Two identical metal spheres gallery charges + q and minus 2 respectively when the spheres are separated by a large distance are the force of the f between them is now the spheres are allowed to touch and then move back to the same separation find the new force of repulsion between them

Dear Student

The force between+q and-2q at a distance r is given as F NowF=K((+q)×(-2q))r2=-2kq2r2When the chages +q and2q comes in contact,then it neutralises the chage+q byq charge and left onlyq.These-q charge will be equally distributed both the sphare.If it is seperated the two sphere again then each sphere will aquire amount of charge ofq2 andq2.Now the new force between them will beFn=K((q2)2)2=14((kq22))=(-F)8So the force will be 1/8 part of previous force and it will be repulsive foce while F was attractive.

Regards

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