Question

The ratio of forces between two small spheres with constant charge in air to that in a medium of dielectric constant K?

Answer 1

Ratio=F1/F2.

where F1=force in air and F2=force in medium

Answer 2

Answer: The correct answer is K:1.

Explanation:

The expression for coulomb's law is as follows;

$F=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}r^{2}}$

Here, F is the force, q_{1},q_{2} are the charges, $\varepsilon _{0}$ is the permittivity of free space and r is the distance.

The expression for force of two small spheres in air is as follows;

$F_{air}=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}r^{2}}$             ........ (1)

The expression for force of two small spheres in medium of dielectric constant is as follows;

$F_{d}=\frac{q_{1}q_{2}}{4\pi K\varepsilon _{0}r^{2}}$          ......... (2)

Divide the equation (1) by the equation (2).

$\frac{F_{air}}{F_{d}}=\frac{K}{1}$

Therefore, the ratio of fores between two smalls spheres with constant charge in air to that in a medium of dielectric constant is K:1.