L.I.H Dielectric: Free, Bound & Total Charge Calculations

[Damascus Road]

Homework Statement

A point charge Q is imbedded in a linear, isotropic, homogeneous dielectric medium of susceptibility $$X_{e}$$.

a.) What is the free charge enclosed by a sphere of radius R centered at Q.

b.) What is the bound charge enclosed by the sphere? How does the result vary with R? Do the calculation 2 ways.

c.)What is the total charge enclosed by the sphere?

Homework Equations

This is part of my problem, I have forgotten my notes at school, so I'm going off of notes I can find online.

The Attempt at a Solution

I'm not sure why there would be any free charge inside the dielectric... gut feeling (as I said.. no notes...)

Pls help :(

 

[mark726]

Thank you for your post. I would be happy to assist you in solving this problem.

First, let's define some terms and equations that will be useful in solving this problem. A point charge is a charge that is concentrated at a single point in space. In this case, the charge is denoted as Q. A dielectric medium is a material that can be polarized by an electric field, and its response to an applied electric field is described by its susceptibility, denoted as Xe.

Now, let's move on to the questions:

a.) To find the free charge enclosed by a sphere of radius R centered at Q, we can use the Gauss's law in integral form:

∮E⋅dA=Qenc/ε0

Where E is the electric field, dA is the differential surface area, and Qenc is the total charge enclosed by the surface. In this case, we are looking for the free charge enclosed, so we can rewrite the equation as:

∮E⋅dA=Qfree/ε0

We can then use the fact that in a linear, isotropic, homogeneous dielectric medium, the electric field is given by:

E=Q/4πε0εrR^2

Where εr is the relative permittivity of the medium. Substituting this into the equation above, we get:

∮(Q/4πε0εrR^2)⋅dA=Qfree/ε0

Since the charge is concentrated at a single point, we can simplify the integral to:

E⋅4πR^2=Qfree/ε0

Solving for Qfree, we get:

Qfree=E⋅4πR^2⋅ε0

b.) To find the bound charge enclosed by the sphere, we can use the equation:

P=χeε0E

Where P is the polarization, χe is the susceptibility, and E is the electric field. We can then use the fact that the bound charge is given by:

Qbound=∫P⋅dV

Where dV is the differential volume. Substituting the equation for P into this, we get:

Qbound=∫(χeε0E)⋅dV

Since the electric field is given by E=Q/