Finding displacement field D for exponential cylindrical charge distribution

[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.

 

[dextercioby]

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


Daniel.

 

[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

 

[dextercioby]

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

Daniel.

 

[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?

 

[dextercioby]

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

Daniel.

 

[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?

 

[dextercioby]

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

Daniel.

 

[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?

 

[dextercioby]

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

Daniel.

 

[robert25pl]

So this is my D

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

I really appreciate your help