Question

Two point charges in air at a distance of $20cm.$ from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity $5$ to obtain the same force of interaction

• A. $8.94\times {10}^{-2}m$
• B. $0.894\times {10}^{-2}m$
• C. $89.4\times {10}^{-2}m$
• D. $8.94\times {10}^{2}m$

Answer 1

The correct option is A $8.94\times {10}^{-2}m$

$F_{air}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}....\left(i\right)$

$F_{medium}=\frac{1}{4\pi {\epsilon }_{0}k}\frac{q_1{q}_2}{d^2}....\left(ii\right)$

Dividing equation (i) by (ii)

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

$d=\sqrt{\frac{r^2}{k}}$

$=\sqrt{80}$

$d=90cm$