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Spherical Capacitors, electric fields.

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    An air-filled capacitor consists of a conducting sphere of radius a, surrounded by a concentric conducting spherical surface of radius b. A charge of 2Q is placed on the inner sphere and –Q on the outer surface. Calculate the electric field strengths E(r), (i) between the two spherical surfaces, i.e. for a < r < b and (ii) for r>b Calculate energy stored in region between 2 surfaces and in region , r> b.

    [in this question you may assume that the dialectric constant of air is equal to 1]

    2. Relevant equations

    Gauss' law for closed surfaces.

    3. The attempt at a solution

    I use the standard method of finding the electric field from each using gauss' law and then sum them, but i find my awnser is:

    [tex] E(r) = \frac{3Q}{4 \pi \epsilon_{0} r^{2}} [/tex]

    When the awnser given is the same but only 2Q rather than 3...
    so it seems that only the centre sphere is contributing to the electric field.. but i think i must be wrong.
  2. jcsd
  3. Jan 14, 2009 #2
    Forgot to mention im using the principle of supersition for the section between the plates.
  4. Jan 14, 2009 #3


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    Well, in between the two conductors, only the charge enclosed by your Gaussian surface contributes to the flux right? And the flux I refer to is the flux of the net electric field through that surface not the flux of just the component due to the inner charge:wink:
  5. Jan 14, 2009 #4
    so to find the flux of the outer sphere and therefore its electric field id have to do a gaussian surface over the outer sphere then subtract the inner spheres electric field and id have equations for both of the spheres fluxes/electric fields, i then just sum these together for the field inbetween?
  6. Jan 14, 2009 #5


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    No, my point was that the E in gauss' law:


    is the total electric field. So when you calculate that E at some location ([itex]a\leq r\leq b[/itex] in this case), You calculate the total electric field at that point. There is no need to add/subtract any other constituent fields.
  7. Jan 14, 2009 #6
    But that means that you could have a tiny sphere of quite small charge in space and at around it have a massive surface that was spherical with a huge charge upon it, taking the gassian surface to just below the larger sphere would only give you the electric field due to the smaller sphere, it compeltely ignores the larger.
    same would happen from 1 almost point charge (dr sphere) and this could be a distance c away from a massively charged object. then taking a gaussian sphere c-d where d<<c youd end up with the electric field from just the small sphere of charge...

    Is this how it really works? its very weird :S
  8. Jan 14, 2009 #7


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    No, it would give you the total electric field. Just because the magnitude of that field depends only on the inner charge, doesn't mean its the field due to just that charge.

    Anyways, the field inside of a uniformly charged spherical shell is zero isn't it? That means the contribution of the outer charge to the total electric field in your eample would be zero anyway.
  9. Jan 14, 2009 #8


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    Taking a spherical shell for your gaussian is only useful if the charge distributions (and hence the elecric field they produce) are spherically symmetric about a common center. So this example would only work if the massive outer charge was a uniform spherical shell centered on the point charge. Otherwise, the fild would not be uniform and radial and gauss law would be useless.
  10. Jan 14, 2009 #9
    i thought it was zero unless it contained a charge inside, which it does :P lol couldve figured it out that way quicker.
    would the same be true of my second example as well? (ie use gauss' law the second larger charge doesnt depend on field)
    so using the q enclosed gives me the required awnser, then for when r>b i do a gaussian surface containing both spheres? or just the outer one?
  11. Jan 14, 2009 #10


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    Well if r>b, then your gaussian does enclose both spheres right:wink: Remember, r is the radius of your gaussian.
  12. Jan 14, 2009 #11
    yup thought so just checking :).

    interlude question :P to express potential energy of a finite charge distribution in empty space in terms of charge density and electrostatic potential V. im then asked to show that the energy may be regarded as distrubuted with energy density:
    [tex] \frac{ \epsilon_{0} E^{2}}{2} [/tex]
    this is after im told to state gauss' law of electrostatics...
    sorry for asking a question in a topic for another question, but this took me by suprise, ive never been asked to derive this, i just assumed that it was this value, would i take my potential energy of the distribution and calculate its derivitive with respect to volume? that would be energy density but how abouts would i do that.. im guessing ihave to relate it to a maxwell equation

    Back on topic.. :

    Energy stored is just the integral of the field by [tex] dV = 4 \pi r^{2} dr [/tex] with limits b and a for the E inside and b and infinity for outside? or is there a speedier way of doing it?
    Last edited: Jan 14, 2009
  13. Jan 15, 2009 #12
    Any ideas? ive read some where energy density is:
    [tex] = \frac{1}{2} E^{2} \epsilon_{0} [/tex]
    but i dont know how to put the limits of distance on it..
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