When two spheres having charges 5Q and -Q are placed at a certain distance, the force acting between them is F. Now they are connected by a conducting wire and again separated from each other. How much force will act between them if the separation now is the same as before?

Question

Accepted Answer

(2) (4F)/5. Given: charges 5Q and -Q at separation r attract with force F; the spheres are joined by a wire and put back at the same separation. Idea: when two identical conducting spheres are joined, the total charge divides equally between them (the question takes the spheres to be identical). Before: F=(k(5Q)(Q))/(r²)=(5kQ²)/(r²) (attractive) Total charge: 5Q+(-Q)=4Q, so each sphere gets (4Q)/2=2Q. After: F'=(k(2Q)(2Q))/(r²)=(4kQ²)/(r²) (now repulsive) (F')/F=4/5⇒ F'=(4F)/5 So the force is (4F)/5, and it is now repulsive.

When two spheres having charges 5Q and -Q are placed at a certain distance, the force acting between them is F. Now they are connected by a conducting wire and again separated from each other. How much force will act between them if the separation now is the same as before?

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