Question

Two equal points charges are fixed at $ x = - a $ and $ x = + a $ on the x-axis another point charge $ Q $ is placed at origin find the change in ‘P.E.’ of particle when it is displaced by a small distance $ x $ along positive x-axis is approximately proportional to:
(A) $ x $
(B) $ {x^2} $
(C) $ {x^3} $
(D) $ {x^4} $

Answer 1

Hint : Potential energy is a scalar quantity and when more than one charge influences a single charge the net potential energy on that single charge will be the sum of potential energy applied by the other charges on that single charge.

Formula used:
 $ \Rightarrow P = \dfrac{{KqQ}}{r} $
Where $ P $ is the electric potential energy, $ K $ is Coulomb constant whose value is $ 9 \times {10^9}N.{m^2}.{C^{ - 2}} $ , $ q $ and $ Q $ are the two charges whose $ P $ is being calculated and $ r $ is the distance between them.

Complete step by step answer
Let the two equal point charges be of charge $ q $ and $ {U_i} $ be the initial potential energy of $ Q $ and $ {U_f} $ be the final potential energy of $ Q $ .
We know that,
 $ \Rightarrow P = \dfrac{{KqQ}}{r} $
Where $ P $ is the electric potential energy, $ K $ is Coulomb constant whose value is $ 9 \times {10^9}N.{m^2}.{C^{ - 2}} $ , $ q $ and $ Q $ are the two charges whose $ P $ is being calculated and $ r $ is the distance between them.
This $ K $ is the short form of $ \dfrac{1}{{4\pi {\varepsilon _o}}} $ where $ {\varepsilon _o} $ is the permittivity in free space i.e. vacuum.
Hence,
 $ \Rightarrow {U_i} = \dfrac{{KQq}}{a} + \dfrac{{KQq}}{a} $
 $ \Rightarrow {U_i} = \dfrac{{2KQq}}{a} $
Let the final position be the situation when the charge $ Q $ has moved a distance $ x $ such that $ x < < a $ .
 $ \Rightarrow {U_f} = \dfrac{{KQq}}{{a + x}} + \dfrac{{KQq}}{{a - x}} $
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{1}{{a + x}} + \dfrac{1}{{a - x}}} \right) $
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{{2a}}{{(a + x)(a - x)}}} \right) $
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{{2a}}{{{a^2} - {x^2}}}} \right) $
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{{2a}}{{{a^2}(1 - {{\left( {\dfrac{x}{a}} \right)}^2})}}} \right) $
Using binomial expansion as $ \dfrac{x}{a} < < 1 $
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{{2a(1 + {{\left( {\dfrac{x}{a}} \right)}^2})}}{{{a^2}}}} \right) $
On further solving,
 $ \Rightarrow {U_f} = KQq\left( {\dfrac{{2({a^2} + {{\left( x \right)}^2})}}{{{a^3}}}} \right) $
 $ \Rightarrow {U_f} = \dfrac{{2KQq{x^2}}}{{{a^3}}} + \dfrac{{2KQq}}{a} $
We want to know the value of,
 $ \Rightarrow \Delta U = {U_f} - {U_i} $
Where $ \Delta U $ is the change in potential energy of the charge $ Q $ .
 $ \Rightarrow \Delta U = \dfrac{{2KQq{x^2}}}{{{a^3}}} + \dfrac{{2KQq}}{a} - \dfrac{{2KQq}}{a} $
 $ \therefore \Delta U = \dfrac{{2KQq{x^2}}}{{{a^3}}} $
Hence we can say that $ \Delta U \propto {x^2} $
Therefore the correct answer is (B) $ {x^2} $ .

Note
The electric P.E. of an arrangement of charges is defined as the work involved in assembling the given system of charges by bringing them close to each other in the given arrangement from an infinite distance. A system of charges always tries to attain an arrangement in which it has minimum electric potential energy.