How Do Different Mediums Affect Coulomb's Law?

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Infinitum
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Hello everyone!

As I understand it, Coulomb's Law gives the electrostatic force of attraction or repulsion between two point charges, in any single medium.

[itex]|F| = \frac{Kq_1q_2}{r^{2}}[/itex]

And for any medium K is [itex]\frac{1}{4\pi\epsilon}[/itex]

My question is, how would different mediums in between the charges effect the Coulombic force?

For example, what would the force be if one particle +Q is kept in water, and the other, say -Q in air? The water surface ending at half the distance between between them.(please see attachment)
Surely the [itex]\epsilon[/itex] won't be the same for this case, as for only water, or only air.

I've referred quite a few books, but they don't explain this kind of situation.
 

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Hello Infinitum! :smile:

Nobody else has answered, so I'll have a go …

I don't think there's any simple solution, you first need to find the field, using the boundary conditions (so it isn't just the straight line between the two charges that matters).

Ultimately, the charges create electric dipoles in the dielectrics, and also I think along the interface, and they have their own Coluomb's law contribution.

The cases you see in textbooks and exam questions are the easy ones where the field is obviously regular (uniform, or cylindrical, or spherical) …

eg a point charge at the centre of a spherical shell. :wink:
 
I see now, I read up an article on how to find the electric field at boundary of two materials, but it's a bit confusing for me to understand, since I don't know a few related concepts. Will try learning those first :-p

Thanks for the reply Tiny-tim. :biggrin: