Question

Two identical non-conducting spherical shells have equal charge $Q$ , which is uniformly distributed on it, are placed at a distance $d$ apart. From where they are released. Find out the kinetic energy of each sphere when they are at a large distance.

Answer 1

The force acting between the two non-conducting spherical shells would be Coulomb's force of repulsion, as they are equal charges of the same nature (positive). The formula of force from Coulomb’s law has to be used here taking the charges to be the same. Thereafter, the formula of the kinetic energy due to this force has to be concluded.
Formula used:
The magnitude of the force acting between the two non-conducting spherical shells, carrying charge $Q$ each and separated by a distance $d$ is given as:
$F = \dfrac{1}{{4{{\pi }}{\varepsilon _0}}} \times \dfrac{{{Q^2}}}{{{d^2}}}$
${\text{K}}{\text{.E}} = \int\limits_d^r {F \cdot {\text{d}}h}$

Complete step-by-step solution:
Let the charge on each of the non-conducting spherical shells be ${Q_1}$ and${Q_2}$, respectively.
According to the question, ${Q_1} = {Q_2} = Q$
Distance of separation between the two spherical shells, $d$
Force of repulsion between the two shells is given as:
$F = \dfrac{1}{{4{{\pi }}{\varepsilon _0}}} \times \dfrac{{{Q^2}}}{{{d^2}}}$
Let the final maximum distance of separation between the shells be $h$
Hence, the final kinetic energy of each of the shell will be equal to the work done in separating the two shells from distance $d$ to distance $h$, respectively i.e
${\text{K}}{\text{.E}} = \int\limits_d^r {F \cdot {\text{d}}h}$
Where, ${\text{d}}h$ is a small displacement.
Completing the integration we get:
${\text{K}}{\text{.E}} = \int\limits_d^h {\dfrac{1}{{4{{\pi }}{\varepsilon _0}}} \times \dfrac{{{Q^2}}}{{{h^2}}} \cdot {\text{d}}h}$
$\Rightarrow {\text{K}}{\text{.E}} = \dfrac{{{Q^2}}}{{4{{\pi }}{\varepsilon _0}}}\int\limits_d^h {\dfrac{{{\text{d}}h}}{{{h^2}}}}$
$\Rightarrow {\text{K}}{\text{.E}} = \dfrac{{{Q^2}}}{{4{{\pi }}{\varepsilon _0}}} \times \left[ {\dfrac{{ - 1}}{h}} \right]_d^h$
$\Rightarrow {\text{K}}{\text{.E}} = \dfrac{{{Q^2}}}{{4{{\pi }}{\varepsilon _0}}} \times \left( {\dfrac{1}{d} - \dfrac{1}{h}} \right)$
So, the kinetic energy of each of the non-conducting spherical shell when separated by a large distance $h$ is given as $\dfrac{{{Q^2}}}{{4{{\pi }}{\varepsilon _0}}} \times \left( {\dfrac{1}{d} - \dfrac{1}{h}} \right)$

Note:

The potential energy of a spherical shell of radius $R$ and charge $Q$ on it is given as ${\text{P}}{\text{.E}} = \dfrac{1}{{4{{\pi }}{\varepsilon _0}}} \times \dfrac{{{Q^2}}}{{2R}}$
The Columb’s formula of force for two different charges is, $F = \dfrac{1}{{4{{\pi }}{\varepsilon _0}}} \times \dfrac{{{Q_1} \times {Q_2}}}{{{d^2}}}$.
That implies, if the amounts of charges increase the force also increases, and if the distance between the two charges increases the force decreases. The term$\dfrac{1}{{4\pi {\varepsilon _0}}}$ is a constant term. Its value is based on the ${\varepsilon _0}$ called the electric permittivity in the air medium.