JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Electrostatics

A uniformly charged ring of radius

3 a

and total charge

q

is placed in

x y

-plane centered at origin. A point charge q is moving towards the ring along the z-axis and has speed v at

z = 4 a

. The minimum value of

v

such that it crosses the origin is:

Answer: (D)

Solution

We use energy conservation between the release point at

z = 4 a

and the origin. The initial potential energy is

U_{i} = \frac{k q ^{2}}{( 4 a ) ^{2} + ( 3 a ) ^{2}} = \frac{k q ^{2}}{5 a}

and at the origin

U_{f} = \frac{k q ^{2}}{3 a}

, with

k = \frac{1}{4 π ϵ _{0}}

. For the charge to just reach the origin all its kinetic energy is spent, so

\frac{1}{2} m v^{2} + \frac{k q ^{2}}{5 a} = \frac{k q ^{2}}{3 a}

. This gives

\frac{1}{2} m v^{2} = k q^{2} (\frac{1}{3 a} - \frac{1}{5 a}) = \frac{2 k q ^{2}}{15 a}

, so

v = \frac{4 k q ^{2}}{15 ma} = (\frac{2}{m} \cdot \frac{2}{15} \cdot \frac{q ^{2}}{4 π ϵ _{0} a})^{1/2}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option