# Dielectric modification of electrostatic equations?

I have learnt that in cases of electrostatic fields inside a dielectric of any source charge, the field is reduced by a factor of K( if K, the dielectric "constant", is taken everywhere to be same). But that means that if there is, say, a fixed (by non-electric forces) charge outside a randomly shaped dielectric block, then the fields experienced by our test charge inside the dielectric will be weaker (by a factor of K) than what it would have experienced had it been outside the dielectric. But should the fields of other charges outside the block be unaffected??

If this is so, we take two opposite charges in vacuum separated by a distance and a dielectric block in between them. The lines of force would be passing through the slab and vacuum both. If the fields outside are not affected than there would be a contradiction since the closed line integral of field wouldn't be zero.

But then how and why would the fields outside the dielectric be changed. The contribution of the induced dielectric charges should be zero at least in the idealised case where infinitely large plates (producing uniform fields) have a dielectric bock (smaller then the separation) between them??

But should the fields of other charges outside the block be unaffected??

Nope. The fields outside are affected. It's not always easy to calculate, though. Basically, inside a (linear) dielectric an induced field is formed in the opposite direction which counteracts the electric field, reducing the net electric field.

Of course, this new field has to be caused by something, in this case it is the charge separation induced on the dielectric by the original field1. This induced charge separation induces a field outside as well (usually), affecting the net field on the outside.

If the fields outside are not affected than there would be a contradiction since the closed line integral of field wouldn't be zero. [...] The contribution of the induced dielectric charges should be zero at least in the idealised case where infinitely large plates (producing uniform fields) have a dielectric bock (smaller then the separation) between them??

Technically, no, as long as the field inside is equally unaffected, the closed line integral will stay zero. Remember, it is a closed integral. You need to pass through the dielectric twice, once in each direction, and the change of field while going from A to B will be cancelled out by the change of field while coming back. This will happen with a large slab, which, as you correctly noted, has a negligible effect on the outside field. However, since the component of distance travelled along the field is the same in this case, and the change in electric field ($\mathbf{E}_{new}-\mathbf{E}_{old}=(K-1)\mathbf E$) is the same throughout, this change cancels itself out in the integral.

1. In the case of a linear dielectric, $\mathbf P=\mathbf E(K-1)$, and we can calculate the bound charge density (i.e., the induced charge) using $\rho_b=-\nabla\cdot\mathbf P$

• Then, the modified equations (using KE instead of E) are valid for fields inside the dielectric only? Moreover, can you give an example of how to calculate the change of field due to dielectric block outside the block for something simple for example a dielectric block beside a positive point charge? – stochastic13 Jul 17 '13 at 14:11
• And also, if the length of the dielectric slab is small in the idealized situation in my question and the fields outside are unaffected, then we can select a field line outside while going and an field line inside while coming back and the integral would be non-zero. – stochastic13 Jul 17 '13 at 14:13
• @SatwikPasani Yeah, only inside the dielectric. I currently don't have the time to give a derivation, you can start working from the footnote on, say, a dielectric sphere and then calculate. I think Griffiths' deals with this. || In case of a smaller slab, the fringe fields come into play -- the field is no longer uniform. And with that, you can't easily predict if the line integral will pan out. You have to do the (rather dirty) calculation in this case. – Manishearth Jul 17 '13 at 15:22