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Homework Help: Gauss's Law/Energy Problem with Concentric Spheres

  1. Oct 14, 2007 #1
    Hi there. I'm new to PF, so please correct me on any mistakes in presenting my question. I'm just hoping to get some direction in doing this problem right.

    1. The problem statement, all variables and given/known data

    Concentric spherical shells of radius a and b, with b > a, carry charge Q and -Q, repsectively, each charge uniformly distributed. Find the energy stored in the electric field of this system.
    (From Purcell, Electricity and Magnetism, 2nd Ed. [I hate this book :yuck:])

    2. Relevant equations

    From Gauss's Law (for a sphere)
    E= [tex] Q/r^2 [/tex]

    U = [tex] \frac{1}{8\pi} \int_{Entire field} E^2 dv[/tex]

    3. The attempt at a solution

    I know I just have to sum up the E field and integrate over the volume of the entire sphere (or rather wherever the field is non-zero?) and that the E field will end up being a constant (or am I completely wrong?)

    E inside inner sphere = 0 (field is zero inside spherical shell of charge)
    E outside inner sphere = [tex]Q/(a^2)[/tex]
    E inside outer sphere = [tex]-Q/b^2 + Q/a^2[/tex] (This one is the one I'm really unsure on)
    E outside outer sphere = 0 (total charge is [tex]Q+(-Q) = 0[/tex])

    And then do I integrate the E field squared in spherical coordinates from a to b? (As I end up with it as a constant I feel like I'm doing something wrong.)
    Any help is greatly appreciated (I'm awful at E&M).

    edit: Forgot to mention this is all in cgs units, not mks/si.
    Last edited: Oct 14, 2007
  2. jcsd
  3. Oct 14, 2007 #2


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    I'm confused... there are 30 concentric spherical shells? or just 2? (one at a and one at b?)

    for a sphere... the field is [tex]E = \frac{Q}{4\pi \epsilon_0 r^2}[/tex]

    And energy is [tex]\frac{\epsilon_0}{2}\int E^2dv[/tex]
  4. Oct 14, 2007 #3
    Ah sorry about that, 1.30 is just the problem number. Just 2 concentric spherical sehlls.

    I "ignored" the constants because I (or rather this book) uses cgs units instead of MKS or SI units.
  5. Oct 14, 2007 #4


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    oh I see... sorry about that.

    In your work you said "E inside outer sphere" and "E outside inner sphere"... they both the same right? The field would just be Q/r^2 between the two shells as far as I can see... just using gauss' law and the charge of the inner shell...

    The field outside the outer sphere and inside the inner sphere are both 0 (both using gauss' law)...

    So you'd integrate [tex] \frac{1}{8\pi} \int_{Entire field} E^2 dv[/tex]

    from a to b... with E being Q/r^2
  6. Oct 14, 2007 #5
    Thanks so much. Now I just have to remember how to integrate a sphere haha. Just to confirm, I think my integral will look like this? I need to retake vector calculus :blushing:

    [tex]E=\frac{-Q}{r^2}, E^2=\frac{Q^2}{r^4}[/tex]
    [tex]U = \frac{1}{8\pi}\int \frac{Q^2}{r^4}dV, dV=r^2sin\phi dr d\phi d\theta[/tex]
    [tex]U = \frac{1}{8\pi}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{a}^{b} \frac{Q^2}{r^2} sin\phi dr d\phi d\theta[/tex]
  7. Oct 14, 2007 #6


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    Looks right to me. you can integrate over dtheta and dphi... and your left with an integral over r.

    You can also use dV = [tex]4\pi r^2 dr[/tex] which is The area of a spherical shell times dr... (same thing as what you have just a quicker method to look at it).

    Why did you use -Q/r^2 instead of Q/r^2 (it doesn't make a difference, but just wondering... since the inner sphere is radius a with charge Q)?
  8. Oct 14, 2007 #7
    Ohhh thank you for that. So [tex]U = \frac{1}{2} \int_{a}^{b} \frac{Q^2}{r^2}dr [/tex] I think.

    Yes, I know it doesn't make a difference in the end, but if it did, it is over the negative charge between the two spheres, right? Since inner sphere has charge Q and between inner and outer sphere the charge is -Q.
    Last edited: Oct 14, 2007
  9. Oct 14, 2007 #8


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    Your expression for U looks good to me.

    For the field being Q/r^2 instead of -Q/r^2

    Well, when you use gauss' law to find the field between the two spheres... you use a gaussian surface (a sphere)...

    Gauss' law says that [tex]\int E\cdot dA = 4 \pi Q_{enclosed}[/tex]

    So what is the charge enclosed in a gaussian sphere between the two given spheres... it is the charge of the inner sphere...

    So field is Q/r^2 (not -Q/r^2).

    I might be making a mistake... let me know if this makes sense.
  10. Oct 14, 2007 #9
    Hmmm okay, I think that makes sense. Thanks again for all your help!
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