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Electric field on a charge some height above the center of a disc

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Given a disc of radius 'R' with a charge 'h' above the center of the disc, find the electric field on the charge.


    2. Relevant equations
    ##E = \frac{kQ}{d^2}##


    3. The attempt at a solution
    I can find the charge density from a ring then integrate over R:

    ##\sigma = \frac{Q}{A}##

    ##Area_{ring} = 2\pi r*dr##

    ##Q = \sigma * 2\pi r*dr##

    Since the horizontal components cancel out, I'll take the cosine angle:

    ##E_y = \frac{kQ}{(h^2 + r^2)} * \frac{h}{\sqrt{h^2 + r^2}}##

    ##E_y = \frac{kh2\pi \sigma dr}{(h^2 + r^2)^{\frac{3}{2}}}##

    ##E_y = kh\pi \sigma \int_{0}^{R} \frac{2r dr}{(h^2 + r^2)^{\frac{3}{2}}}##

    ##E_y = kh\pi \sigma [\frac{-2}{\sqrt{h^2 + r^2}}]##

    Integrate from 0 to R:

    ##E_y = kh\pi \sigma [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##

    Re-substitute ##\sigma = \frac{Q}{A}##:

    ##E_y = kh\pi \frac{Q}{A} [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##

    ##E_y = kh\pi * \frac{Q}{2\pi r*dr} * [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##

    I'm now stuck with this 'dr' in the denominator. Not sure what to do from here
     
  2. jcsd
  3. Sep 16, 2013 #2

    rude man

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    Is the disc charged? Is the charge above the disc very small?
     
  4. Sep 16, 2013 #3

    vela

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    You need to clean up your math. Your carelessness is leading to dumb mistakes.

    Think about what Q and A represent here.

    You can't have lone differentials sitting around. You should have written
    \begin{align*}
    dA &= 2\pi r\,dr \\
    dQ &= \sigma\,dA = 2\pi \sigma r \, dr \\
    dE_y &= \frac{k\,dQ}{h^2+r^2} \frac{h}{\sqrt{h^2+r^2}} = \frac{2\pi hk \sigma r \, dr}{(h^2+r^2)^{3/2}}
    \end{align*} If you have an infinitesimal quantity on one side of the equation, you have to have one on the other side as well.

    Look at those two lines. According to what you wrote,
    $$\frac{kh2\pi \sigma dr}{(h^2 + r^2)^{\frac{3}{2}}} = kh\pi \sigma \int_{0}^{R} \frac{2r dr}{(h^2 + r^2)^{\frac{3}{2}}}$$ because both sides of the equation equal ##E_y##. The integration apparently does absolutely nothing.

    You need to think about what Q and A represent. Q is the charge of what? A is the area of what?
     
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