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Electric field of a disc

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the field of a disc radius a, uniformly charged with surface charge density σ, for x>>a?

    2. Relevant equations


    3. The attempt at a solution
    I've worked out the equation for the field of a disc:
    ##E = ## ##\frac{\sigma x}{2 \epsilon}## ##\frac{\sqrt{a^2+x^2}-x}{x \sqrt{a^2+x^2}}##

    I'm also given a hint, which is that the answer shouldn't be zero. As far as I can tell, if x>>a, this definitely becomes zero! I have looked up my answer for E and it is right. To work out E for x>>a, I essentially ignored all a as irrelevant - is that wrong?

    ##E = ## ##\frac{\sigma x}{2 \epsilon}## ##\frac{\sqrt{x^2}-x}{x \sqrt{x^2}}## is most definitely zero!
     
  2. jcsd
  3. Oct 25, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    For a finite value of ##x## that is much greater than ##a##, ##E## will be small but it won't be zero. You have "over approximated". In this type of problem, you want to keep the lowest order nonzero approximation.

    Instead of approximating ##\sqrt{x^2+a^2}## in the numerator as ##\sqrt{x^2}##, find a more accurate approximation in terms of the small quantity ##a/x##.
    Hint: https://en.wikipedia.org/wiki/Binomial_approximation

    You might find it easier to first simplify your expression for E by cancelling some ##x##'s and splitting the fraction involving the square roots into the difference of two fractions. You will then have only one square root to worry about.
     
  4. Oct 28, 2015 #3
    Simplifying the expression gives

    ##E = \frac{\sigma}{2\epsilon}(1 - \frac{1}{\sqrt{\frac{a^2}{x^2}+1}})##
    And then use a binomial approximation for ##\sqrt{\frac{a^2}{x^2}+1}##.
    Thank you, gave the right answer :)
     
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