Question

9.2.Two point charges in air at a distance of 20 cm from each other

interact with a certain force. At what distance from each other should

these charges be placed in oil to'obtain the same force of interaction?

Answer 1

$F_{air}= \frac{1}{4\piε_0} \frac{q_1q_2}{r_1}.... ...(i)$

$F_{medium}= \frac{1}{4\piε_0k} \frac{q_1q_2}{d^2}.... ...(ii)$

Substituting equation (i) by (ii) we get,

$\frac{1}{r^2}=\frac{1}{k}×\frac{1}{d^2}$

$d= \sqrt{\frac{r^2}{k}} ​$

r=20cm

k=5

$d= \sqrt{\frac{20^2}{5}} =\sqrt{80} = 8.94cm$

$d=8.94×10^{-2}M$