Practice portal › Electric Charge and Properties › Electric Charge and Coulomb's Law

Two identical metal spheres A and B carry the same charge q. When they are a distance r apart, the force acting between them is F. Another identical sphere C, initially uncharged, is first brought into contact with A, then touched to B, and then removed. Now the force acting between A and B at the same distance is

Asked in RS Academy GUJCET booklet · Charge sharing between conductors

Answer: (3) (3F)/8

Step-by-step solution

Idea: when two identical conductors touch, they share their total charge equally.

C touches A: the total is q+0, so A and C each hold q/2.

C touches B: the total is q/2+q=(3q)/2, so B (and C) each hold (3q)/4.

At the same distance the force is proportional to the product of the charges:

(F')/F=((q/2)(3q/4))/(q× q)=3/8

So F'=(3F)/8.

Why the other options are wrong

More Electric Charge and Coulomb's Law questionsAll Electric Charge and Coulomb's Law questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer