# Conducting Sphere shielding an electric field

wil3
Conducting Sphere "shielding" an electric field

Hello. I have a question regarding a problem that I have seen printed many times in various books and which I saw on a test today:

A hollow conducting sphere contains 2 charges, $$q_1\;and\;q_2$$. Outside the sphere are two additional charges, $$q_3\;and\;q_4$$. Which charges determine the properties of the field within the sphere?

The answer, as is often printed, is that just the first two charges contribute to the field.

The traditional explanation of this answer involves Gauss's Law-- only the first two charges are enclosed within the conducting sphere, and so the divergence of the field there thus only relies upon those two sources.

After reflection, I recently realized that this cannot possibly be correct, since equal-field Gaussian surfaces can be arbitrarily produced.

What is the real reason for this? I would prefer a more intuitive, qualitative reason, although math is pretty cool too. I'm guessing that the specific way in which $$q_3\;and\;q_4$$ redistribute the charge on the conducting sphere somehow results in a null field within, but I can't for the life of me figure out how the fact that the sphere is conducting allows for shielding.

Thanks in advance

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## Answers and Replies

wil3

NVM, guys, I figured it out. I'll post it here for the sake of people who later find this thread:

Okay, so lets say I have one really big charge +Q acting on a conducting sphere- because the sphere *conducts* it will push all the positive charge away to the far side of the sphere. Let's think about the sphere now- all the positive charge has been pushed to one side, and all the negative charge stayed put. Thus there's a net electric field within the sphere, induced for the nearby charge.

But hey look! The field induced by the rearrangement of charge on the shell of the sphere is pointing in the OPPOSITE direction as the field due to the original charge, but we still have to include that original +Q field (ignoring the sphere) which we add to the secondary field that is the result of charge rearrangement on the sphere surface.

It just so happens that the vectors sum to zero everywhere within the aegis of the sphere. This is why Gauss's law is able to solve the problem; we can treat the interior as an isolated system electrically, since the field within is the same regardless of whether there is charge outside or not. If there is charge outside, it induces a field within the sphere that cancels itself out.

Curl

You can also show this analytically using the mathematical description of the electric field along with Stokes' Theorem.

wil3

As username Curl, I feel like I shouldn't be surprised that you recommend Stokes' Theorem.

How would I go about doing this? I know that the curl is zero everywhere because of conservative field properties, etc, etc.

Curl

I'm sorry, I meant the divergence theorem.

Write a vector field that represents the E field due to a point charge at the origin. Enclose it with any closed, oriented surface (piecewise smooth, etc etc) and compute the net flux through the surface, and you'll see it is just equal to the charge enclosed.

From the divergence theorem, you will see that the volume integral is the same regardless of the surface since div E times the volume elements add up to a constant (maxwell's first equation).

Thus if the charge is outside, there is no overall flux through the surface.

Also I'll add for reference that since the surface is conducting, it must represent an constant-potential surface.

wil3

No, I totally agree that, by Gauss's law (and thus the Divergence theorem), the net flux through a closed surface which contains no sinks or sources of a conservative field (aka, satisfies the Laplace equation) must be zero.

My problem with this is that this didn't adequately show me that the field is zero, since the divergence theorem would confound the electric field with the differential area element (making it very difficult to solve for a function for the field, which would be necessary to answer this question.)

For example, if the sphere was non-conducting, blindly applying the divergence theorum would be errant- while the flux integral would come out to null over the closed surface, the field within would definitely be affected by the external charges. The question was more about the mechanism by which conductors shield.

wil3

Also, I didn't know that about conductors- It makes sense now that you mention it, but that's still very cool. After all, there cannot be a net field on a conducting surface, since that would cause a voltage by the basic Q=Ed relation. Thanks, Curl!