# A large solid sphere with a uniformly distributed positive charge with smooth narrow turned...

## Question:

A large solid sphere with a uniformly distributed positive charge with smooth narrow turned through its center. A small particle with a negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed v, and passes through the tunnel. Its speed at the center of the sphere will be _____.

## Energy conservation

The energy conservation deals with the condition that the change of kinetic energy and the potential energy of the system in the context of time is always equivalent to the work done by the system.

Let m be the mass of the particle and -q is the charge on the particle.

If {eq}{v_0} {/eq} be the speed of the particles at centre of sphere and {eq}{v_s} {/eq} be the speed surface of the sphere.

Since the particle started from rest loses its potential energy to gain kinetic energy,

Then by conservation of mechanical energy,

{eq}\Delta K + \Delta U = 0 {/eq}

Here, {eq}\Delta U {/eq} is the change in potential energy.

{eq}\begin{align*} \left( {\dfrac{1}{2}mv_s^2} \right) + \left[ { - q\left( {{V_s} - {V_\infty }} \right)} \right] &= 0\\ \left( {\dfrac{1}{2}mv_s^2} \right) &= q\left( {{V_s}} \right)..........................\left( 1 \right) \end{align*} {/eq}

Here, {eq}{V_s} {/eq} is the potential at the surface of sphere.

{eq}{V_s} = K\dfrac{q}{R} {/eq}

Here, R is the radius of sphere and q is the charge on sphere.

Since the potential at the centre of sphere is,

{eq}{V_c} = \dfrac{3}{2}{V_s} {/eq}

Thus at centre of sphere by energy conservation is,

{eq}\left( {\dfrac{1}{2}mv_0^2} \right) = q\left( {{V_c}} \right)..........................\left( 2 \right) {/eq}

Taking ratio of equation 2 from 1 as,

{eq}\begin{align*} \dfrac{{\left( {\dfrac{1}{2}mv_s^2} \right)}}{{\left( {\dfrac{1}{2}mv_0^2} \right)}} &= \dfrac{{q\left( {{V_s}} \right)}}{{q\left( {\dfrac{3}{2}{V_s}} \right)}}\\ \dfrac{{v_s^2}}{{v_0^2}} &= \dfrac{2}{3}\\ \dfrac{{{v_s}}}{{{v_0}}} &= \sqrt {\dfrac{2}{3}} \\ {v_0} &= \sqrt {1.5} {v_s} \end{align*} {/eq}

Thus, the velocity at the center is, {eq}\sqrt {1.5} {v_s} {/eq}. 