Question

The force between two charges placed in air resistance at a distance r apart is F. Then the force between the same two charged if a dielectric having k=4 and thickness $\dfrac{r}{2}$ is introduced between charges is:
A. $\dfrac{4F}{9}$
B. $\dfrac{9F}{4}$
C. $\dfrac{9F}{16}$
D. $\dfrac{16F}{9}$

Answer 1

Hint: Using Coulomb’s law, find the expression for the force between the two charges for the first case. As for the second case where a dielectric is introduced, remember that the dielectric constant is nothing but the relative permittivity of the material. Use this to get the permittivity of the dielectric in terms of permittivity of free space.
Then, determine the effective distance between the two charges with the dielectric introduced and account for the apparent change in the distance over which the dielectric is placed. Then, using Coulomb’s law, find the effective force between the two charges and arrive at an expression in terms of the force experienced by the charges in the first case.

Formula used: Electric force between two charges: $F = \dfrac{1}{4\pi\epsilon}\dfrac{q_1 q_2}{r^2}$,
where $\epsilon$ is the electric permittivity of the medium, $q_1$ and $q_2$ and two charges that are at an effective distance r.

Complete step by step answer:
From Coulomb’s law we know that the electric force between two charges is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the effective distance between them and is given by:
$F = \dfrac{1}{4\pi\epsilon}\dfrac{q_1 q_2}{r^2} $
Now, when the two charges are placed in air resistance at a distance r apart, the force between the two charges is given as:

$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2} $,where $\epsilon_0$ is the electric permittivity of free space or air.
Now, a dielectric of thickness $\dfrac{r}{2}$ is introduced between the two charges. It has a value of $k= 4 \Rightarrow k = \dfrac{\epsilon}{\epsilon_0} = 4 \Rightarrow \epsilon = k\epsilon_0$.
This means that the electric permittivity of the dielectric is $4\epsilon_0$.

Now, the force acting on the charges due to just the dielectric k can be given in general as:
$F = \dfrac{1}{4\pi\epsilon} \dfrac{q_1 q_2}{r^2} = \dfrac{1}{4\pi k\epsilon_0} \dfrac{q_1 q_2}{r^2} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{kr^2} = \dfrac{1}{4\pi \epsilon_0} \dfrac{q_1 q_2}{(\sqrt{k}r)^2}$
Thus, when a dielectric is introduced, it has an effect on the distance between the two charges, wherein if r is the actual distance over which the dielectric is introduced, then it behaves as if the distance that the dielectric occupies is $\sqrt{k}r$.
Therefore, the effective distance between the two charges in our arrangement becomes $\left(\dfrac{r}{2} + \sqrt{k}\dfrac{r}{2}\right)$
And the effective force acting between the two charges is given by:
$F^{\prime} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r_{eff}^2} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{\left(\dfrac{r}{2} + \sqrt{k}\dfrac{r}{2}\right)^2} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{\left(\dfrac{r}{2} + \sqrt{4}\dfrac{r}{2}\right)^2} =\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{\left(\dfrac{r}{2} + 2\dfrac{r}{2}\right)^2} =\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{\left(\dfrac{r}{2} + r\right)^2} $
$\Rightarrow F^{\prime} =\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{\left(\dfrac{3r}{2}\right)^2} =\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2}\dfrac{4}{9} = \dfrac{4}{9}F $

So, the correct answer is “Option A”.

Note: Remember that the apparent change in the actual distance over which the dielectric is introduced is a property of the material that is used as the dielectric. This happens because of the value of the dielectric constant, which gives the relative electrical permittivity of the dielectric material. Note that this apparent change in distance occurs only over the distance between the charges where the dielectric is introduced. The rest of the distance behaves as it was in free space.