Question

A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel through its centre. A small particle with negative charge q, 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 centre of the sphere will be:
a. 0
b. V
c. $v\sqrt 2$
d. $v\sqrt {1.5}$

Answer 1

Apply the law of conservation of energy:
Which is, the sum of kinetic and potential energy must be equal to zero (as no energy is created nor destroyed). Particle is moving from rest. This is the reason potential and kinetic energy concepts come into the solution.
We will also make use of the fact that electric potential at the centre of the sphere is three by two of the potential at the surface of the sphere.
Electric potential is given as: $V = \dfrac{Q}{{4\pi {\varepsilon _0}R}}$

Complete step by step answer:
Let’s calculate the velocity of the negative charged particle present at the centre of the sphere inside the tunnel.
When the particle is at rest (at infinity) its potential is zero.
Potential at infinity = ${V_\infty } = 0$
Potential at the surface of the sphere is: $V = \dfrac{Q}{{4\pi {\varepsilon _0}R}}$
Potential at the centre of the sphere: $V = \dfrac{{3Q}}{{2(4\pi {\varepsilon _0}R)}}$
Let the velocity of the particle at the surface v and at the centre v0.
Kinetic energy of particle at the centre of the sphere is: $\dfrac{1}{2}m{v_0}^2$
Kinetic energy of particle at the centre of the sphere is: $\dfrac{1}{2}m{v^2}$
As per law of conservation of energy kinetic energy is equal to the potential energy.

At the surface of the sphere we have:
$\dfrac{1}{2}m{v^2}$ $= - q\left[ {{V_\infty } - {V_s}} \right] = \dfrac{{qQ}}{{4\pi {\varepsilon _0}R}}$ ( we have brought the charge q from infinity to the surface of the sphere therefore, subtracting the potential of surface from the potential of infinity) ----(1)

At the centre of the sphere we have:
$\dfrac{1}{2}m{v_0}^2$ $= - q\left[ {{V_\infty } - {V_c}} \right] = \dfrac{{3qQ}}{{2(4\pi {\varepsilon _0}R)}}$( when charge is at the centre) ----(2)
On dividing the equation 2 by 1
$\Rightarrow \dfrac{{\dfrac{1}{2}m{v_0}^2}}{{\dfrac{1}{2}m{v^2}}} = \dfrac{{\dfrac{{3qQ}}{{2(4\pi {\varepsilon _0}R)}}}}{{\dfrac{{qQ}}{{4\pi {\varepsilon _0}R}}}}$
$\Rightarrow \dfrac{{{v_0}^2}}{{{v^2}}} = \dfrac{3}{2}$(we have cancelled all the common terms from both numerator and denominator)
$\Rightarrow {v^2}_0 = 1.5{v^2}$
$\Rightarrow {v_0} = v\sqrt {1.5}$ (we have removed the square form both LHS and RHS)

Hence, the correct answer is option (D).

Note: Law of conservation of energy: Total energy of an isolated system remains constant or it is said to be conserved, In other words energy can neither be created nor be destroyed but can be transformed from one form to another. As an example, a particle at rest has potential in it and when it starts moving it has kinetic energy (transformation of energy took place).
Electric potential is defined as the work done in bringing a unit positive charge from infinity to the present location.

We can have the most live and happening example of law of conservation of energy which is hydroelectric power plants. In hydro electric power plant water is stored at the reservoir ( which is at potential energy), then this water is made to fall from a height which possesses kinetic energy and then formation of electrical energy takes place.