The force of repulsion between two point charges is $F$ , when these are at a distance of $1 m$ apart. Now they are replaced by conducting spheres each of radii $25 c m$ having the same charge as that of point charges. The distance between their centres is $1 m$ , then compare the force of repulsion in the two cases.

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Solution

For point charges $F = \frac{k q_{1} q_{2}}{r^{2}} = k q_{1} q_{2}$ , since $r = 1 m$
Since charge on conducting sphere is assumed to be placed at the center of the sphere, hence repulsive force on both sphere is also given by, $F = k q_{1} q_{2}$ , hence force on both sphere will be same as the repulsive force on the both point charges. It does not depend on the radius of the sphere.

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