Question

The force of repulsion between two point charges is $F$ , when these are at distance $0.5 \, m$ apart. Now the point charges are replaced by non conducting spheres of radii $5 \, cm$ each having the same charge as that of the respective point charges. The distance between their centres is again kept $0.5m.$ Then the force of repulsion will

• A. increase
• B. decrease
• C. remain $F$
• D. become $\frac{10 F}{9}$

Answer 1

Answer: (C)

Force between the two point charges at $0.5\text{m}$ is
$F_1=\frac{q_1 q_2}{4 \pi \in_0 \mathrm{r}^2}=\frac{q_1 q_2}{4 \pi \in_0(0.5}$
When the charges are given to the non conducting spheres the charges are assume to be concentrated at the centre, they will behave as point charges & when they are placed at the distace of $0.5\text{m}$
They will experience the same force as the point charges does