Question

Two point charges placed at a certain distance r in air exert a force F on each other. The distance ${r^1}$ at which these charges will exert the same force in the medium if dielectric constant $k$ is given by
A. $r$
B. $\dfrac{r}{k}$
C. $\dfrac{r}{{\sqrt k }}$
D. None of the these

Answer 1

According to the question, the force between the two point charges kept at a distance r in air is the same as the force between the two point charges kept at a distance ${r^1}$ in a dielectric medium. So, we shall use the Coulomb’s law of attraction to obtain relations for the forces both in air and in dielectric medium and then equate them to derive the relation between the distances. We will also make necessary substitutions for the dielectric constant so that the common terms cancel out.

Formula used:
Force exerted by a point charge on another point charge separated by a distance r is given by $F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _r}{r^2}}}$ where ${q_1}\,,\,{q_2}$ are the magnitude of the point charges, r is the distance between the point charges and ${\varepsilon _r}$ is the permittivity of the medium.
The dielectric constant is the ratio of permittivity of the medium with the permittivity of air. Mathematically, it is expressed as $k = \dfrac{{{\varepsilon _r}}}{{{\varepsilon _0}}}$ .

Complete step by step answer:
In air, the force exerted between two point charges is given by,
$F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$
where ${q_1}\,,\,{q_2}$ are the magnitude of the point charges, r is the distance between the point charges.
In a medium with dielectric constant k, the force exerted between two point charges is given by
${F^1} = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _r}{r^1}^2}}$
where ${q_1}\,,\,{q_2}$ are the magnitude of the point charges, ${r^1}$ is the distance between the point charges.

According to the question, the force between the two point charges kept at a distance r in air is the same as the force between the two point charges kept at a distance ${r^1}$ in a dielectric medium.Hence,
$F = {F^1}$
From the above equations, we can say,
$\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}} = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _r}{r^1}^2}}$

Now we know that the dielectric constant is the ratio of permittivity of the medium with the permittivity of air. Mathematically, it is expressed as,
$k = \dfrac{{{\varepsilon _r}}}{{{\varepsilon _0}}}$
This can be rewritten as,
${\varepsilon _r} = k{\varepsilon _0}$
Substituting in the equation, we get
$\dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}} = \dfrac{{{q_1}{q_2}}}{{4\pi k{\varepsilon _0}{r^1}^2}}$

Cancelling out the common terms on both sides,
$\dfrac{1}{{{r^2}}} = \dfrac{1}{{k{r^1}^2}}$
Rearranging the terms, we get,
${r^1}^2 = \dfrac{{{r^2}}}{k}$
Taking square root, we get,
$\therefore {r^1} = \dfrac{r}{{\sqrt k }}$

Hence, option C is the correct answer.

Note: While using the Coulomb’s law of attraction and its mathematical expression, it is a common practice to drop the sign of the point charges while calculating the force. We only take the modulus of both the charges and substitute them in the equations. For the direction of force, we use the simple rule that repels like. This means that if the charges are of the same sign, a repulsive force will act while if the charges are of opposite sign, an attractive force acts between them.