When to Integrate Charge Enclosed for Gaussian Surfaces?

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  • Thread starter Sunbodi
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Hello,

I was looking over my notes and I was trying to figure out when we integrate Q enclosed when Q = ρ*d(volume).

If there's one thing I've learned from physics II you only integrate when a field is non-uniform. I'm just wondering how we know when it's uniform (usually the problem will tell you but sometimes it doesn't).

In my notes for example, my professor did an example where we found Q enclosed of an insulated sphere with a radial charge density of ρ and radius a. It was surrounded by a thick conducting shell of inner radius b, outer radius c, and total charge Q. We then were asked to find q where r is less than a. In this part, we integrated ρ*d(volume).

Why is it for that problem we integrated but when I look over one of our past problems we didn't integrate:

A plasma, a hot gas of positively-charged ions, is confined to a very long, thin conducting cylindrical shell of inner radius a and outer radius b. The plasma within this conducting tube has a uniform volume charge density ρ0. (a) In terms of the variables of the problem (e.g., ρ0, a, b, ǫ0, and r), what is the field in the plasma tube (r < a) at some distance r from the axial center of the plasma tube?



Please don't solve the problem unless you're bored and want to. I'm just trying to figure out why we integrate in the first problem but not the second.


Lastly Thanks!

Really appreciate your help! :)
 

Answers and Replies

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This will sound like a tautology, but the short answer is you integrate when you can't think of a way not to. I'm not trying to be flip. I think at a very fundamental level that is the answer you are seeking.

Another way to put it: if you are trying to sum up a total amount (or a subset of the total) of something then integrating is the default. To avoid integrating you look at symmetry or note uniformities that let you reason the answer without integrating. Can't spot an obvious solution? Then integrate.

The most common feature that avoids integration is when the functional form is constant over one or more dimensions. (Eliminating having to do one or more integrals is often what determines the best coordinate system for the integration)

Suppose the charge density had been uniform in the inner sphere. Since the charge density is constant, you know that the charge enclosed by some volume is the charge density times the volume with no need to integrate. That's the same answer you would have gotten had you integrated, but the integral would have been trivial. In this case you have a radially varying charge density. How can you determine the charge enclosed in a sphere of some radius R less than a? You don't have a constant density, so you can't just multiply by the volume. You have no choice but to integrate over r. But why just over r? If it hadn't been radially symmetric you would have had to integrate over the angular dimensions, too. Noting that the charge density was radially symmetric allowed you to skip two dimensions of integration and integrate spherical shells in r only. The other two integrations are in there, but you just happen to know the answer.

In a sense you are always integrating in all dimensions. You often don't even notice that symmetry is letting you skip right to the answer of one or two of the dimensions. Integrating spherical shells just seems like the thing to do!

Some other features that allow you to skip integration are antisymmetry making the integral obviously zero, periodicity making the integral a sum, or perhaps zero by inspection. However the usual case is that the function is constant over one or more dimensions.
 

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