Question

two like charges when placed in a medium of dielectric constant of2.5 at adistance of 10 cm repel each other with a force of 1 N.find the force between them when placed5cm apart in air​

Answer 1

Given

• two like charges when placed in a medium of dielectric constant of 2.5 at a distance of 0.1 m repel each other with a force of 1 N.

To Find

The force between them when they are placed 0.05 m apart in air.

Working formula

Force between charge particles placed in air

$\dfrac{Kq_1q_2}{r^2}$

• Where, K = $9 \times 10^9$ and r is the distance between the charged particles.

Force between charge particles placed in a dielectric medium

Here, if the distance between the particles is r, we use the same formula as that of air ,but the new distance is given as $\sqrt{E_or}\:,E_o$ is the dielectric constant of the medium.

Calculations

0.1 m distance in a dielectric medium of constant 2.5 gives a distance $\sqrt{2.5 \times 0.1}\:=\:0.5$ m distance in air.

Hence,when placed 0.5 m apart , force = 1 N. Thus, we have to find force when they are placed 0.05 m apart.

$\dfrac{F_1}{F_2}\:=\: {(\dfrac{r_2}{r_1})}^2 \\ \\ \\ => \: \dfrac{1}{F_2}\:=\: {(\dfrac{0.05}{0.5})}^2 \\ \\ \\=> \: \dfrac{1}{F_2} \:=\:\dfrac{1}{100} \\ \\ \\ => \: F_2\:=\:100\:N$

Hence, the force when they are placed in air is 100 N