Two charges, each equal to q , are kept at $\mathrm{x}=-\mathrm{a}$ and $\mathrm{x}=\mathrm{a}$ on the x -axis. A particle of mass m and charge $\mathrm{q}_{0}=\frac{\mathrm{q}}{2}$ is placed at the origin. If charge $\mathrm{q}_{0}$ is given a small displacement ( $\mathrm{y} \ll$ a) along the y -axis, the net force acting on the particle is proportional to:

JEE Main · 2013 · 07 Apr 2013 · Offline · Electrostatics

Two charges, each equal to q , are kept at

x = - a

and

x = a

on the x -axis. A particle of mass m and charge

q_{0} = \frac{q}{2}

is placed at the origin. If charge

q_{0}

is given a small displacement (

y ≪

a) along the y -axis, the net force acting on the particle is proportional to:

(A) $- y$
(B) $\frac{1}{y}$
(C) $- \frac{1}{y}$
(D) $y$ ✓ Correct

Answer: (D)

Solution

The charge sits at the midpoint, and when it is nudged a small distance

y

along the perpendicular axis each of the two fixed charges repels it. By symmetry the components along the line of charges cancel while the components along

y

add, giving a net force

F = 2 \cdot \frac{k q ( q /2 )}{a ^{2} + y ^{2}} \cdot \frac{y}{a ^{2} + y ^{2}}

directed along

y

. For a tiny displacement

y ≪ a

the distance

a^{2} + y^{2} \approx a

, so

F \approx \frac{k q ^{2}}{2 a ^{3}} y

, which grows in direct proportion to

y

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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JEE Main · 2013 · 07 Apr 2013 · Offline · Electrostatics

Solution

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