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Electric field on the surface of uniformly charged spherical shell

  1. Apr 6, 2013 #1
    We can easily find from gauss's theorem(or otherwise) the field inside and outside a uniformly charged spherical shell.But i was wondering what would be the field on the surface of the shell.
     
  2. jcsd
  3. Apr 6, 2013 #2

    BruceW

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    Homework Helper

    interesting question. You could probably calculate it by using Coulomb's law, and integrating over the charge distribution? I was thinking of one other possible way... Although I'm not certain it is correct. Um, so outside the shell, we have:
    [tex]\vec{E} = \frac{Q}{4 \pi \epsilon_0 r^2} \hat{r} [/tex]
    Where Q is the total charge, given by: [itex]Q=4 \pi a^2 \sigma [/itex] Where 'a' is the radius of the shell, and sigma is the charge per area on the shell. Right, so we knew that already. But now, if we imagine we are at a point very close to the the surface of the shell (but just outside it), then the part of the shell right next to this point will look approximately flat, right? So the electric field coming from that very nearby part of the shell will be:
    [tex]\vec{E} = \frac{\sigma}{2 \epsilon_0} \hat{r} [/tex]
    So, I was thinking that, if we then go onto the shell itself, we are going to 'lose' this contribution to the electric field. So to calculate the electric field on the shell, we just do that - take away this contribution. So my prediction is that the electric field on the shell is:
    [tex]\vec{E} = ( \frac{Q}{4 \pi \epsilon_0 r^2} - \frac{\sigma}{2 \epsilon_0} ) \hat{r} [/tex]
    And now, writing out Q fully, and recognising that r=a when we are 'on' the shell, we get:
    [tex]\vec{E} = \frac{\sigma}{2 \epsilon_0} \hat{r} [/tex]
    OK. This is my prediction... I am not at all sure if it is correct though.

    Edit: and a quick sanity check, to make sure that it makes sense: we can do the same thing to get from being 'on the shell' to being 'inside the shell', so we will add a similar contribution, but with the electric field going radially inward this time, so then the electric field will equal zero inside the shell. (Which agrees with what we get from Gauss' law).
     
  4. Apr 6, 2013 #3

    Jano L.

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    I agree with BruceW, this is very clever way to find the electric field acting on the charges on the surface. One possible "check" is to calculate electrostatic energy ##W## of the sphere as a function of its volume ##V## and find the pressure necessary to hold it together:

    $$
    p = -\frac{dW}{dV}
    $$

    The force of pressure balances the electrostatic force on the charged patch of surface ##dS##:

    $$
    pdS = \sigma E dS
    $$

    From these two equations, E can be found and the result is the same as BruceW indicated, that is, the charges on the surface experience electric field that is one half of that just above the surface.
     
  5. Apr 6, 2013 #4
    Amazing!!! Thanks for both of your help.
     
  6. Apr 7, 2013 #5
    As BruceW said I calculated it using coulomb's law by integrating over the charge distribution and result comes to be same as his.
     
  7. Apr 7, 2013 #6

    Jano L.

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    That seems like a difficult calculation. If you managed to do it, then congratulations!
     
  8. Apr 7, 2013 #7
    I did it like this.
    Let us consider an elemental circular ring on the shell with its centre lying on the line joining the considered point and the centre of the sphere.Its radius is asinθ and area is (2∏asinθ)(adθ) and charge on it is dq = σ2∏a2sinθdθ.So field on the considered point due to this elemental ring is
    (1/4∏ε0)(dq.z/(z2+R2)3/2) where
    z=a(1-cosθ) and R=asinθ
    which on simplification comes to be (σ/4ε0)cos(θ/2)dθ
    so field at required point can be found by integrating it from θ=0 to θ=∏ and answer comes same as found by BruceW.
    The procedure is same as for finding field at any point outside(or inside) a uniformly charged spherical shell using Coulomb's law.
     
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