Gauss's Law with non-uniform E-field

  • Thread starter flyingpig
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  • #36
flyingpig
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Actually a better question would be, why wouldn't it be different?
 
  • #37
Doc Al
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When inside the charged sphere, as r changes so does the total charge contained within r. But once you are outside of the sphere, the total charge remains fixed.
 
  • #38
flyingpig
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Why? Shouldn't it get weaker?
 
  • #39
flyingpig
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Someone come back
 
  • #40
Doc Al
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Why? Shouldn't it get weaker?
Shouldn't what get weaker? (If you're responding to my last post, realize that I was talking about charge not field strength.)
 
  • #41
flyingpig
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Shouldn't what get weaker? (If you're responding to my last post, realize that I was talking about charge not field strength.)

Well the density is non-uniform

[tex]\rho = \frac{Q}{V}[/tex]

Volume can't really change so

[tex]\rho\; \alpha \;Q[/tex]

so as radius goes up, density goes down and since they are proportional, charge drops?
 
  • #42
Doc Al
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Well the density is non-uniform

[tex]\rho = \frac{Q}{V}[/tex]

Volume can't really change so

[tex]\rho\; \alpha \;Q[/tex]

so as radius goes up, density goes down and since they are proportional, charge drops?
I have no idea what you're talking about.

Within the sphere, obviously the total charge within your Gaussian sphere increases with radius; Outside the sphere, there's no additional charge so the total charge remains fixed.
 
  • #43
flyingpig
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[tex]Q_{en} = \int \rho dV[/tex]

I am going to take a while shot at this one. Since we are looking at r > R, that means we only care about the field r > R

I think it would be

[tex]Q_{en} = \int_{R}^{r} \rho dV[/tex]

But r > R not greater or EQUAL to. In other words, even though our Gaussian surface encloses the whole surface, it only cares about outside of r > R
 
  • #44
Doc Al
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I think it would be

[tex]Q_{en} = \int_{R}^{r} \rho dV[/tex]
No. The relevant charge is everything contained within the Gaussian surface. Since the charge only extends to R, we have (for a Gaussian surface with r > R):

[tex]Q_{en} = \int_{0}^{R} \rho dV[/tex]
 
  • #45
flyingpig
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But [tex]Q_{en} = \int_{0}^{R} \rho dV[/tex] is for a Gaussian surface that has radius R.

WOuld it be...

[tex]Q_{en} = \int_{0}^{R} \rho dV + \int_{R}^{r} \rho dV[/tex]

That includes inside and outside!?

EDIT: wait, that's no different from my first integral then...

EDIT2: nvm, you are right. My other integral encloses no charge because from R to r, it is just space. I keep thinking my integral is summing field, not charge.
 
  • #46
flyingpig
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wait would the integral change if the density was uniform?
 
  • #47
flyingpig
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Actually I should show you the final answer first to repay your time you put into helping me

[tex]2\pi l\int_{0}^{R} re^{ar} dr = 2\pi l \frac{e^{aR} (aR - 1) + 1}{a^2}[/tex]

[tex]\oint \vec{E} \cdot d\vec{A} = 2\pi l \frac{e^{aR} (aR - 1) + 1}{\epsilon_0 a^2}[/tex]

[tex]\vec{E} = \frac{e^{aR} (aR - 1) + 1}{\epsilon_0ra^2} \hat{r}[/tex]
 

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