Question

Two point charges exert force of $F$ Newton when separated by a distance of $10cm$ in air. The distance for which the force between them will be same if the medium them has ${\epsilon }_r=4$, is:-

• A. 10 cm
• B. 20 cm
• C. 5 cm
• D. 2.5 cm

Answer 1

The correct option is C 5 cm

$F=\frac{q_1{q}_2}{4\pi {ϵ}_o{ϵ}_r{r}^2}$

for air ${ϵ}_r=1$

$\frac{1}{1\times {10}^2}=\frac{1}{4\times r^2}$

$r^2=25$

$r=5cm$