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joshanders_84
Nov16-04, 07:56 PM
This problem is also giving me a lots and lots of trouble, and I don't even know where to begin.

A point charge q is imbedded in a solid material of dielectric constant K.

A) Use Gauss's law as stated in equation \oint{K \vec{E} \cdot \vec{A}} \;=\; \frac{Q_{free}}{\epsilon_{0}} to find the magnitude of the electric field due to the point charge q at a distance d from the charge.

B) Use your result from part (a) and Gauss's law in its original form as given in equation \oint{\vec{E} \cdot \vec{d A}}\; =\; \frac{Q_{encl}}{\epsilon_{0}} to determine the total charge (free and bound) within a sphere of radius d centered on the point charge q.

C) Find the total bound charge within the sphere described in part (b).

I obviously need A before doing B and C, but I don't even know how to get it. I didn't fully understand Gauss law in the first place, so now I am even more confused. Any help appreciated, thanks.

gnome
Nov16-04, 08:22 PM
I think you'll have to do a better job with those equations so people can help you without spending all night trying to decode the question.

dduardo
Nov16-04, 08:25 PM
Here is a hint:

E and D are in the radial direction. That means that there will never be a tangential component. Now all you have to figure out is what the normal component is. This is just a simple boundary condition problem.

joshanders_84
Nov16-04, 08:30 PM
what do you mean a boundary condition problem? I am confused a bit, I'm afraid to say.

those two equations, which I thought would show up properly, are:
integral KEA = Q(free)/epsilon_0
and
integral EdA = Q(enclosed)/epsilon_0

dduardo
Nov16-04, 08:41 PM
Forget about the formulas and just think for a moment. In the case when you have a charge in free space what is the electric field at some point due to the charge? Once you can answer that question using Gauss's Law you'll be ready to move on to the second part which is: how is an electric field affected by a dielectric boundary.

joshanders_84
Nov16-04, 08:51 PM
The electric field from some point is given by a formula, right? Specifically Kq/(r^2)

dduardo
Nov16-04, 08:55 PM
yes, now what is the relationship between En inside the dielectric and En inside free space.

joshanders_84
Nov16-04, 08:57 PM
the dielectric has a charge built up on the sides of it? And this would effect E?

dduardo
Nov16-04, 09:06 PM
Ok, so what is the charge density?

joshanders_84
Nov16-04, 09:11 PM
I honestly am not sure of the charge density. I know this sounds dumb but I don't know...?

dduardo
Nov16-04, 09:17 PM
Q = \int \rho dV where rho is the charge density

dduardo
Nov16-04, 09:22 PM
This may help:

http://www.shef.ac.uk/physics/teaching/phy205/lectures_8_and_9.htm

Specifically: Gauss’s Law in dielectrics, D-field