Gauss's Law: a nonconductive wall

In summary, the conversation discusses the use of Gauss's Law and the equation for the electric field to calculate the electric field around a nonconducting wall with a uniform charge density. The key to the solution is to use a Gaussian cylindrical surface, and the distance from the wall does not affect the result. The conversation also addresses the concept of flux and surface charge density, and how they are related to the electric field. In real-life situations, the electric field around a finite plane is not constant and is only a good approximation for points close to the plane. Lastly, the conversation clarifies some misunderstandings and corrects some mistakes made in the calculations.
  • #1
Breedlove
27
0

Homework Statement



A nonconducting wall carries charge with a uniform denisty of 8.g µC/cm2. What is the electric field 7.00cm in front of the wall? Explain whether your result changes as the distance from the wall is varied.

Homework Equations



I'm still really new, I just registered, and trying to find the integral sign and stuff is proving to be pretty difficult
[tex]\int[/tex]EdotdA=Qenclosed/A

E=k[tex]\frac{q}{r2}r[/tex]unit vector

So Gauss's Law and the equation for the electric field, I think, are relevant. The relationship [tex]\sigma[/tex]=Q/A I think is important too. Let's throw in Coulomb's Law for good measure.

F=kqq0/(r2)r unit vector

The Attempt at a Solution



Okay, so from my understanding of Gouss's Law, you can use it to calculate the electric field around a closed symmetrical surface that surrounds the thing you want to find the electric field of. Rather, Gauss's Law is a way of counting up the electric field lines. I think I understand how to use Gauss's Law to find the electric field at a point around a sphere of charge; you can just make a Gaussian Sphere around it.

Anyway, I guess my major source of confusion is that you can't really encapsulate a wall, like if it was a rod, or if we could interpret it as a long line of charge, then we could put a cylinder around it. I think I gathered from another post that
E=[tex]\sigma[/tex]/(2[tex]\epsilon[/tex]0)

I think that this is the key, but I don't know how this came about. How can this be derived from Gauss's Law or the equation for electric field?

I'm generally pretty confused, and I think I'm going to try to firm up my understanding of applying Gauss's law while I hope that someone replies.
 
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  • #2
Welcome to PF,

Perhaps this will help?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1

I think the key is to what distance your Gaussian cylindrical surface can be taken and how close to en edge you are at such that the walls of the cylinder can be taken not to have any contribution to flux through the end caps.

For instance at 100 m from the wall? How big is the wall?
 
  • #3
That link was super helpful! I now understand why the flux = E2A, but am still a little fuzzy how that equals [tex]\sigma[/tex]A/[tex]\epsilon[/tex]0 I was thinking because flux=Qenclosed/[tex]\epsilon[/tex]0, and maybe you could say that the charge density is equal to the charge enclosed? Am I getting the ideas mixed up? Because if Qenclosed = [tex]\sigma[/tex], then you could say
E=[tex]\sigma[/tex]/[tex]\epsilon[/tex]02A

and since the integral of dA=1, it would be then simplified to

E=[tex]\sigma[/tex]/2[tex]\epsilon[/tex]0

Okay... yay! I'm pretty happy now. Sweet! Excellent! Am I right? Is my thought process right? Is it okay to assume that when they are talking about charge density they are talking about Qenclosed?

Oh, and they didn't give me any dimensions of the wall, I think the wall is highly similar to the flat sheet of charge depicted in the picture in the link. From this calculation, I think, the distance from the wall doesn't matter.

Bah, that seems weird to me. I mean, what about the inverse square relationship? Like, do things really give off fields that are not dependent upon distance? Like field lines, shouldn't field lines be tapered or something? Maybe my understanding of field lines is off.

Thank you so much for the help with the problem! Thanksthanksthanks :)
 
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  • #4
Breedlove said:
That link was super helpful! I now understand why the flux = E2A, but am still a little fuzzy how that equals [tex]\sigma[/tex]A/[tex]\epsilon[/tex]0 I was thinking because flux=Qenclosed/[tex]\epsilon[/tex]0, and maybe you could say that the charge density is equal to the charge enclosed? Am I getting the ideas mixed up? Because if Qenclosed = [tex]\sigma[/tex],

Close but not quite. [itex] \sigma [/itex] is a surface charge density so it must be multiplied by an area to give a charge. So Q enclosed = [itex] \sigma A [/itex] where A is the area of the material that is enclosed by your gaussian surface

then you could say
E=[tex]\sigma[/tex]/[tex]\epsilon[/tex]02A

and since the integral of dA=1, it would be then simplified to

No, the integral of dA is just A, the area enclosed.

Now, the A will cancel out and you will recover the result you had before. It was correct but the two steps above were incorrect (basically, you made two small mistakes that canceled one another by pure luck)

To adress your other question: for an infinite plane surface (uniformly charged), the E field is indeed constant, it does not depend on the distance. But of course, this is an unphysical situation. In real life, planes are always finite in which case the answer you get is only approximative and is a good approximation only for points at distances much less than the dimensions of the plane (points where the plane effectively looks almost infinite)
 
  • #5
crap... the integral of dA is A. bahhh well...I think I was on the right track. woahhh I think I was thinking about that integral way incorrectly. It's the integral of E2AdA right, so that gives E2A2/2 so EA2. bahhh I'm back at square one. It's alright I think I've got some sort of handle on something or other :-/
 
  • #6
Bah! Willickers I forgot that [tex]\sigma[/tex]=Q/A ! Okay okay hmm okay. Sorry about my last response, I was working on it and wasn't paying attention when you replied. Thanks for catching my mistakes!

Woah! Okay it works now! disregard that crazy stuff with the integral bit, I don't know what that was.
 
  • #7
Breedlove said:
crap... the integral of dA is A. bahhh well...I think I was on the right track. woahhh I think I was thinking about that integral way incorrectly. It's the integral of E2AdA right, so that gives E2A2/2 so EA2. bahhh I'm back at square one. It's alright I think I've got some sort of handle on something or other :-/

No, it is the integral

[tex] \int \vec{E} \cdot \vec{dA} = \int E \cos \theta dA [/tex]

In your case, theta is 0 so you get ismply [itex] \int E dA = E \int dA [/itex]

You were completely correct except that you did not have the factors of A where they should have been. But they cancel out.
 
  • #8
That makes complete sense. I should've just looked and relied on the definitions instead of extrapolating weird junk, yeah that makes total sense now. Thanks so much!
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the distribution of electric charge to the resulting electric field. It states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface.

2. How does Gauss's Law apply to a nonconductive wall?

In the case of a nonconductive wall, Gauss's Law tells us that the electric field must be perpendicular to the surface of the wall. This means that the electric field lines will either be directed towards or away from the wall, depending on the distribution of charge on the other side of the wall.

3. Can Gauss's Law be used to calculate the electric field inside a nonconductive wall?

No, Gauss's Law only applies to the electric field at the surface of the wall. To calculate the electric field inside the wall, additional information and equations would be needed.

4. How is Gauss's Law useful in understanding the behavior of electric fields?

Gauss's Law allows us to determine the strength and direction of the electric field at a given point, based on the distribution of charge in the surrounding space. This is useful in predicting the behavior of electric fields and how they interact with different materials and objects.

5. Can Gauss's Law be applied to non-stationary electric fields?

Yes, Gauss's Law can be applied to both stationary and non-stationary (changing with time) electric fields. However, for non-stationary fields, additional equations and considerations must be taken into account.

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