Two conducting spheres a distance d away with charges +q and -q

[Jaccobtw]

Homework Statement

Two conducting spheres are a distance d away with charges +q and -q. Is the attractive force equal to, greater than, less than $k_e q^2/d^2$, or zero

Relevant Equations

Coloumb's Law

you can treat the center of two conducting sphere's like two point charges. Therefore it should be equal to $k_e q^2/d^2$, but the answer is greater than $k_e q^2/d^2$. Can someone explain how? Thank you

 

[Gordianus]

The surce charge density in both spheres isn't uniform. Thus, we can't say the charge distribution can be replaced by a points charges at the spheres ' centers. Opposite charges tend to attract (they're closer) and this results in a larger force.

 

[Delta2]

Yes I think @Gordianus is right. I 'll try to explain a bit in more detail what he means:

Because the spheres are conductors, when we bring them close together their surface charges interact and more charge of each sphere is moved toward the areas of the spheres that are close together. So their surface charge density isn't uniform anymore and you can't treat the sphere as point charges. Because there are these accumulated charges positive on one sphere and negative on the other which are distanced apart much less than $d$, the attractive force is bigger.

If the spheres were insulators then their charges wouldn't interact, the surface charge density on each sphere would remain uniform and you could treat the spheres as point charges.