Question

A positive point charge $ q $ is fixed at the origin. A small dipole with a dipole moment $ \vec p $ is placed along the x-axis far away from the origin with $ \vec p $ pointing along the positive x-axis. Find
(A) The kinetic energy of the dipole when it reaches a distance $ d $ from the origin, and
(B) The force experienced by the charge $ q $ at this moment

Answer 1

Hint: Here we will find the potential of the initial and final positions of the dipole. Then we will find the kinetic energy by multiplying the charge present at the origin with the difference of these potentials. Next for the force experienced by the charge $ \;q $ , we will simply multiply the charge present at the origin with the electric field due to the dipole at the origin.

Formula used:
 $ V = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{r^2}}}\cos \theta $
 $ \vec E = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{2\hat p}}{{{r^3}}} $ .

Complete step by step solution
The formula for the potential due to a dipole at a distance $ r $ is
 $ V = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{r^2}}}\cos \theta $ ,
where $ \theta $ is the angle between the line joining the dipole and the point charge at the origin with the direction of $ \vec p $ which is already given to be pointing along the positive x-axis.
So to find the kinetic energy of the dipole as required when the dipole reaches a distance $ r $ from infinity, we find the potential difference between the points and multiply it with the point charge present at the origin. Thus the potential difference is $ {V_d} - {V_\infty } $ , which can be found using the formula as, $ {V_d} - {V_\infty } = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{d^2}}}\cos \theta - \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{\infty ^2}}}\cos \theta $ .
Here we will substitute the value of $ \theta = 180{\;^\circ } $ and get the following result,
 $ {V_d} - {V_\infty } = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{d^2}}}\cos {180^\circ } - 0 $
 $ \Rightarrow \Delta V = - \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{d^2}}} $ .
This is the potential difference between the dipole’s initial and final position. Thus the kinetic energy which can be obtained from the potential difference can be given as,
 $ K.E. = q \times |\Delta V| $
 $ \Rightarrow K.E. = q \times |\Delta V| = \dfrac{q}{{4\pi {\varepsilon _0 }}}\dfrac{{\vec p}}{{{d^2}}} $
Thus the kinetic energy of the dipole when it reaches a distance $ d $ from the origin has been obtained.

Now we need to find the force experienced by the charge $ q $ at this moment, i.e. when the dipole is at a distance $ d $ from the origin.
We know that the electric field due to a dipole on its axis is,
 $ \vec E = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{2\hat p}}{{{r^3}}} $ .
Since the dipole is on the x-axis this electric field at a distance $ d $ from the dipole is given by
 $ \vec E = \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{2p}}{{{d^3}}}\hat i $
Thus the force experienced by the charge $ q $ is this field times the charge itself.
Thus the force
 $ F = q\vec E $
 $ \Rightarrow F = q \times \dfrac{1}{{4\pi {\varepsilon _0 }}}\dfrac{{2p}}{{{d^3}}}\hat i $
This is the required force which is being experienced by the point charge at the origin.

Note
Since the dipole moment is along the positive $ x $ direction, the negative charge of the dipole is closer to the origin, since the direction of a dipole is from the negative to the positive charge. Thus it experiences an attractive force due to which it moves closer to the origin.