From Charge Density -> Electric Field(r) using Gauss


by zSanityz
Tags: charge, density, electric, fieldr, gauss
zSanityz
zSanityz is offline
#1
Apr26-10, 12:33 PM
P: 13
1. The problem statement, all variables and given/known data
"Consider a charge density distribution in space given by [rho] = [rho]_0 * e^(-r/a), where [rho]_0 and a are constants. Using Gauss' Law, derive an expression for the electric field as a function of radial distance, r. Sketch the E vs. r graph.

Was a question on a quiz that I messed up, curious now on how to do it properly so I'm more prepared for the quiz, and teacher is out for awhile.


2. Relevant equations
These are the equations I know that seem relevant, probably not all useful though:
[rho] = q/V or [delta]q = [rho] * [delta]V
[del] dot E = [rho] / [epsilon]_0 or [line integral]E * [delta]A = Q/[epsilon]_0


3. The attempt at a solution
I have no idea how to start this thing, and what I should be pursuing. I know it should end up as some sort of hyperbole, since that's the relation between electric field and radial distance(?). Any showing of how to start this, or preferably the whole process in a somewhat understandable manor would be great, I will be checking the forums every couple minutes for conversation.

I'm also very unclear on del, I feel like that's the path to go but don't really understand how it works.

Thank you!
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elect_eng
elect_eng is offline
#2
Apr26-10, 02:43 PM
P: 370
Quote Quote by zSanityz View Post
1. The problem statement, all variables and given/known data
"Consider a charge density distribution in space given by [tex]\rho = \rho_0 e^{-r/a}[/tex], where [tex] \rho_0 [/tex] and [tex]a[/tex] are constants. Using Gauss' Law, derive an expression for the electric

field as a function of radial distance, [tex]r[/tex]. Sketch the [tex]E [/tex]vs. [tex]r[/tex] graph.
For any radius [tex]r [/tex], the charge density outside that radius will contribute nothing to the electric field [tex]E[/tex], while the charge inside that radius will act as a point source with total charge Q equal to the integrated charge density over the volume inside that radius. These are direct results of Gauss' Law and the spherical symmetry of the problem.
zSanityz
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#3
Apr26-10, 03:11 PM
P: 13
Quote Quote by elect_eng View Post
For any radius [tex]r [/tex], the charge density outside that radius will contribute nothing to the electric field [tex]E[/tex], while the charge inside that radius will act as a point source with total charge Q equal to the integrated charge density over the volume inside that radius. These are direct results of Gauss' Law and the spherical symmetry of the problem.
Sorry about the delay, was having some connection issues. :X

And that all does sound very familiar, not sure I completely grasped it all though, if someone doesn't mind going through it or doing a step or two in the right direction (I learn best through example) that might be a bigger help for me personally, or just a simpler explanation.

zSanityz
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#4
Apr26-10, 04:05 PM
P: 13

From Charge Density -> Electric Field(r) using Gauss


I played around with it algebraically and brought in Coulomb's Law, E = 1/(4*[pi]*[epsilon]_0)*q/r^2, and [rho] = q/V, and got [rho] = (E*4*[pi]*[epsilon]_0*r^2)/V, and using volume of a circle = 4/3[pi]r^3, I was able to eliminate a lot, ending up with [rho] = (3*E*[epsilon]_0)/r.

And bringing it back to the original equation and isolating E, you get E = ([rho]_0*r)/(3*[epsilon]_0*e^(r/a)), meaning the graph would be dictated by r*e^(-r) ?

Let me know if any of this is wrong, I feel like I just butchered this problem by oversimplifying it. :X
elect_eng
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#5
Apr26-10, 04:06 PM
P: 370
Quote Quote by zSanityz View Post
Sorry about the delay, was having some connection issues. :X

And that all does sound very familiar, not sure I completely grasped it all though, if someone doesn't mind going through it or doing a step or two in the right direction (I learn best through example) that might be a bigger help for me personally, or just a simpler explanation.
Sorry, if the explanation is not clear. Sometimes these ideas are easier to express mathematically. Try the following.

Express Gauss's Law as follows:

[tex]\oint \vec D \cdot d\vec s = \int \rho dv=Q[/tex]

Consider a homogeneous medium with spherically symmetric charge density, and choose the surface of integration to be a sphere with radius R.

[tex]E_r = {{\int \rho dv}\over{4 \pi \epsilon R^2}}[/tex]

Notice that the field at any radius only depends on the charge density that is inside that radius. This is because the effects of the outside charges perfectly cancel. You could prove this yourself by doing the integration over exterior spherical shells of charge, but Gauss' Law already tells you this directly. The logic is a little subtle, but it depends on the fact that there is spherical symmetry in this case.

So now you have a formula for which you can substitute in the exact function for the charge density. Just integrate it out and you will have the function E(R), which is really the E(r) you are looking for.
zSanityz
zSanityz is offline
#6
Apr26-10, 04:15 PM
P: 13
Quote Quote by elect_eng View Post
Sorry, if the explanation is not clear. Sometimes these ideas are easier to express mathematically. Try the following.

Express Gauss's Law as follows:

[tex]\oint \vec D \cdot d\vec s = \int \rho dv=Q[/tex]

Consider a homogeneous medium with spherically symmetric charge density, and choose the surface of integration to be a sphere with radius R.

[tex]E_r = {{\int \rho dv}\over{4 \pi \epsilon R^2}}[/tex]

Notice that the field at any radius only depends on the charge density that is inside that radius. This is because the effects of the outside charges perfectly cancel. You could prove this yourself by doing the integration over exterior spherical shells of charge, but Gauss' Law already tells you this directly. The logic is a little subtle, but it depends on the fact that there is spherical symmetry in this case.

So now you have a formula for which you can substitute in the exact function for the charge density. Just integrate it out and you will have the function E(R), which is really the E(r) you are looking for.
Thank you! :D That's a much more understandable way to show it for me (math major).
I was looking over what you just explained and looking back at what I did the post beforehand and I think that's actually what I attempted to do, but I'm not 100% sure I did it all right, and is it okay to substitute in 4/3[pi]r^3 for V?
elect_eng
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#7
Apr26-10, 04:23 PM
P: 370
Quote Quote by zSanityz View Post
... is it okay to substitute in 4/3[pi]r^3 for V?
No, you have to do the integration out. If [tex]\rho[/tex] were constant then you could factor it out of the integral and use [tex]\rho V[/tex] with V being the volume of the chosen sphere. However, in this case [tex]\rho[/tex] is constant for changes in the [tex]\phi[/tex] and [tex]\theta[/tex] coordinates, but not the radial coordinate. Just plug into the formula for charge density and do the integral out slowly.
zSanityz
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#8
Apr26-10, 04:41 PM
P: 13
Quote Quote by elect_eng View Post
No, you have to do the integration out. If [tex]\rho[/tex] were constant then you could factor it out of the integral and use [tex]\rho V[/tex] with V being the volume of the chosen sphere. However, in this case [tex]\rho[/tex] is constant for changes in the [tex]\phi[/tex] and [tex]\theta[/tex] coordinates, but not the radial coordinate. Just plug into the formula for charge density and do the integral out slowly.
Ahh, alright, I'll try that! Thanks.

Though this is my worst subject in physics (though you can probably tell? :( ) and I'm REALLY bad at integrating in physics, great at it in math, but have trouble grasping it in physics for some reason. I also have to leave for a couple hours, so (if you don't mind!) could you leave the answer here so I could check my results after I get back and work through it.

And thanks for all the help on this, I hope my denseness in this area hasn't made it too difficult.

<3's
elect_eng
elect_eng is offline
#9
Apr26-10, 05:12 PM
P: 370
Quote Quote by zSanityz View Post
I also have to leave for a couple hours, so (if you don't mind!) could you leave the answer here so I could check my results after I get back and work through it.
Well, that assumes that I have actually worked it out, which I have not.

It's quite simple, but should be worked out slowly to avoid mistakes.

It's well in your wheelhouse if you are a math major.
zSanityz
zSanityz is offline
#10
Apr26-10, 05:23 PM
P: 13
Quote Quote by elect_eng View Post
Well, that assumes that I have actually worked it out, which I have not.

It's quite simple, but should be worked out slowly to avoid mistakes.

It's well in your wheelhouse if you are a math major.
Lol, well like I said, I haven't quite grasped integrating in physics. Math is easy since variables there don't usually actually stand for anything. I have made some mistakes in the past in this area, and usually learn from them best by seeing the correct answer and being able to compare mine to it, finding where my mistake was (and I never make that mistake again), which is why I asked for yours. :p

But since you don't have it, I understand if you don't feel like working through it for me (especially if it's long), but if you are up for it (or anyone else on this forum who happens to check out this thread) I would be very appreciative if you would post the final result (and any major step if you feel like that as well -though I can usually figure those out when I see the final result)

Thanks!! <3's


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