Question

Two point charges ${{q}_{1}}$ and ${{q}_{2}}$ are placed at a distance of 50 m from each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is:
A. 16.6 m
B. 22.3 m
C. 28.0 m
D. 25.0 m

Answer 1

We are given two point charges kept in air and the distance of separation between them. It is said the same charges are placed in oil and the force between the charges remains the same. We are also given the relative permittivity of oil. Since it is said that the force between the charges are equal in the two mediums we can equate the force experienced in two mediums. By substituting the known values and eliminating the common terms we get the solution.

Formula used: Force experienced between two charges, $F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Complete step by step answer:
In the question it is said that two point charges ${{q}_{1}}$ and ${{q}_{2}}$ are placed in air and the distance of separation between them is 50 m.
Let ‘${{r}_{1}}$’ be the distance of separation in air, then
${{r}_{1}}=50m$
It is said that the same point charges are placed in oil.
The relative permittivity of oil is given to us,
${{\varepsilon }_{1}}=5$, were ‘${{\varepsilon }_{1}}$’ is the relative permittivity of oil.
We know that the force between two charges is given by the equation,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$, were’ ‘F’ is force, ‘$\dfrac{1}{4\pi {{\varepsilon }_{0}}}$’ is a constant, ‘${{q}_{1}}$’ and ‘${{q}_{2}}$’ are the two charges, ‘${{\varepsilon }_{0}}$’ is the relative permittivity of the medium and ‘$r$’ is the separating distance between the two charges.
Now let us consider the case when the charges are placed in the air.
The force between the two charges,
${{F}_{1}}=\dfrac{1}{4\pi \varepsilon }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}$
Here the medium is air; we know that the relative permittivity of air is 1.
Therefore, we have
$\Rightarrow {{F}_{1}}=\dfrac{1}{4\pi }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}$
Now let us consider the case when the charges are placed in oil.
The force between the two charges will be,
${{F}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{1}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}$
Since relative permittivity of oil is given, ${{\varepsilon }_{1}}=5$ we get
$\Rightarrow {{F}_{2}}=\dfrac{1}{4\pi \times \left( 5 \right)}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}$
In the question it is said that the force between the two charges is the when in both mediums.
Hence we can equate ${{F}_{1}}$ and ${{F}_{2}}$
$\Rightarrow {{F}_{1}}={{F}_{2}}$
$\Rightarrow \dfrac{1}{4\pi }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}=\dfrac{1}{4\pi \times \left( 5 \right)}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}$
$\Rightarrow \dfrac{1}{{{r}_{1}}^{2}}=\dfrac{1}{5\times {{r}_{2}}^{2}}$
$\Rightarrow 5\times {{r}_{2}}^{2}={{r}_{1}}^{2}$
$\Rightarrow {{r}_{2}}^{2}=\dfrac{{{r}_{1}}^{2}}{5}$
$\Rightarrow {{r}_{2}}=\sqrt{\dfrac{{{r}_{1}}^{2}}{5}}$
We know that ${{r}_{1}}=50m$, by substituting this in the above equation, we get
$\Rightarrow {{r}_{2}}=\sqrt{\dfrac{{{50}^{2}}}{5}}$
$\Rightarrow {{r}_{2}}=\sqrt{\dfrac{2500}{5}}$
$\Rightarrow {{r}_{2}}=\sqrt{500}$
$\Rightarrow {{r}_{2}}=22.36m$
Therefore the distance of separation between the two charges in oil is 22.36 m.

So, the correct answer is “Option B”.

Note: The force experienced between two charges is given by Coulomb’s law.
According to Coulomb’s law the electrical force between two charges separated by a distance is directly proportional to the product of the magnitude of the two charges and is inversely proportional to the square separating distance between the charges.
This proportionality is equated by the constant $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$ which is $=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$