Question

Two point charges $q_1$ and $q_2$ are placed at a distance of 50 m from each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is :

• A. 16.6 m
• B. 22.3 m
• C. 28.0 m
• D. 25.0 cm

Answer 1

The correct option is B 22.3 m

Here, $F_{air}=\frac{q_1{q}_2}{4\pi {ϵ}_0{d}_{air}^2}$ and

$F_{oil}=\frac{q_1{q}_2}{4\pi {ϵ}_r{ϵ}_0{d}_{oil}^2}$

According to question, $F_{air}=F_{oil}$

or $\frac{q_1{q}_2}{4\pi {ϵ}_0{d}_{air}^2}=\frac{q_1{q}_2}{4\pi {ϵ}_r{ϵ}_0{d}_{oil}^2}$

or ${ϵ}_r{d}_{oil}^2=d_{air}^2$

or $d_{oil}^2=\frac{(50{)}^2}{5}=500$

or $d_{oil}=22.3m$