Homework Help: Help with Maxwell’s equation in integral form

1. Mar 12, 2005

robert25pl

Electric field due to a cylindrical charge distribution using Gauss' law.
Charge is distributed with density $$\rho_{0}e^{-r^{2}}$$ C/m^3 in cylindrical region r < 1. Find D (displacement flux density vector) everywhere.

I did used this equation
$$\int_{s}D\cdot\,dS=\int_{V}\rho\*d\upsilon$$

Since this is a cylindrical charge distribiution I used Gaussian surface in the shape of a cylinder.

$$\int_{s}D\cdot\,dS=\rho\*l$$

So if I understand good the D=0 inside cylinder. therefore r>R is valid.
The surface area is $$2\pi\*rL$$.

I'm having a problem to set up the the equation or I'm doing everything wrong?
Thanks for any help and recommendation.

2. Mar 12, 2005

dextercioby

1.U need to evaluate the charge...What's the charge in the cylinder...?

Daniel.

3. Mar 12, 2005

robert25pl

$$\int_{s}D\cdot\,dS=\int_{V}\rho\*d\upsilon$$
So fom this I need to evaluate volume integral?

$$\int_{s}D\cdot\,dS=Q_{V}$$

$$Q_{V}=\int_{V}\rho_{0}e^{-r^{2}}$$

But what limits should I used for x,y,z (0 and 1 for all three)?
Thanks

4. Mar 12, 2005

dextercioby

Nope,you need the cilindrical coordinates $$r,\varphi,z$$.What's the volume element in cilindrical coordinates...?

Daniel.

5. Mar 12, 2005

robert25pl

$$Q_{V}=\int_{V}\rho_{0}e^{-r^{2}}$$

$$Q_{V}=\int_{r=0}^{r}\int_{\phi=0}^{2\pi}\int_{z=0}^{l}\rho_{0}e^{-r^{2}}dr\,d\phi\,dz$$

Is that correct?

6. Mar 12, 2005

dextercioby

U need another "r" in the volume element.And the cilinder has radius 1 (see text of the problem)...

Daniel.

7. Mar 12, 2005

robert25pl

$$Q_{V}=\int_{r=0}^{1}\int_{\phi=0}^{2\pi}\int_{z=0}^{l}\rho_{0}e^{-r^{2}}r dr\,d\phi\,dz$$

That is not easy integration, but I think I got it.

$$Q_{V}=l\pi\rho_{0}(e-1)e^{-1}}$$

Is this is correct what would be next step?

8. Mar 12, 2005

dextercioby

It is correct.Now u have to apply Guass' theorem which gives the flux of the induction $\vec{D}$...

Daniel.

9. Mar 12, 2005

robert25pl

if I understand well D depend on r only, so:

$$\int_{s}D\cdot\,dS=\int_{\phi=0}^{2\pi}\int_{z=0}^{l}D_{r}(r)a_{r}\cdot{r}\,d\phi\,dz\,a_{r}=$$

$$=2\pi\,rlD_{r}(r)$$

And this I should compare to Qv and find D, right?

10. Mar 12, 2005

dextercioby

It's the same cyclinder of radius unity...That "r" is 1...

Daniel.

11. Mar 12, 2005

robert25pl

So this is my D

$$D=\frac{\rho_{0}(e-1)e^{-1}}{2r} a_{r}$$

I really appreciate your help

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