Gauss' Law: Calculating charge enclosed

  • Thread starter henry3369
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  • #1
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Homework Statement


I've been doing a few Gauss' Law problems and I'm slightly confused about calculating charge enclosed by a nonconducting sphere.

So I have done 2 problems that involve finding the electric field inside nonconducting spheres:
1. Charge is uniformly distributed
2. Charge per unit volume given by the function ρ(r) = pr, where p is some constant.

I know how to solve both, but I'm not sure why I have to take different approaches to finding charge enclosed.

Homework Equations


EA = Qenc

The Attempt at a Solution


For (1), I was able to calculate Qenclosed using the fact that the charge density is Q/V, where V is the volume of the entire sphere, then multiplying it by the volume of the inner sphere, dV. Then I rearranged for E and got kQr/R3.

My question is, why can't I multiply the the volume of the inner sphere by the function ρ(r), to find the charge enclosed? Instead, I have to integrate ρ(r)*dV = ρ(r)*(4πr2) from 0 to r.

Both Q/V (1) and ρ(r) (2) provide the charge per unit volume, I don't understand why finding Qenclosed differs between the two.
 

Answers and Replies

  • #2
squelch
Gold Member
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In the first situation, the charge per unit area in each "shell" within the sphere is precisely the same as every other shell, because the charge density is uniform throughout. Since you know the charge density is uniform, if you know the volume of the sphere it's a very simple matter to get the total charge by multiplying the charge density by the volume of the sphere.

In the second situation the charge isn't uniformly distributed -- it's more like a gradient of charge. The charge is more densely concentrated on the outermost parts of the sphere and less densely concentrated on the center, so it's not as simple as considering the charge density to be some constant you can just multiply by the volume. The charge per unit area at each "shell" an infinitesimal width "dA" within the sphere, however, is constant, so you can multiply the charge per unit area by the area of the shell to find the charge contained in each infinitesimal shell. I hope that's fairly clear without the use of diagrams.

Knowing that, you'd need to sum the total charge enclosed on the area of each thin shell within the sphere to find the total charge enclosed within the volume of your sphere -- which requires you to use integration.

That's the difference, I hope it helps.
 

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