Question

A charge $ + q$ is fixed at each of the points \[x = {x_0},x = 3{x_0},x = 5{x_0},\] ....... up to infinity on x axis and a charge $\left( { - q} \right)$ is fixed on each of points of \[x = 2{x_0},x = 4{x_0},x = 6{x_0},\] ....... up to infinity, here ${x_0}$ is a positive constant. If the potential due to a charge $Q$ at a point which is at a distance $r$ from it is given by $\dfrac{Q}{{4\pi {\varepsilon _0}r}}$ , then find the potential at the origin due to above system of charges.
A) $0$
B) $\infty $
C) $\dfrac{{q{{\log }_e}2}}{{4\pi {\varepsilon _0}{x_0}}}$
D) $\dfrac{q}{{8\pi {\varepsilon _0}{x_0}{{\log }_e}2}}$

Answer 1

Hint: Electric potential at a point is equal to the work done per unit positive charge in carrying it from infinity to that point in the electric field. Electric potential due to a point charge $Q$ at a distance $r$ from it is given by $\dfrac{Q}{{4\pi {\varepsilon _0}r}}$ .
The electric potential is a scalar quantity. So, to find the potential at the origin due to the above system of charges, simply add the potential due to each charge with proper sign.
Use ${\log _e}\left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} - ......{\text{ upto }}\infty $ (if required)

Complete step by step answer:
Let us first discuss the potential due to a point charge.
Electric potential at a point is equal to the work done per unit positive charge in carrying it from infinity to that point in the electric field. Electric potential due to a point charge $Q$ at a distance $r$ from it is given by $\dfrac{Q}{{4\pi {\varepsilon _0}r}}$.
As given in the question that a charge $ + q$ is fixed at each of the points \[x = {x_0},x = 3{x_0},x = 5{x_0},\] ....... up to infinity on x axis and a charge $\left( { - q} \right)$ is fixed on each of points of \[x = 2{x_0},x = 4{x_0},x = 6{x_0},\] ....... up to infinity, where ${x_0}$ is a positive constant.
We know that the electric potential is a scalar quantity. So, to find the potential at the origin due to the above system of charges, we can simply add the potential due to each charge with proper sign.
Therefore the total potential at origin due to the system of charges is given by
${\text{V}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{{x_0}}} + \dfrac{1}{{3{x_0}}} + \dfrac{1}{{5{x_0}}} + ......} \right] - \dfrac{q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{2{x_0}}} + \dfrac{1}{{4{x_0}}} + \dfrac{1}{{6{x_0}}} + ......} \right]$
On simplifying we have
${\text{V}} = \dfrac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} + .......} \right]$ ……(i)
Now, we can use the expansion of ${\log _e}\left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} - ......{\text{ upto }}\infty $
On substituting $x = 1$ in the above expansion we get
${\log _e}2 = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} + .......{\text{ upto }}\infty $
Now, substituting this in equation we have
${\text{V}} = \dfrac{{q{{\log }_e}2}}{{4\pi {\varepsilon _0}{x_0}}}$ which is the required potential.

Hence, option (C) is correct.

Note: As electric potential work done per unit positive charge, its SI unit is $J/C$ or $Volt$ and its dimension is $\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$ . By convention, the electric potential is assumed to be zero at infinity.
The electric potential energy is the amount of energy gained by an object when it is moved against the electric field. The electric potential for any charge is obtained by dividing the potential energy by the quantity of charge.