The force of repulsion between two point charges is $F,$ when these are $1 m$ apart. Now the point charges are replaced by conducting spheres of radii $5 c m$ having the charge same as that of point charges. The distance between their centres is $1 m,$ then the force of repulsion will:

increase

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decrease

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remain same

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become

\frac{10 F}{9}

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Solution

The correct option is C remain same
From Coulombs law
$F α \frac{q_{1} q_{2}}{r^{2}}$
As the charges and distance are same in either cases, the force between the charges remains constant.
Hence, the force of repulsuion in the second case equal to $F$

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