $-x$ ), not $\dfrac q{8\pi\varepsilon_0R^2}$ . False.
(B) $U=k\left[\dfrac{q^2/9}{2R}-\dfrac{2q^2/9}{R}-\dfrac{2q^2/9}{\sqrt3R}\right]\ne0$ . False.
(C) $F_{CB}=\dfrac{k(2q/3)(q/3)}{3R^2}=\dfrac{2kq^2}{27R^2}=\dfrac{q^2}{54\pi\varepsilon_0R^2}$ . True.
(D) All charges are at distance R from O and sum to zero: $V_O=0$ . False.

(C)

Topics:Coulomb's law and superposition Electric potential Potential energy of charge systems

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IIT-JEE 2008 · Paper 2, Physics question 23