# Solving for Electric Field with Gauss' Law

• robert25pl
In summary, the conversation discusses finding the electric field using Gauss' law in differential form for a charge density of ρ = ρ0r^3 in spherical symmetry and then in cylindrical coordinates. The speaker suggests applying Gauss' law on a sphere and cylinder, and using the given charge density to solve for Qencl. They also provide an example for finding the electric field at different distances from the center of a sphere. Finally, the conversation touches on using D = εE to find E and solving for Q.
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

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 $$a < r \leq b$$

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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## 1. What is Gauss' Law and how is it used to solve for electric field?

Gauss' Law is a fundamental law in electromagnetism that relates electric fields to the distribution of electric charges. It states that the electric flux through a closed surface is proportional to the enclosed electric charge. This law can be used to solve for electric field by using the formula E = Q/ε0A, where Q is the enclosed charge, ε0 is the permittivity of free space, and A is the area of the closed surface.

## 2. What is the difference between using Gauss' Law and Coulomb's Law to solve for electric field?

Coulomb's Law is used to calculate the electric field at a specific point due to a point charge or a collection of point charges. Gauss' Law, on the other hand, allows us to calculate the electric field at any point in space due to a distribution of charges, without having to consider each individual charge. Gauss' Law is generally easier to use for more complex distributions of charges.

## 3. Can Gauss' Law be used for all types of charge distributions?

Yes, Gauss' Law can be used for any type of charge distribution, whether it is a point charge, a line of charge, a sheet of charge, or a more complex distribution. However, the shape of the closed surface used in the calculation may vary depending on the type of charge distribution.

## 4. How do we choose the appropriate closed surface for a Gauss' Law calculation?

The closed surface used in a Gauss' Law calculation should be chosen such that the electric field is constant and perpendicular to the surface. This makes the calculation simpler and ensures that the electric flux through the surface is proportional to the enclosed charge. The choice of surface also depends on the symmetry of the charge distribution.

## 5. Is it necessary to know the electric field in order to use Gauss' Law to solve for electric field?

No, it is not necessary to know the electric field beforehand. Gauss' Law can be used to find the electric field at a point even if the field is not known. This is because the electric flux through the closed surface is directly related to the enclosed charge, allowing us to solve for the electric field using the formula E = Q/ε0A.

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