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Two identical spheres carrying charges 2Q and -Q are placed at a certain distance, and the force acting between them is F. They are now connected by a conducting wire and then separated again. If the separation is the same as before, the force acting between them will be

Asked in RS Academy GUJCET booklet · Charge sharing between conductors

Answer: (4) F/8

Step-by-step solution

Given: identical spheres with 2Q and -Q; force F at separation r.

Before: F=(k(2Q)(Q))/(r²)=(2kQ²)/(r²) (attractive).

Connecting identical spheres shares the net charge equally: each gets (2Q+(-Q))/2=Q/2.

After: F'=(k(Q/2)(Q/2))/(r²)=(kQ²)/(4r²) (now repulsive).

(F')/F=(1/4)/2=1/8, so F'=F/8.

Why the other options are wrong

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