Question #71216

two identical metal sphere carry charges +q and -2q .when the spheres are separated by a large distance r.the force btw them is F .now the sphere are allowed to touch and then moved back to the same separation .find new F

Expert's answer

The force between +q and -2q at a distance r is given as F

Now F= K ((+q)×(-2q))/r^2 = -2kq^2/r^2

When the chages +q and –2q comes in contact, than it neutralises the chage +q by –q charge and lefts only –q. These -q charge will be equally distributed both the sphare. If it is seperated the two sphare again than each sphare will aquire a mount of charge of –q/2 and –q/2 .

Now the new force between them will be

F_n= K ((-q^2/2)(-q^2/2))/r^2 = 1/4((kq^2)⁄r^2 ) = (-F)/8

So the force will be 1/8 part of previous force and it will be repulsive foce while F was attractive.

Now F= K ((+q)×(-2q))/r^2 = -2kq^2/r^2

When the chages +q and –2q comes in contact, than it neutralises the chage +q by –q charge and lefts only –q. These -q charge will be equally distributed both the sphare. If it is seperated the two sphare again than each sphare will aquire a mount of charge of –q/2 and –q/2 .

Now the new force between them will be

F_n= K ((-q^2/2)(-q^2/2))/r^2 = 1/4((kq^2)⁄r^2 ) = (-F)/8

So the force will be 1/8 part of previous force and it will be repulsive foce while F was attractive.

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