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two spherical conductors B &C having equal radii and carrying equal charges with them repel eachother with a force F when kept at a small distance apart . a third spherical conductor A having same charge as that of B but uncharged is brought in contact with B and then with C and finally removed away from both . what is the new force of repulsion between B & C ?

Shivam Chopra
45 Points
6 years ago
Let charge on B & C = q.
So, F = force b/w both conductors = K * q * q / r2.  where, K = 1/(4πϵ)
Also, charge on A = 0.
So, when A is touched with b , charge gets distributed equally on both conductors               So, new charge on both = (q + 0)/2 = q/2.
Both A & B have q/2 charge.
Now, A & C are touched together.
So, new charge on both = (q/2 + q)/2 = 3q/4.
So, now charge on B = q/2, charge on C = 3q/4.
Force = ( K * q/2 * 3q/4 ) / r2 = ( K * 3q2 ) / 8r2 = 3F/8
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