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  4. 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

Q. 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

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Solution:

Solution
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

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