Solving for Electric Field with Gauss' Law

AI Thread Summary
To find the electric field using Gauss' law in differential form with a charge density of ρ = ρ₀r³ in spherical and cylindrical coordinates, the discussion emphasizes applying Gauss' law effectively. The approach involves calculating the enclosed charge (Q_encl) for different regions defined by the radius (r) and using the relationship E = D/ε. For spherical coordinates, the total charge within a sphere of radius a is given, and for r > a, the enclosed charge can be calculated using integration. In cylindrical coordinates, the electric flux is expressed as E multiplied by the surface area, leading to a similar calculation for the electric field. The conversation concludes that solving for Q and substituting into the equations will yield the electric field.
robert25pl
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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
 
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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
 
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So for 0\leq r\leq a Qencl=0

for a &lt; 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
 
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It's E 4 \pi r^2 =~electric~flux~=~EA= \frac{q}{\epsilon_0}


I'm sure you can take it from here
 
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
 
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