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Spherical charge distribution

  1. Jun 19, 2012 #1
    good evening!

    i am trying to calculate the electric field of a spherical charge distribution ρ=ρ[itex]_{0}[/itex]e[itex]^{-kr}[/itex], where r is the radial distance. i am a little bit embarressed,but i have to say that i am not comfortable with spherical coordinates in practical calculations. i would appreciate if somebody could help me thru this:

    the field is independent of the angels, so it is sufficient to look at the z-axis. i choose r=z the distance should be given by:

    [itex]
    \mid r-r'\mid^{2}=z^{2}+r'^{2}-2zr'cos\theta'
    [/itex]

    now, to caluclate the coulombintegral i need the differencevector between the source and the field points. how would i do that?

    thanks in advance for your help!
     
  2. jcsd
  3. Jun 19, 2012 #2

    phyzguy

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    Try using Gauss' law. Since everything is spherically symmetric, E is a function of r only. so 4*pi*r^2 *E(r) = (Charge inside r) / ε0. To calculate the charge inside r, you just have to integrate (4*pi*r^2 rho(r)) from 0 to r. Try it!
     
  4. Jun 19, 2012 #3

    gabbagabbahey

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    Are you sure? Even if you are an atheist, could it not be that someone named Angel manufactured this charged sphere? :tongue:

    Is there a reason you are trying to use Coulomb's Law to solve this problem? Surely using Gauss Law would be easier.

    In any case, to calculate the difference vector, you just write down the two vectors and take their difference: [itex]\mathbf{r}=z\hat{\mathbf{z}}[/itex], [itex]\mathbf{r}'=r'\hat{\mathbf{r}}[/itex] [itex]\implies \mathbf{r} -\mathbf{r}' = z\hat{\mathbf{z}} -r'\hat{\mathbf{r}}[/itex]. Then just express [itex]\hat{\mathbf{r}}[/itex] in terms of the Cartesian unit vectors (and [itex]\theta'[/itex] and [itex]\phi'[/itex] of course).
     
  5. Jun 19, 2012 #4
    haha! sorry can't hide my atheism :)

    about using gauss's law: since the charge distribution is not conifned to a specific region of space, this would not be a good idea since i would not include the chargedensity outside the sphere ...or am i missing something? (also i would like to practice these things, since as i mentioned, i am not good at those calculationes)

    i attached a pdf...would this be correct?

    i think i hear an angle sing :)
     

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  6. Jun 19, 2012 #5

    gabbagabbahey

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    What's the problem with some of the charge being outside your Gaussian sphere? Gauss' law involves only the charge enclosed (which will be a function of the radius of your Gaussian sphere), and is useful as long as you can argue that the electric field has appropriate symmetry (spherically symmetric in this case).

    It looks more or less correct, but a little confusing with you using [itex]r[/itex] in the numerator and [itex]z[/itex] (and even a [itex]z'[/itex]?) in the denominator all to represent the distance to the field point... pick one variable and stick to it. You've also dropped your unit vector, the [itex]2\pi[/itex] you get from integrating [itex]\phi'[/itex] and the exponent of the denominator in the last step, and [itex]r-r'\cos\theta \neq (r-r')\cos\theta[/itex]

    And as long as the angels don't have the phone-booth, I'm sure it will work out fine.
     
  7. Jun 20, 2012 #6

    phyzguy

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    You're missing something. The field inside a uniform spherical charge shell is zero, so the field is only determined by the charge inside - the charge outside doesn't matter.
     
  8. Jun 20, 2012 #7
    Gauss' law involves charge enclosed by the Gaussian Surface and flux linked with that surface. of course the field involved in flux calculation is net. But in any case due to symmetry field will be same at same radial distances and symmetry tell us that filed will be radial. These aspects will make cosθ = 1, and take E out of integral in calculation of flux.

    There flux linked with a Gaussian Sphere of radius r will be E 4∏ r^2

    calculation of charge enclosed
    Take a spherical shell of radius x and . Charge enclosed by that surface can be calculated as follows.

    dq/dv = ρ
    dq = ρ dv
    dv is the volume of shell of radius x and thickness dx.
    ∴ dv = 4∏ (x^2) dx
    ∴ dq = ρ 4∏ (x^2) dx
    q = ∫ρ 4∏ (x^2) dx
    limits should be taken from o to r.
    Perform integral, use gauss law & get the answer.
     
  9. Jun 20, 2012 #8
    good morning!

    thanks for your answers everbody!

    gabbagabbahey: you are right of course, r=z, and of course there is a 2∏ missing. and the last thing you wrote, i meant (r-r')cosθ', thanks! i droped the unit vector, because there is only a field in z direction, i should have given E a subscribt z.

    what i mean about gauss's law is the following: i want to calculate the e-field in an arbitrary point in space. if i use gauss's law i would leave some of the charge unaccounted for. for example if i had to point charges and would integrate a surface arround the first charge wich doesn't include the second, i would get the e-field of the first point charge, but not the field due to both charges of course.

    thanks!
     
  10. Jun 20, 2012 #9
    Please consider this argument:
    According to the charge density equation, charge is continuously distributed. The result you will get in terms of r which is a variable. In final result you can put any value of r (from o to infinity) you get the result.
    As far as if we are talking about only this distribution we are doing right. You do not worry about all points when u talk in terms of variable.
     
  11. Jun 20, 2012 #10
    ok, if i take r=a then i account for all the charge inside the sphere with radius a. you are saying, that the charge outside the sphere doesn't contibute to the field at a? (of course it doesn't if i use guass's law because the net flux throu the sphere of charge outside the sphere is zero). yes the distribution is continous but i don't see how this would make it ok to neglect the charge wich is at positions at r>a. if i would only be interessted in the field of the charge wich is enclosed in the sphere r=a then this would be fine with me...
     
  12. Jun 20, 2012 #11
    This is what Shell theorem says. The forces at any point inside the shell will cancel out. You can derive it by using Gauss's law or by direct integration over a spherical shell setting a<R.

    Both methods will yield the same result that there is no gravity inside a hollow Earth.

     
  13. Jun 20, 2012 #12
    i am perfectly aware of what gauss's law is, about that it is applicable to mass distributions as well as charge distributions. but this is not a cahrged spherical shell it is a charge distribution over all of space. if i have two charges and exclude one from the surface, this does not mean, that this charge gives no contribution to the electric field in some point of space, it just means that it does not contribute to the surface integral.

    if you calculate the gravitational field of the earth with gauss's law, it will give you the gravitational field the mass of the earth produces, this does not mean that the net gravitational field in the solarsystem is just that of the earth.

    thank you!
     
  14. Jun 20, 2012 #13
    for example lets say you have this charge distribution ρ=ρ[itex]_{0}[/itex]e[itex]^{-kr}[/itex] this time not all over space but for r<= a and a singel point charge q at x=a+c where c>0. now if you use gauss's law for the spherical charge distribution, this does not mean you can forget about the point charge, does it?
     
  15. Jun 20, 2012 #14

    phyzguy

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    Don't forget that the equations of electromagnetism are linear, so in this case, you can calculate the field due to the spherical charge distribution, and separately calculate the field due to the point charge. Then the resulting field is just the sum of these two.

    Similarly, since the field inside a spherical shell is zero, I can decompose a spherically symmetric charge distribution into an infinite sum of infinitesimally thin shells. Since the field inside each shell is zero, the field inside any spherically symmetric charge distribution is zero. So the field at radius r only depends on the charge interior to r, and charge exterior to r can be ignored. Of course, this is only the case if the charge distribution is spherically symmetric.
     
  16. Jun 20, 2012 #15

    gabbagabbahey

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    But should you not have [itex]r-r'\cos\theta'[/itex] instead?

    [tex]\mathbf{r}-\mathbf{r}' = r\hat{\mathbf{z}}-r'\left(\sin\theta'\cos\phi'\hat{\mathbf{x}} + \sin\theta'\sin\phi'\hat{\mathbf{y}}+\cos\theta' \hat{\mathbf{z}}\right)[/tex]

    No. Gauss' Law tells you that in cases where you know that the Electric field is spherically symmetric (so that [itex]\int\mathbf{E}\cdot d\mathbf{a} = 4\pi r^2 |\mathbf{E}|[/itex] where [itex]r[/itex] is the radius of your Gaussian surface, and the distance to your field point) then the field at [itex]r[/itex] must be due entirely to the charge enclosed in your sphere of radius [itex]r[/itex]. The field due to just the charges outside your Gaussian sphere is evidently zero in such cases.

    But in this case, there would be no symmetry that would allow you to conclude that [itex]\int\mathbf{E}\cdot d\mathbf{a} = A|\mathbf{E}|[/itex], so all you could say from Gauss' law is that the flux is due only to the charges enclosed in your Gaussian surface, and this much is always true: the field line from charges outside the surface will both enter and exit the surface resulting in zero net flux, while the field line from sources inside the surface only exit. You would not be able to determine the field from that knowledge.

    You could use a combination of Gauss' law and the superposition principle to calculate the field in this case though. All you would need to do is apply Gauss' law assuming only one charge is present for each of your two point charges, and carefully add (remember this is vector addition and the unit vector away from charge one is not the same as the unit vector away from charge two) the two E-fields together[/QUOTE]
     
  17. Jun 20, 2012 #16
    my point exactly. the field extends to infinity and i want the field due to all charges. i've painted a picture, maybe this helps to clarify this. i have the feeling i am beeing missunderstood. (of course the distribution is continous, etc...)

    yes, thats right!
     

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  18. Jun 20, 2012 #17

    gabbagabbahey

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    You are missing the key point here. The [itex]\mathbf{E}[/itex] in Gauss' Law is the field due to all the charges. And since, for spherically symmetric fields (like the one in this problem), you can say [tex]\int\mathbf{E}\cdot d\mathbf{a} = 4\pi r^2 |\mathbf{E}|[/tex] simply by choosing a concentric spherical surface of radius [itex]r[/itex] as your Gaussian surface, Gauss' law tells you [itex]|\mathbf{E}| = \frac{1}{4\pi\epsilon_0}\frac{Q_{enc}}{r^2}[/itex]. The field due to all the charges at distance [itex]r[/itex] can thus be determined entirely by the charges inside the sphere of radius [itex]r[/itex]. The field due to the charges outside must add up to zero.

    This should not be a complete surprise to you. You are no doubt familiar with the fact that the field inside a uniformly charged spherical shell is zero. Now, consider that what you have in this problem is an infinite number of infinitesimally thin shells of thickness [itex]dr[/itex], that each carry a charge spread uniformly over their surface, that decreases for each successive surface like [itex]\text{e}^{-kr}[/itex]. The field due to each spherical shell that is outside your field point is then zero, and by superposition, that means that the net field due to the all the shells/charge further from the origin than your field point, is zero.

    If you still have doubts, I suggest you finish your direct application of Coulomb's law, and then compare your result to what you get using Gauss' law. You should find the results are equal, and that Gauss' Law was a much easier method. This is why Gauss' Law is so useful in cases where the field has certain types of symmetry.
     
  19. Jun 20, 2012 #18

    vela

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    The other posters understand what you're saying, but you're not getting what they're saying.

    One approach is to use Gauss's law and only worry about the charge inside the Gaussian surface. If you have spherical symmetry, this calculation is straightforward, and you can find the field at points on the surface. When you place a single point charge outside of a sphere of charge, you break the spherical symmetry and complicate the calculation. In particular, the electric field will no longer be spherically symmetric, so you can't easily find the electric field at points on the surface by simply applying Gauss's law. At best, you can make a statement about the total flux through the surface. The total flux will depends only on the charge inside the surface. The flux due to the charge outside will integrate to 0.

    The other approach some of the others have alluded to is to go ahead and calculate the electric field due to the charge outside the Gaussian surface. If you do this with a spherically symmetric charge distribution, you'll find the field due to that outside charge is 0.
     
  20. Jun 21, 2012 #19
    nicly done, very good argument! i am convinced, thank you for the insight gabbagabbahey!
    thanks to all of you for the discussion and excuse my stubbornness!
     
  21. Jun 22, 2012 #20
    Glad that you get that. Always try to use Gauss's law if possible. If Gauss Law can't resolve the problem, direct integration would be also very difficult or nearly impossible except for some trivial cases.
    For instance, in your problem, if instead, [tex]\rho=\theta\rho_0[/tex], find E along z direction, you would have a terrible integral to crack.
     
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