# Explain to me how this works

If two point charges are separated by distance then according to Coulomb's law, the attractive force between them is fve = kvexq^2/d^2...

where is Coulomb's constant. But what if the charges were distributed on spheres instead of at points? Two charged conducting spheres with charges and are separated by center-to-center distance

The attractive force between them is ?

the answer to this question in the weekly is plugging in the answers to the question. the attractive force between is equal to the equation, but the answer is wrong and say the attractive for is greater than. so explain to me. how fve = kvexq^2/d^2 =/< fve

this is illogical

if the force was greater than, the distance would be decreasing, right? less than, and they would be increasing in distance because + and - q would want to balance out.

im guessing kve is the matter or space bettween them, medium between them or resistance. which confuses me because it's multiplied.

maybe im just really stupid, but as far as i know how equations work. equal to is equal to and not great than or less than. Note by Bo Treat
3 years, 6 months ago

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say the attractive for is greater than. so explain to me. how fve = kvexq^2/d^2 =/< fve

The important point is that the spheres are conducting, so charge can freely move on the sphere. The charge arrange such that the potential is minimized, so the effective distance is less than $d$, say $d_{sphere} < d$, so the force is now $\dfrac{kq^2}{d_{sphere}^2} > \dfrac{kq^2}{d^2}$.

if the force was greater than, the distance would be decreasing, right? less than, and they would be increasing in distance because + and - q would want to balance out.

The question asks for the force at one instant. Sure, the spheres attract each other because they have opposite signs, and will start moving towards each other.

- 3 years, 6 months ago