Electric Flux through a Gaussian Surface

[phisci]

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 $$\phi$$= Q_enc/$$\epsilon$$Since the enclosed net charge is the charge -q and the induced charge around the cavity is +q, the net Q_enc 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.}

 

[Doc Al]

If I consider a Gaussian sphere just inside the inner surface of the sphere, due to symmetrical properties of the sphere, the electric flux $$\phi$$= Q_enc/$$\epsilon$$

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 Q_enc 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 Q_enc.

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.

 

[phisci]

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.

 

[Doc Al]

So my Gaussian sphere should be the sphere enclosing -q? inside the cavity?

Yes.

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.

 

[Doc Al]

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.

 

[phisci]

Ah, i understand now. Thank you so much for your time and your quick response, appreciate it! :)