Find Free Charge Enclosed by Sphere of Radius R

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SUMMARY

The discussion centers on calculating the free charge enclosed by a sphere of radius R centered at a point charge Q embedded in a dielectric medium with relative permittivity ε_r. The key equations referenced include Gauss' Law, specifically ∇ · D = ρ_f and ∮ D · dA = Q_free,enclosed. The participants conclude that the only free charge present is Q itself, as the dielectric does not contribute additional free charge, and any induced charge is bound, not free.

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  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric displacement field (D) and its relation to free charge
  • Knowledge of the distinction between free charge and bound charge
  • Concept of dielectric materials and their properties
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zephyr5050
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Homework Statement



A point charge ##Q## is embedded in a dielectric medium with relative permittivity ##\epsilon_r##. What is the free charge enclosed by a sphere of radius ##R## centered at ##Q##.

Homework Equations



##\nabla \cdot \mathbf D = \rho_f##

##\oint \mathbf D \cdot d \mathbf A = Q_{free,enclosed}##

The Attempt at a Solution



My initial attempt would be to use ##\nabla \cdot \mathbf D = \rho_f##. For a single charge ##Q## at the origin, the displacement field should be

##\mathbf D = \frac{Q}{4 \pi r^2} \hat{\mathbf r}##

which can be found from Gauss' Law. The only issue is, that requires knowing the free charge enclosed in the gaussian surface, which is what the problem asks for to begin with. It's completely circular. I just can't figure a way to get the free charge enclosed in a sphere, especially without any information about electric fields or electric potentials. Any guidance on how to approach this problem?
 
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hi zephyr5050! :smile:

hint: what is the total charge enclosed by the sphere of radius R ? :wink:
 
tiny-tim said:
hi zephyr5050! :smile:

hint: what is the total charge enclosed by the sphere of radius R ? :wink:

Hello tiny-tim. Thanks for the response.

As it turns out, that is part c of the problem, after asking what the bound charge density is. Obviously the total charge enclosed should be ##Q## plus any other free charge (although I don't believe there is one) and whatever bound charge density is induced. Conceptually I know that a bound charge density is induced around the charge ##Q##, but I still don't see how I can use the total charge enclosed without also knowing the bound charge density.
 
To help answer the question, how do you distinguish between "free" and "bound" charge? Is Q a free charge? Does the dielectric material itself have any free charge?
 
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TSny said:
To help answer the question, how do you distinguish between "free" and "bound" charge?

Well, the free charge would be anything which can move freely around the system, whereas the bound charge is any charge confined to an atom, but which is creating an electric field due to the fact that it has been pulled away from the rest of it's atom.

TSny said:
Is Q a free charge?

I believe it should be considering it is not confined to an atom. And as far as I can reason, it should be the only free charge.

TSny said:
Does the dielectric material itself have any free charge?

Well obviously not. I can certainly qualitatively reason that the only free charge should be that which is given by the problem. That charge can't induce any other free charge in the dielectric. It can merely polarize the dielectric and create a bound charge near the free charge. But that doesn't seem to bring me any close to quantitatively calculating the free charge.
 
zephyr5050 said:
I can certainly qualitatively reason that the only free charge should be that which is given by the problem. That charge can't induce any other free charge in the dielectric. It can merely polarize the dielectric and create a bound charge near the free charge.

Yes. Good.

But that doesn't seem to bring me any close to quantitatively calculating the free charge.

I would think it does. :redface:
 
But that seems absurdly simple. I don't even have to do any math for that. I suppose that is the answer, but it just feels like its not enough to satisfactorily answer the question.
 
Yes, it's simple. But it is a check on your understanding.
 
Alright, thanks for your help TSny, and you too tiny-tim.
 

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