Gauss's Law with non-uniform E-field

[Doc Al]

Sorry at the end i meant

$$\frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{aR} \pi R^2}$$

Like I evalutated the non-uniform density at R for e^ar

What's that a calculation of? And why would your answer have Q in it--you are given the charge density as a function of a and r. I would not accept an answer that had Q in it.

 

[flyingpig]

This stuff $$\frac{[2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{aR} \pi R^2$$

is supposed to cancel itself out (dimension analytically)

 

[flyingpig]

Just wondering, does that mean part ii will be the same answer?

 

[Doc Al]

Just wondering, does that mean part ii will be the same answer?

No. Why would you think that?

 

[flyingpig]

Because the gaussian surface is now bigger...?

 

[flyingpig]

Actually a better question would be, why wouldn't it be different?

 

[Doc Al]

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.

 

[flyingpig]

Why? Shouldn't it get weaker?

 

[flyingpig]

Someone come back

 

[Doc Al]

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.)

 

[flyingpig]

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

$$\rho = \frac{Q}{V}$$

Volume can't really change so

$$\rho\; \alpha \;Q$$

so as radius goes up, density goes down and since they are proportional, charge drops?

 

[Doc Al]

Well the density is non-uniform

$$\rho = \frac{Q}{V}$$

Volume can't really change so

$$\rho\; \alpha \;Q$$

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.

 

[flyingpig]

$$Q_{en} = \int \rho dV$$

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

$$Q_{en} = \int_{R}^{r} \rho dV$$

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

 

[Doc Al]

I think it would be

$$Q_{en} = \int_{R}^{r} \rho dV$$

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):

$$Q_{en} = \int_{0}^{R} \rho dV$$

 

[flyingpig]

But $$Q_{en} = \int_{0}^{R} \rho dV$$ is for a Gaussian surface that has radius R.

WOuld it be...

$$Q_{en} = \int_{0}^{R} \rho dV + \int_{R}^{r} \rho dV$$

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.

 

[flyingpig]

wait would the integral change if the density was uniform?

 

[flyingpig]

Actually I should show you the final answer first to repay your time you put into helping me

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

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

$$\vec{E} = \frac{e^{aR} (aR - 1) + 1}{\epsilon_0ra^2} \hat{r}$$