- #36

flyingpig

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Actually a better question would be,

*why wouldn't it be different?*You are using an out of date browser. It may not display this or other websites correctly.

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- 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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- #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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Shouldn'tWhy? Shouldn't it get weaker?

- #41

flyingpig

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Shouldn'twhatget 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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I have no idea what you're talking about.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?

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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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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No. The relevant charge isI think it would be

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

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

- #45

flyingpig

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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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[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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