JEE Main · 2011 · 10 Apr 2011 · Offline · Electrostatics

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Two identical charged spheres suspended from a common point by two massless strings of length I are initially a distance

d (d ≪ l)

apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance

x

between them,

(A) $v \propto x^{- 1}$
(B) $v \propto x^{1/2}$
(C) $v \propto x$
(D) $v \propto x^{- 1/2}$ ✓ Correct

Answer: (D)

Solution

With the strings nearly vertical, equilibrium of one sphere balances the Coulomb repulsion against the horizontal pull,

\frac{k Q ^{2}}{x ^{2}} = m g \cdot \frac{x}{2 l}

, so

Q^{2} \propto x^{3}

and

Q \propto x^{3/2}

. The charge leaks at a steady rate, so

\frac{d Q}{d t}

is constant. Differentiating

Q \propto x^{3/2}

in time,

\frac{d Q}{d t} \propto x^{1/2} \frac{d x}{d t}

; since the left side is constant,

\frac{d x}{d t} \propto x^{- 1/2}

. The approach speed therefore varies as

v \propto x^{- 1/2}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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JEE Main · 2011 · 10 Apr 2011 · Offline · Electrostatics

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