Two spherical conductor A and B having equal radii and carrying equal charge s in them repel each other with a force F when placed apart at some distance.A third uncharged spherical conductor having same radii as that A is brought in contact with A and then B and finally removed away from both .The new force of repulsion between A and B is

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Let spherical conductor B and C each is of charge q and kept at a distance r apart from each other and electrostatic force of repulsion acting between them is F, given by :
F = 14 π ε0 (q) (q)r2
When an uncharged conductor A is brought in contact with B, charge on both A and B will become q/2.
When this conductor A ( now having charge q/2) is touched with C, charges on both A and C will become 3q/4.
Finally, now conductor B and conductor C will have charge q/2 and 3q/4 respectively.
Therefore, new force of repulsion between B and C is given by :
F' = 14 π ε0 (q/2) (3q/4)r2F' = 14 π ε0 q2r2 × 38F' = 3 F8
Hence, proved.

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