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Gauss's Law Problem

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A spherical drop of water has radius R = 5 * 10^-4 cm. The drop has an even surface charge distribution.

    Consider an element dA of the surface. This element is exposed to a force from the rest of the surface elements.

    The effective field in the surface is equal to the mean value of the effective field just outside and just inside the surface.

    Use Gauss's Law to compute the effective field in the surface.

    3. The attempt at a solution

    I've done several Gauss's Law exercises before, but in this one I'm not even sure I understand the question.
    What bothers me is the line about the "effective field" being the "mean value" of the field just inside and just outside the surface. In one problem I just did, I showed that the electric field inside an even spherical surface distribution is zero, so how can the field on the surface be the "mean" of 0 and a field slightly less than on the surface?

    Am I misreading something?

    Appreciate any help on this one!
     
  2. jcsd
  3. Mar 12, 2013 #2

    TSny

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    Good.
    I'm not sure I understand the last part your question. You know the field just inside the surface is 0. What is the field just outside the surface? You want to take the mean (average) of the fields just inside and just outside.

    Did they give you the amount of charge on the drop?
     
  4. Mar 12, 2013 #3
    Thank you for your reply!

    They only write the charge per area as a sigma, so I have to solve for the electric field in terms of sigma.

    So, I basically have to compute the charge just outside the sphere and divide it by 2? Still, I don't understand what they mean with an "effective field"... I figured the field just outside the sphere would be just a tiny amount smaller than the field in the surface, not twice as big!
     
  5. Mar 12, 2013 #4

    TSny

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    I think you mean't "field" instead of "charge" here. But, yes, that's right. All you'll need to do is divide the field just outside the drop by 2.
    Surface charge densities are interesting in that they cause a finite discontinuous jump in the field as you pass from one side of the surface to the other. If you replaced the infinitesimally thin surface layer of charge by a small but finite thickness of charge, then the field would change continuously but rapidly withing the charge layer (from zero on the inner surface of the layer of charge to some finite value at the outer surface of the layer of charge.) The "effective" value of the field within the layer of charge is taken to be the average of the fields at the inner and outer surfaces of the layer. When taking the limit of an infinitesimal thin layer of surface charge, then you get essentially a discontinuous jump in the field. But you still take the effective field to be the average of the fields just inside and outside the surface layer of charge.

    So, you'll need to find an expression for the field just outside the drop.
     
  6. Mar 13, 2013 #5
    I got the field just outside the sphere to equal sigma / epsilon, which means the field in the surface should equal sigma / 2 epsilon.

    Your explanation was interesting, and I tried googling an explanation for why this is true, but without any luck. Is the electric field anywhere always equal to the mean of the EF just outside and just outside the field in question? Say, if this was a solid sphere with a uniform volume distribution instead of an empty shell? When I think about it that last question is obviously true haha, but is it true in general?

    Calculating the eletric field on the surface this way was not really hard to do, but I don't think I quite get why it works... Is there a way to find the field on the surface without first calculating the field just outside?

    Thanks a lot!!
     
  7. Mar 13, 2013 #6

    TSny

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    Last edited by a moderator: May 6, 2017
  8. Mar 13, 2013 #7

    collinsmark

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    Hi gralla55,

    I don't recommend putting much effort into this "mean value of the effective field just outside and just inside the surface." This is the first time I've seen it defined this way. Usually, I see the field defined on the outside and also defined on the inside, but not so much at the very surface itself.

    If it helps though, consider the simple, unit Heaviside step function (http://en.wikipedia.org/wiki/Heaviside_step_function). Why is it that H(0) = 1/2? Well, that's a good question. But if you define
    [tex] H(x) = \left\{ \begin{array}{rl}
    0 &\mbox{ if $x<0$} \\
    1 &\mbox{ if $x>0$}
    \end{array} \right. [/tex]
    and leave H(0) undefined for now, then break it apart through Fourier decomposition, then finally put it back together again (also Fourier decomposition), taking whatever limits need to be taken, I think you'll find that the H(0) = 1/2. Is it a big deal? No, I don't think so. I wouldn't recommend putting too much effort into this.

    What *is* important though is that electric field can be, and often is discontinuous. In the case of this problem, it jumps from [itex] \frac{\sigma}{\epsilon_0} [/itex] down to zero all within an infinitesimal length. You can expect electric fields to change a lot over small spaces, in some situations.

    On a related note, what is *not* usually discontinuous (and is instead usually continuous) is electric potential. If you find sudden discontinuities in your electric potential functions, it's time to raise an eyebrow.
     
    Last edited: Mar 13, 2013
  9. Mar 13, 2013 #8

    TSny

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    The nice thing about the "effective field", ##E_{eff}##, at the surface is that it provides a direct way to calculate the electrostatic force per unit area, ##f##, on the surface charge: ##f = \sigma E_{eff}##.
     
  10. Mar 14, 2013 #9
    Thanks a lot, that really helped!
     
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