Gauss' Law - find magnitude of the electric field

In summary: It will ask you for all the information needed for us to help you.In summary, we have a slab of insulating material with thickness 2d and parallel faces oriented in the yz-plane. The slab has a uniform positive charge density ρ and the y- and z-dimensions are much larger than d. We need to find the magnitude of the electric field at points 0≤x≤d and x≥d. Using Gauss's Law, we can determine the charge enclosed within a Gaussian surface and use that to calculate the electric field. We can use a cylinder or box as our Gaussian surface, as long as we account for the vector dot product in the equation. For the second part, we can use
  • #1
NickPA
4
0
A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x=−d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density ρ.

find magnitude of the electric field due to the slab at the points 0≤x ≤d.

and magnitude of the electric field due to the slab at the points x≥d


Equations: Gauss's Law (too hard to type)



So i was working through this and am stuck about the area and charge. I doesn't give charge and am i just to assume that the height of the cylinder i choose is just arbitrary z?
 
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  • #2
NickPA said:
A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x=−d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density ρ.

find magnitude of the electric field due to the slab at the points 0≤x ≤d.

and magnitude of the electric field due to the slab at the points x≥dEquations: Gauss's Law (too hard to type)

For clarity, Gauss' Law is

[tex] \oint \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

So i was working through this and am stuck about the area and charge. I doesn't give charge

The charge enclosed within the Gaussian surface is the volume of the Gaussian surface times the charge density, ρ, for the situation where the Gaussian surface is completely within the material. For the situation where one end of the Gaussian surface sticks out of the material, the volume in question is intersection of the Gaussian surface and the material (the charge is the volume of material that happens to be within the Gaussian surface, multiplied by ρ).

By the way, you could use a cylinder for your Gaussian surface. But you could also use a "box" for your Gaussian surface if you would rather stick with Cartesian coordinates. It's totally up to you. (Either way is fine. It won't affect the final answer).

Keep in mind the vector dot product in the left hand side of the equation. Only certain sides of the cylinder or box will have a non-zero [itex] \vec E \cdot \vec{dA} [/itex] product. Those sides of the cylinder/box that have [itex] \vec E [/itex] perpendicular to [itex] \vec{dA} [/itex] do not contribute, and you can ignore them. [Edit: you may also ignore any sides of the cylinder/box that have zero [itex] \vec E [/itex] at those particular sides.]

and am i just to assume that the height of the cylinder i choose is just arbitrary z?

I think you mean, 'x'. (At least for the first part of the problem, where 0≤x ≤d.)
 
  • #3
Put a rectangular gaussian box with the left side at x < -d and the right at 0 < x < d. You know the E field at x < -d and you know the contained charge. Only E flux is out of the side at 0 < x < d. That gives you E(0<x<d).

Part b is obvious if you figured out the E field at x < -d.
 
  • #4
rude man said:
Put a rectangular gaussian box with the left side at x < -d and the right at 0 < x < d. You know the E field at x < -d and you know the contained charge. Only E flux is out of the side at 0 < x < d. That gives you E(0<x<d).
It's even easier if you put one side of the Gaussian box at x = 0, invoking symmetry to determine the E at x = 0, then let the other side vary between 0 and an arbitrary x. But that way still works too. Either way. :smile:
 
  • #5
collinsmark said:
It's even easier if you put one side of the Gaussian box at x = 0, invoking symmetry to determine the E at x = 0, then let the other side vary between 0 and an arbitrary x. But that way still works too. Either way. :smile:

Quite so, cm. Quite so! :thumbs:
 
  • #6
Can someone help me to solve this question:
.Three concentric hollow metallic spherical shells of radii r1,r2 and
r3, where r1 < r2 < r3, carry charges +2Q,−3Q and +Q, respectively.
Determine the charge on the inner and the outer surface of each sphere
 
  • #7
sarah2201 said:
Can someone help me to solve this question:
.Three concentric hollow metallic spherical shells of radii r1,r2 and
r3, where r1 < r2 < r3, carry charges +2Q,−3Q and +Q, respectively.
Determine the charge on the inner and the outer surface of each sphere
It's best to start a new thread to post your question.

When doing that, you should use the template that will be provided for you.
 

Related to Gauss' Law - find magnitude of the electric field

1. What is Gauss' Law and why is it important in the study of electric fields?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net charge enclosed by that surface. It is important because it allows us to calculate the electric field at a point due to a distribution of charges, without needing to know the individual contributions of each charge.

2. How is Gauss' Law related to Coulomb's Law?

Gauss' Law is a more general form of Coulomb's Law. While Coulomb's Law only applies to point charges, Gauss' Law can be used to calculate the electric field for any charge distribution.

3. What is the equation for Gauss' Law and how is it used to find the magnitude of the electric field?

The equation for Gauss' Law is ΦE = ∫S E ⋅ dA = Qenc0, where ΦE is the electric flux through a closed surface S, E is the electric field, dA is a differential area element, Qenc is the net charge enclosed by the surface, and ε0 is the permittivity of free space. To find the magnitude of the electric field at a point, we use the equation E = ΦE/A, where A is the area of the surface and E is the magnitude of the electric field.

4. What are the assumptions made in using Gauss' Law to find the electric field?

The main assumptions made in using Gauss' Law to find the electric field are that the charge distribution is static (not changing with time) and that the electric field is constant over the surface being considered. Additionally, the surface must be a closed surface and the charge enclosed must be stationary.

5. Can Gauss' Law be applied to any charge distribution?

Yes, Gauss' Law can be applied to any charge distribution, as long as the assumptions mentioned in the previous question are met. It is a powerful tool for calculating the electric field and is used extensively in the study of electromagnetism.

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