Electric Flux through a Gaussian Surface

AI Thread Summary
The discussion centers on calculating the electric flux through a Gaussian surface just inside a hollow conducting sphere with a uniform surface charge density and a charge placed inside its cavity. The initial misunderstanding arises from incorrectly considering both the internal charge and the induced charge when determining the net enclosed charge. It is clarified that only the charge -q inside the cavity contributes to the Gaussian surface's enclosed charge, leading to a net flux of q/ε. The ambiguity in the term "just inside" is addressed, emphasizing that it refers to the cavity rather than the conducting material itself. Ultimately, the correct interpretation leads to understanding that the electric flux is indeed q/ε, aligning with Gauss's law.
phisci
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Hi all, I need your help with a question. Here goes,

Homework Statement



There is a hollow conducting sphere with a uniform surface charge density of +\sigma. A charge -q, is placed inside the cavity of the hollow sphere. What is the electric flux through a spherical surface just inside the inner surface of the sphere?



Homework Equations


Gaussian sphere.


The Attempt at a Solution


From my understanding, for a conductor with uniform surface charge density, there is no electric field inside the conductor, assuming electrostatic situation. When a charge -q is placed inside the cavity of the hollow sphere, it will induce positive charges amounting to +q around the cavity to cancel out the effect of -q and hence the net electric field within the conductor is zero. If I consider a Gaussian sphere just inside the inner surface of the sphere, due to symmetrical properties of the sphere, the electric flux \phi= Qenc/\epsilonSince the enclosed net charge is the charge -q and the induced charge around the cavity is +q, the net Qenc is 0, hence the electric flux is 0.

However the answer to this question is q/\epsilon Where have I gone wrong in my reasoning? Thank you so much!

Sorry if the post seems messy, I'm not too sure how to post the Greek Symbols with the subscripts. The 0 should be a subscript 0 for epsilon. My apologies again.
 
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phisci said:
If I consider a Gaussian sphere just inside the inner surface of the sphere, due to symmetrical properties of the sphere, the electric flux \phi= Qenc/\epsilon
That's Gauss's law.

Since the enclosed net charge is the charge -q
Right. Stop right there.
and the induced charge around the cavity is +q,
True, but not relevant.
the net Qenc is 0, hence the electric flux is 0.
Why are you adding those two charges? The only charge enclosed by the Gaussian surface is -q. The induced surface charge is outside the Gaussian surface, so it doesn't contribute to Qenc.

I suspect the problem is that you are interpreting 'just inside the inner surface' as being within the conducting material. But they don't mean that. If the inner radius of the hollow conductor is R, then the radius of your Gaussian surface should be a little less, R - δ.
 
So my Gaussian sphere should be the sphere enclosing -q? inside the cavity?

What would then be the net flux if the Gaussian Sphere is enclosing the entire cavity. The Gaussian sphere having a radius of R+\delta? Do I need to consider the induced charges around the cavity now? Thanks for the quick reply.
 
Last edited:
phisci said:
So my Gaussian sphere should be the sphere enclosing -q? inside the cavity?
Yes.
phisci said:
What is the electric flux through a spherical surface just inside the inner surface of the sphere?
By 'just inside' they mean within the cavity. But the language is ambiguous and should have been clearer.
However the answer to this question is q/\epsilon
I'd say that answer has the wrong sign.
 
phisci said:
What would then be the net flux if the Gaussian Sphere is enclosing the entire cavity. The Gaussian sphere having a radius of R+\delta? Do I need to consider the induced charges around the cavity now?
In that case you must include all the enclosed charge include the charge on the inner surface. Just as you did in your first post. The flux through that Gaussian surface will be zero. Of course, this should require no calculation if you recall that the electrostatic field within a conductor must be zero.
 
Ah, i understand now. Thank you so much for your time and your quick response, appreciate it! :)
 
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