Question

A spherical conductor of radius R is charged with Q units of charge. The escape velocity of a particle of charge -q and mass m from surface of conductor is:
a) $\sqrt{\dfrac{Qq}{4 \pi \epsilon_{o} R}}$
b) $\sqrt{\dfrac{Qq}{2 m \pi \epsilon_{o} R}}$
c) $\sqrt{\dfrac{Qq}{ \epsilon_{o} m R}}$
d) $\sqrt{\dfrac{Qq}{4 \pi \epsilon_{o} m R^{2}}}$

Answer 1

We have a spherical conductor of charge Q and radius R and a charge of -q and mass m. We will find the escape velocity of the -q. We will use the concept that the loss of the kinetic energy is equal to the gain of the potential energy.

Complete step-by-step solution:
We have a spherical conductor having charge Q and radius R.
Another charge -q of mass m escapes from the surface of the conductor, we have to find its escape velocity.
The loss in kinetic energy is equal to gain in potential energy.
Loss in kinetic energy = Gain in Kinetic energy
$\dfrac{1}{2} m v^{2} - \dfrac{1}{2} m 0^{2} = 0 - \dfrac{-kQq}{R}$
$\implies \dfrac{1}{2} m v^{2} - 0 = 0 - \dfrac{-kQq}{R}$
$\implies \dfrac{1}{2} m v^{2} = \dfrac{kqQ}{R}$
$\implies v^{2} = \dfrac{2kqQ}{mR}$
$\implies v = \sqrt{ \dfrac{2kQq}{mR}}$
$\implies v = \sqrt{ \dfrac{2Qq}{4 \pi \epsilon_{o} mR}}$
$\implies v = \sqrt{\dfrac{Qq}{2 m \pi \epsilon_{o} R}}$
Option (b) is correct.

Note: The common repulsion of excess positive charges inside a spherical conductor distributes them evenly on its surface. The resulting electric field is at right angle to the surface and zeroes inside. Outside the conductor, the field is the same as a point charge at the centre, similar to the excess charge.