Question

A particle $A$ of charge $q$ is placed near a uniformly charged infinite plane sheet with surface charge density $\sigma $ then it experiences force ${F_1}$ while when the particle near a metal plate with surface charge density $\sigma $ then it experiences a force ${F_2}$ Then $\dfrac{{\left| {{F_1}} \right|}}{{\left| {{F_2}} \right|}}$ is equal to
A. $1$
B. $2$
C. $\dfrac{1}{2}$
D. $3$

Answer 1

Hint: In order to solve this question we need to understand electric field and force. Electric field is the line of forces around a charge on which when charge is placed it experiences a force either towards or opposite in direction to it. Gauss law states that net flux through any closed surface (surface enclosing volume) is net charge enclosed divided by permittivity of free space.

Complete step by step answer:
From Gauss law, we can consider a pill box outside a plane sheet of charge and by calculating electric field due to surface charge distribution on plane sheet at some distance is given by, $E = \dfrac{\sigma }{{2{\varepsilon _0}}}$
So Force Experience by charge q when placed in front of plane sheet is given by,
${F_1} = qE$
Putting values we get,
${F_1} = \dfrac{{q\sigma }}{{2{\varepsilon _0}}}$
But when gauss law is considered around a metal or conductor electric field is given by,
$E = \dfrac{\sigma }{{{\varepsilon _0}}}$
So Force Experience by charge q when placed in front of plane sheet is given by,
${F_2} = qE$
Putting values we get,
${F_2} = \dfrac{{q\sigma }}{{{\varepsilon _0}}}$
$\Rightarrow \dfrac{{\left| {{F_1}} \right|}}{{\left| {{F_2}} \right|}} = \dfrac{{q\sigma }}{{2{\varepsilon _o}}} \times \dfrac{{{\varepsilon _0}}}{{q\sigma }}$
$\therefore \dfrac{{\left| {{F_1}} \right|}}{{\left| {{F_2}} \right|}} = \dfrac{1}{2}$

Hence, the correct option is C.

Note: It should be remembered that electric field in both of the cases is uniform otherwise if it is non uniform then there could be uniform force acts on it. Also the electric field never penetrates inside a good conductor, it is evident from the fact that penetration depth calculated using Maxwell equations comes out to be negligible and it is very less in case of a good conductor.