# What force is acting on the charge in the dielectric?

For example I have a dielectric solid with a small charged ball in it. And I have external electric field $E$. So what force is acting on this ball?

The field in dielectric is $\frac{E}{\epsilon}$, so the force should be $\frac{Eq}{\epsilon}$.

On the other hand. If I remove a small piece of dielectric then the field in the hole will be $E$. Now I put charge in this hole and the force is $Eq$. The hole is just for charged ball and there is no free space.

So what is the force?

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The force on the system as a whole (dielectric+ball) will certainly be $Eq$ by conservation arguments. I believe that the forces on the dielectric will cancel out, and the force on the ball will be $Eq$; after all, the ball is not "inside" the dielectric but merely surrounded by it. –  Harry Johnston Jan 5 '12 at 22:11
I think that I don't get the difference between "inside" and "surrounded". For example what if I have "balls" of some salt in the water? Water will dissolve it because the electric force of connection between "balls" will decrease because of big $\epsilon$. So it's obviously "inside" case. –  Ximik Jan 5 '12 at 22:29
I've never heard it said that salt dissolves in water because water is a dielectric. Do you have a reference for that? –  Harry Johnston Jan 6 '12 at 0:56
It seems like you might be getting confused by the definition of "in the dielectric," as part of the phrase "the field in the dielectric is $\epsilon_0 E/\epsilon$." This means that the electric field in regions occupied by dielectric material is $\epsilon_0 E/\epsilon$.
If you remove a piece of the dielectric material to make a hole, then that hole is no longer "inside the dielectric" for purposes of this definition, because the space of the hole is not occupied by dielectric material. Similarly, if you put a charged ball inside the dielectric, then the space occupied by the charged ball is not "inside the dielectric" for purposes of this definition. The charged ball would be subject to an electric field of $E$ (assuming its own electric susceptibility is negligible), and would experience a force of $qE$.
I got what you tried to explain, but than in all cases of a charged material inside a dielectric, the force on it would have to be $E$ only and never $E/K$? how can this be correct? –  Satwik Pasani Jul 22 '13 at 16:52