Question

Two point charges in air at a distance of 20cm from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction
${\text{A}}{\text{. 8}}{\text{.94}} \times {\text{1}}{{\text{0}}^{ - 2}}m \\\ {\text{B}}{\text{. 0}}{\text{.894}} \times {\text{1}}{{\text{0}}^{ - 2}}m \\\ {\text{C}}{\text{. 89}}{\text{.4}} \times {\text{1}}{{\text{0}}^{ - 2}}m \\\ {\text{D}}{\text{. 8}}{\text{.94}} \times {\text{1}}{{\text{0}}^2}m \\\$

Answer 1

We know the distance between charges in air and can write the expression for Coulomb’s force in air. Then by writing the expression for Coulomb’s force in medium and comparing these two expressions, we can get the required answer.

Formula used:
The Coulomb’s force between two charges ${q_1}$ and ${q_2}$ separated by a distance r in a medium having permittivity $\in$ is given as
$F = \dfrac{1}{{4\pi \in }}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Relation between permittivity of a medium and that of a vacuum is given as
$\in = { \in _r}{ \in _0}$

Complete step-by-step answer:
We are given two charges. Let the charges be ${q_1}$ and ${q_2}$. When they are placed in the air, then we can write the expression of Coulomb's law in terms of the permittivity of the air or vacuum in the following way. Let the force between the two charges in air be F then we have
$F = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ …(i)
Here r is the distance between two charges in air whose value is given as
$r = 20cm$
Now when the two charges are placed in the oil then the permittivity of the medium between the two charges changes. We are given the relative permittivity of the oil from which we can calculate the permittivity of the medium in terms of the permittivity of the vacuum. It is given as
$\in = { \in _r}{ \in _0} = 5{ \in _0}$
Now we can write the expression for Coulomb’s law in the oil in the following way.
$F' = \dfrac{1}{{4\pi \in }}\dfrac{{{q_1}{q_2}}}{{r{'^2}}}$ …(ii)
Now by dividing expression (i) by expression (ii), we get
$\dfrac{F}{{F'}} = \dfrac{{\dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}}}{{\dfrac{1}{{4\pi \in }}\dfrac{{{q_1}{q_2}}}{{r{'^2}}}}} = \dfrac{ \in }{{{ \in _0}}}\dfrac{{r{'^2}}}{{{r^2}}}$
If the forces are same in air and oil then we have
$1 = \dfrac{ \in }{{{ \in _0}}}\dfrac{{r{'^2}}}{{{r^2}}} \\\ \Rightarrow r{'^2} = \dfrac{{{ \in _0}}}{ \in }{r^2} \\\ \Rightarrow r' = \sqrt {\dfrac{{{ \in _0}}}{ \in }} r \\\$
Inserting the known values, we get
$r' = \sqrt {\dfrac{{{ \in _0}}}{{5{ \in _0}}}} \times 20 = \dfrac{{20}}{{\sqrt 5 }} = 8.94cm = 8.94 \times {10^{ - 2}}m$
This is the required answer.

Therefore, the correct answer is option A.

Note:
The permittivity of a medium tells us how much of the electric lines of force can pass through the medium. As the permittivity of a medium increases, the magnitude of the force between two charges decreases if they are kept at the same distance in both mediums.