Question

Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increases.

Answer 1

Hint: From the formula of force between two point charges, we can find the situation in the free space and in a different medium. From there we can get the dielectric constant and how it affects the force between two charges in any medium.

Formula used: In this solution we will be using the following formula,
 $\Rightarrow F={\displaystyle \frac{1}{4\pi {\epsilon }_o}}{\displaystyle \frac{Q_1{Q}_2}{R^2}}$
where $F$ is the force,
 ${\epsilon }_o$ is the permittivity in free space,
 $Q_1$ and $Q_2$ are the two charges and $R$ is the distance between the two charges.

Complete step by step solution:
According to the question, we first need to calculate the force between the two point charges in free space. It is given by the formula,
 $\Rightarrow F={\displaystyle \frac{1}{4\pi {\epsilon }_o}}{\displaystyle \frac{Q_1{Q}_2}{R^2}}$
Now when the charges are placed at the same distance but in a different medium, then the permittivity value used in the formula changes. Let the permittivity of this medium be $\epsilon$ then the force between these point charges in this medium will be
 $\Rightarrow F^{\prime }={\displaystyle \frac{1}{4\pi \epsilon }}{\displaystyle \frac{Q_1{Q}_2}{R^2}}$
Now for this medium, the relative permittivity is given as,
 $\Rightarrow {\epsilon }_r={\displaystyle \frac{\epsilon }{{\epsilon }_o}}$
So we can write, $\epsilon ={\epsilon }_o{\epsilon }_r$
Now the relative permittivity of the medium with respect to free space is also called dielectric constant. The dielectric constant of a medium is denoted by the letter $K$ . Therefore, we can write ${\epsilon }_r=K$ . So substituting we get,
 $\Rightarrow F^{\prime }={\displaystyle \frac{1}{4\pi K{\epsilon }_o}}{\displaystyle \frac{Q_1{Q}_2}{R^2}}$
On arranging the above equation we can get,
$\Rightarrow F^{\prime }={\displaystyle \frac{1}{K}}\left({\displaystyle \frac{1}{4\pi {\epsilon }_o}}{\displaystyle \frac{Q_1{Q}_2}{R^2}}\right)$
Now we can write $F={\displaystyle \frac{1}{4\pi {\epsilon }_o}}{\displaystyle \frac{Q_1{Q}_2}{R^2}}$
So we get, $F^{\prime }={\displaystyle \frac{1}{K}}F$
So as the dielectric constant increases, then the force between two charges decreases accordingly.

Note:
The relative permittivity or the dielectric constant of a medium is defined as the ratio of the permittivity of that medium to the permittivity in vacuum. It is the property of the material of the medium and affects the Coulomb force between the charges placed in that medium.