Suppose I have two charged solid conducting spheres, both given a charge $Q$ and are separated by a distance a. The force between them should be $\frac{Q^2}{4\pi\epsilon_0 a^2}$. Is it correct? Because in many places, I have seen the answer is derived using the method of images. I do not understand why such an approach is necessary.
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2$\begingroup$ Can you give a link to what you are talking about? For problems involving the method of images, usually you are looking at a charged particle near a grounded, conducting a surface. i.e. the potential on the conductor is specified, not the total charge on that conductor. You might just be getting assumptions mixed up. $\endgroup$– BioPhysicistCommented Jun 29, 2021 at 19:17
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1$\begingroup$ Your result would be correct if you are talking about charged "points", i.e zero-radius spheres... or anyway spheres which are much smaller that their relative distance a. If you have a sphere of finite size then you need to take into account how surface charge deviates form the uniform distribution, due to the presence of the other charged sphere. This modifies the force from your limit result (for instance, you get a dipole moment and interaction). Image charges can help you to do the calculations, they work well with spherical conductors. $\endgroup$– SteCommented Jun 29, 2021 at 23:19
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$\begingroup$ @Ste can you please give an idea of how to attack this problem using image charges? $\endgroup$– Uranium238Commented Jun 30, 2021 at 11:06
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3$\begingroup$ The formula is only true for a homogeneously charged non-conductor. What you saw is the effect that if you hold a charge A close to a conductor, it will pull opposite charges towards the charge A, leading to an attraction. This can be calculated with the methods of images. $\endgroup$– lalalaCommented Jul 3, 2021 at 13:13
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