A point charge inside infinite medium of dielectric

Click For Summary

Homework Help Overview

The problem involves a point charge situated within an infinite medium of dielectric material characterized by a relative permittivity. The objective is to determine the electric field vector and the potential function at any point in space, with the condition that the potential is zero volts at infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Gauss' Law and the choice of a Gaussian surface to leverage the symmetry of the problem. There is uncertainty about the implications of the infinite dielectric medium on the electric field and potential. Questions arise regarding the appropriateness of using a spherical Gaussian surface and the location of its center.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the application of Gauss' Law in terms of the displacement field. Some guidance has been offered regarding the use of a spherical surface, but clarity on the reasoning behind this choice is still being sought.

Contextual Notes

Participants are navigating the complexities of the problem, including the effects of the infinite dielectric medium and the definitions of relevant physical quantities. There is a lack of consensus on the initial steps to take in solving the problem.

Ready2GoXtr
Messages
74
Reaction score
0
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 \epsilon<sub>r</sub>. <--- 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.

Homework Equations


D = \epsilon * E + P
P = \epsilon<sub>0</sub> * \chi<sub>e</sub> vector E
\epsilon[\tex] = \epsilon&lt;sub&gt;0&lt;/sub&gt; * \epsilon&lt;sub&gt;r&lt;/sub&gt;<br /> \chi&lt;sub&gt;e&lt;/sub&gt; = \epsilon&lt;sub&gt;r&lt;/sub&gt; - 1<br /> <br /> <br /> well its not letting me put it in right so I am going to enter them in with () next to them <br /> D(vector) = epsilon*E(vector) + P(vector)<br /> P(vector) = epsilon(sub 0)*chi(sub e)*E(vector)<br /> epsilon = epsilon(sub 0)*epsilon(sub r)<br /> chi(sub e) = epsilon(sub r) - 1<br /> Electric Field of Point Charge = k*q/r^2<br /> Electric Field of Sphere = q/(4*pi*epsilon(sub0)*r^2)<br /> <br /> <h2>The Attempt at a Solution</h2><br /> 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 wouldn&#039;t its electric field be nothing?
 
Last edited:
Physics news on Phys.org
Just use Gauss' Law for \textbf{D}...can you think of a Gaussian surface that will exploit the symmetry present?
 
what do you mean by exploit?
 
"Exploit" is just another word for "use"
 
I feel that a sphere represents a good shape.
 
Ready2GoXtr said:
I feel that a sphere represents a good shape.

Okay, but why? And where will the center of the sphere be?
 
The center of the sphere will be located on the point charge
 
so would my answer just be q/(4*pi*epsilon(sub0)*r^2)
?
 
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?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
879
  • · Replies 1 ·
Replies
1
Views
994
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
905
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K