Question

Let ${E_1}(r)$ , ${E_2}(r)$ and ${E_3}(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$ , an infinitely long wire with constant linear charge density $\lambda$ and an infinite plane with uniform surface charge density $\sigma$ . If ${E_1}({r_0}) = {E_2}({r_0}) = {E_3}({r_0})$ at a distance ${r_0}$ , then
(A) $Q = 4\sigma \pi r_0^2$
(B) ${r_0} = \dfrac{\lambda }{{2\pi \sigma }}$
(C) ${E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$
(D) ${E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_3}\left( {\dfrac{{{r_0}}}{2}} \right)$

Answer 1

Hint Equate equal expressions in pairs. Find the relationship between the charge distributions at distance ${r_0}$ and at distance $\dfrac{{{r_0}}}{2}$ i.e. for point charge, find the relationship between ${E_1}({r_0})$ and ${E_1}\left( {\dfrac{{{r_0}}}{2}} \right)$ . Likewise for the other charge distributions.

Formula used: ${E_1}\left( {{r_0}} \right) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}}$ where $Q$ is charge, ${\varepsilon _o}$ is the permittivity of free space and ${r_0}$ is the distance from the charge
$\Rightarrow {E_2}({r_0}) = \dfrac{\lambda }{{2\pi {\varepsilon _0}{r_0}}}$ and ${E_3}({r_0}) = \dfrac{\sigma }{{2{\varepsilon _o}}}$ , where $\lambda$ is the line charge density and $\sigma$ is the surface charge density.

Complete step by step answer
Firstly, we write the equation of an electric field by a point charge $Q$ at a distance ${r_0}$ . This is given by
$\Rightarrow {E_1}({r_0}) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}}$ where ${\varepsilon _o}$ is the permittivity of free space.
Similarly, we write the equation of an electric field for an infinite line charge of density $\lambda$ at a distance ${r_0}$ . This is given by
$\Rightarrow {E_2}({r_0}) = \dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}}$
Finally,we write the equation of an electric field for an infinite surface charge of density $\sigma$ at a distance ${r_0}$ . This is written as
$\Rightarrow {E_3}({r_0}) = \dfrac{\sigma }{{2{\varepsilon _o}}}$
According to the question, all three equations are equal. Thus, we could equate any pair so as to simplify and compare with the options.
For option A, we equate ${E_1}$ and ${E_2}$ . Doing this we get,
$\Rightarrow \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} = \dfrac{\sigma }{{2{\varepsilon _o}}}$
To make $Q$ subject of the formula, multiply both sides by $4\pi {\varepsilon _o}r_0^2$ , cancel out ${\varepsilon _o}$ , and divide numerator and denominator by 2.
Hence, we get
$\Rightarrow Q = 2\sigma \pi r_0^2$ .
This is not the same as option A. Thus, option A is incorrect.
For option B, equating ${E_1}$ and ${E_3}$ we get
$\Rightarrow \dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}} = \dfrac{\sigma }{{2{\varepsilon _o}}}$
Cross multiplying and cancelling 2 and ${\varepsilon _o}$ , we get
$\Rightarrow \lambda = \sigma \pi {r_0}$
Rearranging the above equation we get,
$\Rightarrow {r_0} = \dfrac{\lambda }{{\sigma \pi }}$ which is not equal to option B.
Therefore, option B is incorrect.
For option C, we need to find ${E_1}\left( {\dfrac{{{r_0}}}{2}} \right)$ and ${E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$ .
We do this by simply replacing ${r_0}$ with $\dfrac{{{r_0}}}{2}$ in ${E_1}({r_0}) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}}$ .
This gives
$\Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{Q}{{4\pi {\varepsilon _o}\dfrac{{r_0^2}}{4}}} = 4\dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} \\\ \Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_1}({r_0}) \\\$
Similarly,
$\Rightarrow {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{\lambda }{{2\pi {\varepsilon _o}\dfrac{{{r_0}}}{2}}} = 2\dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}} \\\ \Rightarrow {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}({r_0}) \\\$
This in turn, implies that
$\Rightarrow \dfrac{1}{2}{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = {E_2}({r_0}) = {E_1}({r_0})$ .
Substituting $\dfrac{1}{2}{E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$ for ${E_1}({r_0})$ in ${E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_1}\left( {{r_0}} \right)$ we get,
${E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$
This is equal to option (C).
Hence, the correct is option (C).

Note
Alternatively, in option C, from ${E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {{r_0}} \right)$ we can say that
${E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2 \times \dfrac{1}{4} \times {E_1}\left( {\dfrac{{{r_0}}}{2}} \right)$ (since ${E_2}({r_0}) = {E_1}({r_0})$ ).
Simplifying further we get,
${E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$ .