Solving for Electric Field with Gauss' Law

[robert25pl]

I have to find the electric field everywhere using Gauss’ law in differential form. Charge density is $$\rho = \rho_{0}r^{3}$$ for a<r<b and 0 otherwise in spherical symmetry and then in cylindrical coordinates

$$\nabla \cdot D=\rho$$
I have look for D and then just get E = D/epsilon. D is where I need help. Thanks

 

[marlon]

Well, just apply Gauss' law on a sphere and on a cilinder. this should not be that difficult. You don't need this D. You will only need to be careful when the radial coordinate r is inside the sphere or cilinder. However, you have been given a charge density, so that is no problem.

What have you done so far ?

marlon

Just as an example : Suppose you have a sphere of radius a in which there is an uniform chargedensity. The total charge Q is then equal to $$\frac{4 \pi a^3}{3} \rho$$ Then Gauss's law says : $$E 4 \pi r^2 = \epsilon_0 Q$$

This yields the electric field at distance r from the center of the sphere. Keep in mind that r must be BIGGER then a in this case

 

[robert25pl]

So for $$0\leq r\leq a$$ Qencl=0

for $$a < r \leq b$$

Qencl = $$\int_{a}^{r} \rho dv$$


for r>b
Qencl = $$\int_{a}^{b} \rho dv$$
and I just have to solve that,
Is this ok?

so for cylindrical I have to used 2pi*r*L

 

[GCT]

It's $$E 4 \pi r^2 =~electric~flux~=~EA= \frac{q}{\epsilon_0}$$


I'm sure you can take it from here

 

[robert25pl]

To get E I just have to solve Q and substitute to the equation.
But isn't D = epsilon*E and I can get E from that