Question

Let ${E_1}\left( r \right)$, ${E_2}\left( r \right)$ and ${E_3}\left( r \right)$ 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 ${E_1}\left( {{r_0}} \right) = {E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$ at a given 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

We can use the formulae for the electric field due to a point charge at a distance, electric field at a distance from an infinitely long wire with constant linear charge density and electric field $E$ at a distance from an infinite plane with uniform surface charge density. Rewrite these formulae for the distance ${r_0}$. Solve these equations and derive the relations for the quantities given in the options and check which of the options is correct.

Formulae used:
The electric field $E$ due to a point charge $q$ at a distance $r$ is
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$ ……. (1)
The electric field $E$ at a distance $r$ from an infinitely long wire with constant linear charge density $\lambda$ is
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda }}{r}$ ……. (2)
The electric field $E$ at a distance $r$ from an infinite plane with uniform surface charge density $\sigma$ is
$E = \dfrac{\sigma }{{2{\varepsilon _0}}}$ ……. (3)

Complete step by step answer:
We have given that ${E_1}\left( r \right)$ is the electric field at a distance $r$ from the point charge $Q$, ${E_2}\left( r \right)$ is the electric field from an infinitely long wire with constant linear charge density $\lambda$ and ${E_3}\left( r \right)$ is the electric field from an infinite plane with uniform surface charge density $\sigma$.From equations (1), (2) and (3), we can write
${E_1}\left( r \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}$
$\Rightarrow {E_2}\left( r \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda }}{r}$
$\Rightarrow {E_3}\left( r \right) = \dfrac{\sigma }{{2{\varepsilon _0}}}$
Also at a distance ${r_0}$, we have
${E_1}\left( {{r_0}} \right) = {E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$

Let us consider the relation
${E_1}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$
Substitute $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{r_0^2}}$ for ${E_1}\left( {{r_0}} \right)$ and $\dfrac{\sigma }{{2{\varepsilon _0}}}$ for ${E_3}\left( {{r_0}} \right)$ in the above equation.
$\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{r_0^2}} = \dfrac{\sigma }{{2{\varepsilon _0}}}$
$\Rightarrow Q = 2\sigma \pi r_0^2$ …… (4)
Hence, the option A is incorrect.

Let us consider the relation
${E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$
Substitute $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda }}{{{r_0}}}$ for ${E_2}\left( {{r_0}} \right)$ and $\dfrac{\sigma }{{2{\varepsilon _0}}}$ for ${E_3}\left( {{r_0}} \right)$ in the above equation.
$\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda }}{{{r_0}}} = \dfrac{\sigma }{{2{\varepsilon _0}}}$
$\Rightarrow {r_0} = \dfrac{\lambda }{{\pi \sigma }}$ …… (5)
Hence, option B is incorrect.

From equation (4), we can write
$\Rightarrow \sigma = \dfrac{Q}{{2\pi r_0^2}}$
Substitute $\dfrac{Q}{{2\pi r_0^2}}$ for $\sigma$ in equation (5).
$\Rightarrow {r_0} = \dfrac{\lambda }{{\pi \left( {\dfrac{Q}{{2\pi r_0^2}}} \right)}}$
$\Rightarrow Q = 2\lambda {r_0}$
Let us rewrite the equation of ${E_1}\left( r \right)$ for the distance $\dfrac{{{r_0}}}{2}$.
${E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{{\left( {\dfrac{{{r_0}}}{2}} \right)}^2}}}$
$\Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{r_0^2}}$
Substitute $2\lambda {r_0}$ for $Q$ in the above equation.
$\Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda {r_0}}}{{r_0^2}}$
$\Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{\lambda }{{2\pi {\varepsilon _0}{r_0}}}$ …… (6)
Let us rewrite the equation of ${E_2}\left( r \right)$ for the distance $\dfrac{{{r_0}}}{2}$.
${E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2\lambda }}{{\dfrac{{{r_0}}}{2}}}$
$\Rightarrow {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{\lambda }{{\pi {\varepsilon _0}{r_0}}}$ …… (7)
From equations (6) and (7), we can write
$\therefore{E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right)$

Hence, the correct option is C.

Note: The formulae for the electric fields used in the solutions should be correct. The students should be careful while doing the calculations in the solution. If these calculations go wrong then we will not end with the correct derivations for the charge, distance or relations between the electric fields. Hence, the correct step by step calculations should be done in order to avoid the wrong answers.