# A point charge inside infinite medium of dielectric

1. Nov 2, 2009

### Ready2GoXtr

A point charge inside infinite dielectric material

1. The problem statement, all variables and given known data
A point charge is an infinite medium of dielectric material having a relative permittivity $$\epsilonr$$. <--- epsilon(sub r). Find the electric field vector and the potential function at any point in space, assuming that the potential is zero volts at infinity.
2. Relevant equations
D = $$\epsilon$$ * E + P
P = $$\epsilon0$$ * $$\chie$$ vector E
$$\epsilon[\tex] = [tex]\epsilon0$$ * $$\epsilonr$$
$$\chie$$ = $$\epsilonr$$ - 1

well its not letting me put it in right so im gonna enter them in with () next to them
D(vector) = epsilon*E(vector) + P(vector)
P(vector) = epsilon(sub 0)*chi(sub e)*E(vector)
epsilon = epsilon(sub 0)*epsilon(sub r)
chi(sub e) = epsilon(sub r) - 1
Electric Field of Point Charge = k*q/r^2
Electric Field of Sphere = q/(4*pi*epsilon(sub0)*r^2)

3. The attempt at a solution
Im not quiet sure what my first step would be. I would think that a point charge inside a dielectric medium would have a reduced electric field, but it is infinite so wouldnt its electric field be nothing?

Last edited: Nov 2, 2009
2. Nov 2, 2009

### gabbagabbahey

Just use Gauss' Law for $\textbf{D}$....can you think of a Gaussian surface that will exploit the symmetry present?

3. Nov 2, 2009

### Ready2GoXtr

what do you mean by exploit?

4. Nov 2, 2009

### gabbagabbahey

"Exploit" is just another word for "use"

5. Nov 2, 2009

### Ready2GoXtr

I feel that a sphere represents a good shape.

6. Nov 2, 2009

### gabbagabbahey

Okay, but why? And where will the center of the sphere be?

7. Nov 2, 2009

### Ready2GoXtr

The center of the sphere will be located on the point charge

8. Nov 2, 2009

### Ready2GoXtr

so would my answer just be q/(4*pi*epsilon(sub0)*r^2)
?

9. Nov 6, 2009

### gabbagabbahey

Not quite...What is Gauss' Law in terms of $\textbf{D}$? Do you see why Gauss' Law in terms of $\textbf{E}$ isn't helpful here?

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