How to get zero electric field inside spherical shell without using Gauss law


by chemfirus
Tags: coulomb's law, electric field, gauss law, surface integral
chemfirus
chemfirus is offline
#1
Sep25-09, 05:31 PM
P: 3
Now, i am trying to prove that electric field inside spherical shell whose constant surface charge density [tex]\rho[/tex] is zero. Usually, it is proved by using Gauss law (there is no charge enclosed inside gaussian surface so there is no electric field inside the sphere). But, whe i am trying to solve schaum electromagnetics problem (actually, it is not my homework and i don't have to do it) no. 2.53 i was asked to prove it. Because in the chapter is not explained gauss law yet, i try to prove it. Next is my step:

1) I would like to prove it in spherical coordinate. So, i need to use next equation:
d2 = R12 + R22 - 2*R1*R2[cos[tex]\theta[/tex]1*cos[tex]\theta[/tex]2 + sin[tex]\theta[/tex]1*sin[tex]\theta[/tex]2*(cos([tex]\phi[/tex]1-[tex]\phi[/tex]2))]..............(1)
Equation (1) is used to find distance between two points (or vector), [tex]\vec{R1}[/tex] and [tex]\vec{R2}[/tex], in spherical coordinate. I would use it as R in coulomb's law.

2) The next step is to model the equation in coulomb's law:
([tex]\rho[/tex]/(4*[tex]\pi[/tex]*[tex]\epsilon[/tex]))*[tex]\oint[/tex](R12sin[tex]\theta[/tex]1/(R12 + R22 - 2*R1*R2*cos[tex]\theta1[/tex])) d[tex]\theta[/tex]1d[tex]\phi[/tex]1.......(2)
In equation (2), i have made [tex]\theta[/tex]2 = 0 because i put the evaluated point at [tex]\theta[/tex] = 0 (or simply at z axis in cartesian coordinate). R1 is distance between dS and centre point, R2 is distance between evaluated points with centre points. So, distance between dS and evaluated point got by using equation (1).

I have tried to integrate the equation (2) and i can't prove that the result is zero. I got a natural logaritmic equation at first integral (integrate with d[tex]\theta[/tex]1). When i substitute upper and lower limit with [tex]\pi[/tex] and 0 i don't get zero result. I don't try the second integral because i am not sure i would get the result i want.
Suddenly, i aware that i don't use unit vector yet in equation (2). It makes the distance not become (R12 + R22 - 2*R1*R2*cos[tex]\theta1[/tex])(3/2) like equation we usually met. But, i don't know what is the vector in spherical coordinate. How to write it and how to evaluate it? I stop to use spherical coordinate and try to use cartesian coordinate. But, it is extremely difficult even only for evaluating dS . I get a form which i can't use my integral technique. I really...really stuck now. You need to know that two equations above is equations i derive alone. So, ther is possibility to be wrong.

Which is wrong from my steps? Is the equation? Or model? Could you help me?
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kanato
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#2
Sep25-09, 07:37 PM
P: 416
Your equation doesn't even give 0 for [tex]R_2 = 0[/tex]. The reason is that you are only summing up the magnitude of the electric field at the point R_2, you are not including the vector nature of the electric field. The correct starting equation is

[tex]
E(\vec{r}) = \frac{1}{4\pi \epsilon} \int \frac{\rho(\vec{r'}) (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^3} d^3 r'
[/tex]

Note the extra factor of [tex]\vec{r} - \vec{r'}[/tex] has its magnitude canceled by the extra factor in the denominator so that only the direction of r-r' is kept.

It's a bit easier to calculate the potential and show that it is independent of R2 if R2 < R1.
chemfirus
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#3
Sep25-09, 08:18 PM
P: 3
Quote Quote by kanato View Post
Your equation doesn't even give 0 for [tex]R_2 = 0[/tex]. The reason is that you are only summing up the magnitude of the electric field at the point R_2, you are not including the vector nature of the electric field. The correct starting equation is

[tex]
E(\vec{r}) = \frac{1}{4\pi \epsilon} \int \frac{\rho(\vec{r'}) (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^3} d^3 r'
[/tex]

Note the extra factor of [tex]\vec{r} - \vec{r'}[/tex] has its magnitude canceled by the extra factor in the denominator so that only the direction of r-r' is kept.

It's a bit easier to calculate the potential and show that it is independent of R2 if R2 < R1.
That's what i mean in last sentences. I don't include the vector in the equation. When iwant to use it, i don't understand how to. Is the magnitude of [tex]\vec{a}[/tex][tex]\theta[/tex] is [tex]\theta[/tex]1 and there is no magnitude in [tex]\vec{a}[/tex][tex]\phi[/tex]? In other words, the miss vector numerator in equation (2) above is

(R1 - R2)[tex]\vec{a}[/tex]r + ([tex]\theta[/tex]1 - 0)[tex]\vec{a}[/tex][tex]\theta[/tex] ?

I still confuse with the equation you wrote. Is it surface integral or volume integral? Whati is d3r' means? And, if you would, could you evaluate your integral and show me the prove. I need concrete solution. Anyway, thanks for helping me.

kanato
kanato is offline
#4
Sep26-09, 11:05 AM
P: 416

How to get zero electric field inside spherical shell without using Gauss law


That's what i mean in last sentences. I don't include the vector in the equation. When iwant to use it, i don't understand how to.
In spherical coordinates, [tex]\vec{r} = r \hat{r}[/tex]. You have to be careful when subtracting two vectors like that because they each have a different [tex]\hat{r}[/tex] direction. So you have to write it in cartesian coordinates. In that case, [tex]\vec{r} = r \sin \theta \cos\phi \, \hat{x} + r \sin\theta \sin\phi \, \hat{y} + r\cos\theta \, \hat{z}[/tex]. Then you do the subtraction, keeping in mind that [tex]\vec{r}[/tex] and [tex]\vec{r'}[/tex] each have their own [tex]\theta,\phi[/tex] components. In this case it will be simple, because you restrict [tex]\hat{r}[/tex] to the z axis, so when you integrate [tex]\phi[/tex] the x and y components will drop out.

I still confuse with the equation you wrote. Is it surface integral or volume integral? Whati is d3r' means? And, if you would, could you evaluate your integral and show me the prove. I need concrete solution. Anyway, thanks for helping me.
Yes, it's a volume integral. If you have a thin shell of charge, then you define [tex]\rho(\vec{r}) = \sigma \delta(r - R)[/tex] where you use the Dirac delta function. If you are not familiar with the Dirac delta, then the result is that it reduces to:
[tex]
E(\vec{r}) = \frac{1}{4\pi\epsilon}\int \frac{\sigma (\vec{r} - R \hat{r'})}{|\vec{r} - \vec{r'}|^3} R^2 \sin \theta' \, d\theta' \, d\phi'
[/tex]
which is a surface integral. I'm not going to do the whole thing for you, since (a) it's a good practice problem for you to work through yourself, and (b) I don't feel like it.

(Btw, when you use tex on these forums, it's best to use it for the whole equation rather than just a few special symbols. It's a lot easier to read that way.)
chemfirus
chemfirus is offline
#5
Sep26-09, 07:52 PM
P: 3
Quote Quote by kanato View Post
In spherical coordinates, [tex]\vec{r} = r \hat{r}[/tex]. You have to be careful when subtracting two vectors like that because they each have a different [tex]\hat{r}[/tex] direction. So you have to write it in cartesian coordinates. In that case, [tex]\vec{r} = r \sin \theta \cos\phi \, \hat{x} + r \sin\theta \sin\phi \, \hat{y} + r\cos\theta \, \hat{z}[/tex]. Then you do the subtraction, keeping in mind that [tex]\vec{r}[/tex] and [tex]\vec{r'}[/tex] each have their own [tex]\theta,\phi[/tex] components. In this case it will be simple, because you restrict [tex]\hat{r}[/tex] to the z axis, so when you integrate [tex]\phi[/tex] the x and y components will drop out.



Yes, it's a volume integral. If you have a thin shell of charge, then you define [tex]\rho(\vec{r}) = \sigma \delta(r - R)[/tex] where you use the Dirac delta function. If you are not familiar with the Dirac delta, then the result is that it reduces to:
[tex]
E(\vec{r}) = \frac{1}{4\pi\epsilon}\int \frac{\sigma (\vec{r} - R \hat{r'})}{|\vec{r} - \vec{r'}|^3} R^2 \sin \theta' \, d\theta' \, d\phi'
[/tex]
which is a surface integral. I'm not going to do the whole thing for you, since (a) it's a good practice problem for you to work through yourself, and (b) I don't feel like it.

(Btw, when you use tex on these forums, it's best to use it for the whole equation rather than just a few special symbols. It's a lot easier to read that way.)
Thank you very much! I have just done it! Like your suggest, i use cartesian coordinate vector instead of spherical coordinate vector. This is the final equation i evaluate:

[tex]\frac{\rho}{4\pi\epsilon}[/tex][tex]\oint[/tex][tex]\frac{R_1^2\sin\theta_1}{R_1^2+R_2^2-2R_1R_2\cos\theta_1}[/tex][tex]\frac{(R_1\sin\theta_1\cos\phi_1)\widehat{a_x}+(R_1\sin\theta_1\sin\phi _1)\widehat{a_y}+(R_1\cos\theta_1-R_2)\widehat{a_z}}{\sqrt{R_1^2+R_2^2-2R_1R_2\cos\theta_1}}[/tex]d[tex]\theta_1[/tex]d[tex]\phi_1[/tex]

Its quite difficult for me to evaluate the integral. Because of symmetry reason i think components in [tex]\hat{a_x}[/tex] and [tex]\hat{a_y}[/tex] is zero when i evaluate it. So, i choose to pass it. I just evaluate the [tex]\hat{a_z}[/tex] component. But, it is still difficult for first time. Fortunately, thanks to god, i could do it and get zero result in the end. Many things i have got rom this question.

And, special thanks for Kanato for helping me...
kanato
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#6
Sep27-09, 02:35 AM
P: 416
Yeah, the x and y components vanish. Do the [tex]d\phi_1[/tex] integral first, you will see that gives zero.
sophiecentaur
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#7
Sep27-09, 09:15 AM
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There is a more 'arm waving' but valid method, I think.
Take any point inside the sphere and draw a chord through it. Take an elemental square 'pyramid' about the chord. The two ends of this bow-tie will meet the surface of the sphere and form identical shaped surfaces. The areas will be proportional to the square of the distances. The forces on a unit charge will be inversely proportional to the distance and proportional to the area intersected. And the two forces will be exactle opposite in direction, producing a resultant of zero. If you integrate over a whole sphere (around the point in question) you will get zero.
I suppose it would be better to use an elemental pyramid of d phi and d theta rather than dx and dy but the same argument would apply, I think.
Minimal Maths makes it easier for me!


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