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Electric field due to a uniformly charged disc

  1. Apr 12, 2012 #1
    At the end of the derivation, it is given
    The electric fiel due to elementary ring at the point P is dE = [2∏rσdrx]/[4∏epsilon zero (x^2 +r^2)^(3/2)

    To find the total E due to disc is given by

    ∏σx/4∏ε(2rdr)/(x2 + r2)3/2

    I am stuck with the calculus done here to arrive at the solution. Please help me.

    Thank You!

  2. jcsd
  3. Apr 12, 2012 #2
    Do you mean

    [tex]\mathrm{d}E=\dfrac{2\cdot\pi\cdot r\cdot\sigma\cdot x\cdot\mathrm{d}r}{4\cdot\pi\cdot\epsilon_0\cdot\left(x^2+r^2\right)^{\frac32}}[/tex]

    and you want to know how to determine that

    [tex]E=\dfrac{\pi\cdot\sigma\cdot x}{4\cdot\pi\cdot\epsilon_0}\cdot\displaystyle\int\left(\dfrac{2\cdot r\cdot\mathrm{d}r}{\left(x^2+r^2\right)^{\frac32}}\right)[/tex]

    Also, why isn't the first TeX parsing?

    Simple. Integrate both sides.
  4. Apr 12, 2012 #3
    I can't see your formula!!!!
  5. Apr 12, 2012 #4
    Yep, for some reason, the TeX isn't parsing. It should! It does on AoPS!
  6. Apr 12, 2012 #5

    WOuld you mind if you tell me how to solve the integration part?

    Thank You
  7. Apr 12, 2012 #6
    Divid [itex](x^2+r^2) \;[/itex] by [itex]r^2\;[/itex] to get [itex] 1+(\frac x r)^2\;[/itex] and use trig function to substitute. and see what happen. Something like [itex]\tan \theta=\frac x r [/itex]. [itex]1+\tan^2\theta=\sec^2\theta[/itex]
    Last edited: Apr 12, 2012
  8. Apr 12, 2012 #7
    Thanks Yungman!!

    I am gonna solve that way.
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