Question

two point charges repel each other with a force F when placed in water of dielectric constant 81 .what will be the force on them when placed the same distance apart in air?

Answer 1

81 times greater i.e. 81F

Answer 2

The force between point charges in air is 81 times the force exhibited by them when it is placed in water.

Explanation:

The force between point charges is,

$F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} \epsilon_{w} r^{2}}$

Where,

$\varepsilon_{0}$ is the permittivity of free space

$\epsilon_{w}$ is the dielectric constant of water

So,

$F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}(81)}$

On rearranging, we get,

$\frac{q_{1} q_{2}}{r^{2}}=F \times 4 \pi \varepsilon_{0} \times 81 \rightarrow(1)$

The force between point charges in air,

$f=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}} \rightarrow(2)$

So, substitute eqn (1) in eqn (2)

$f=\frac{F \times 4 \pi \varepsilon_{0} \times 81}{4 \pi \varepsilon_{0}}$

On cancelling, we get,

$\therefore f=81 F$