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Elecrtic field of Charged disc

  1. May 17, 2005 #1
    Electric field of Charged disc

    Find the electric field at a distance 'z' above a uniformly charged disk of radius R.

    I have solved this problem. Can someone just clarify if my solution is right?
    I couldn't attach a diagram..sorry!!

    Solution: Using conventional notations,
    [itex]\sigma[/itex] is the charge density on the circular surface. dq is the differential

    charge and dA is the differential area.
    Dividing the disc into flat concentric rings of thickness dx and considering one such ring

    of radius x : [tex]dA = 2\pi xdx[/tex]
    Hence [tex]dq = \sigma 2\pi xdx[/tex]
    Magnitude of the electric field is given by(perpendicular components cancel each other here):[tex]dE\cos \theta[/tex]
    [tex]\cos \theta = \frac{z}{r}[/tex]
    [tex]r = \sqrt{z^2 + x^2}[/tex]

    [tex]E = \int dE\cos \theta[/tex]

    [tex]E = \frac{\sigma z}{4\epsilon_0} \int_{0}^{R} \frac{2x}{(z^2 + x^2)^{3/2}}dx[/tex]

    On solving after the necessary substitutions the answer I got is:
    [tex]E = \frac{\sigma}{2\epsilon_0}[/tex][tex][1 - \frac{z}{\sqrt{z^2 + r^2}}][/tex]

    Is my answer correct?
    How will my answer modify if [itex]R\rightarrow \infty[/itex]?
    Also check the modifications for [itex]z >>R[/itex]
     
    Last edited: May 18, 2005
  2. jcsd
  3. May 17, 2005 #2

    dextercioby

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    So you found that

    [tex] E(z,R)=\frac{\sigma}{2\epsilon_{0}}\left(1-\frac{z}{\sqrt{z^{2}+R^{2}}}\right) [/tex] (1)

    (I corrected your typo).

    It's perfect.Take [itex] R\rightarrow +\infty [/itex] and then u'll find

    [tex] E_{\mbox{charged plane}}=\frac{\sigma}{2\epsilon_{0}} [/tex] (2)

    which is exactly the result u'd be getting if u were computing (2) using Gauss' theorem.

    Daniel.
     
  4. May 17, 2005 #3

    SpaceTiger

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    Try doing a second-order Taylor expansion around R=0. Does the result look familiar?
     
  5. May 17, 2005 #4

    dextercioby

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    After that,to recover something known,take

    [tex] \sigma=\frac{q}{\pi R^{2}} [/tex]

    Daniel.
     
  6. May 18, 2005 #5
    Thank you dextercioby for correcting my result :smile:
    I understood the first part of my question for [itex]R\rightarrow \infty[/itex]

    For the second part, z>>R
    [tex] E=\frac{\sigma}{2\epsilon_{0}}[1-\frac{z}{\sqrt{z^2}}][/tex]
    Wouldn't this mean E=0?
     
    Last edited: May 18, 2005
  7. May 18, 2005 #6
    For the second part, z>>R

    [tex] E=\frac{\sigma}{2\epsilon_{0}}[(1-\frac{z}{\sqrt{z^2}}][/tex]
    Wouldn't this mean E=0?[/QUOTE]

    You missed a second-order term here! :smile:
     
  8. May 18, 2005 #7
    I'm sorry, I'm not familiar quite familiar with Taylor Expansion in this context :frown:
    Can you help me on this?
     
  9. May 18, 2005 #8

    SpaceTiger

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    For a given z:

    [tex]E(R)=\frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})[/tex]

    The Taylor expansion, to second order, is given by:

    [tex]E\simeq E(0)+E'(0)R+\frac{1}{2}E''(0)R^2[/tex]
     
    Last edited: May 18, 2005
  10. May 18, 2005 #9

    dextercioby

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    Another way,using something (hopefully) familiar

    [tex] E (z,R)=\frac{\sigma}{2\epsilon_{0}}\left[1-\frac{1}{\sqrt{1+\left(\frac{R}{z}\right)^{2}}}\right] [/tex]

    To the second order,

    [tex] \frac{1}{\sqrt{1+\left(\frac{R}{z}\right)^{2}}}\simeq 1-\frac{1}{2}\left(\frac{R}{z}\right)^{2} [/tex]

    ,using the famous (hopefully for you,too)

    [tex] \frac{1}{\sqrt{1+x^{2}}}\simeq 1-\frac{1}{2}x^{2} [/tex]

    for [itex] x\rightarrow \frac{R}{z} <<1 [/itex].

    Daniel.
     
    Last edited: May 18, 2005
  11. May 19, 2005 #10
    I evaluated the series using Taylor's formula for R=0. Please let me know if they are correct.
    E(0) = 0
    [tex] E'(R) = \frac{\sigma z}{2\epsilon_0} (z^2 + R^2)^{-3/2} R[/tex]
    Wouldn't this mean E'(0) R = 0 ?

    [tex] E"(R) = \frac{\sigma z}{2\epsilon_{0}}[/tex] [tex]-3(z^2 + R^2)^{-5/2} R[/tex]

    Meaning (1/2)E"(0)R2 =0 ?

    Edit: My codes aren't working? :cry: Can someone look into this?
     
    Last edited: May 19, 2005
  12. May 19, 2005 #11
    Dextercioby-- I think your answer is more comprehensive :smile:
     
  13. May 19, 2005 #12

    dextercioby

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    Completing

    [tex] f(x)=:\frac{1}{\sqrt{1+x^{2}}} [/tex] (1)

    To the third order

    [tex] f(x)\simeq f(0)+\frac{1}{1!}\left\frac{df(x)}{dx}\right|_{x=0} x+\frac{1}{2!}\left\frac{d^{2}f(x)}{dx^{2}}\right|_{x=0} x^2 [/tex] (2)

    One can easily show that

    [tex] f(0)=1 [/tex] (3)

    [tex] \left\frac{df(x)}{dx}\right|_{x=0} =0 [/tex] (4)

    [tex] \left\frac{d^{2} f(x)}{dx^{2}}\right|_{x=0} =-1 [/tex] (5)

    Going with (3)-(5) in (2),u find exactly

    [tex] \frac{1}{\sqrt{1+x^{2}}}\simeq 1-\frac{1}{2}x^{2} [/tex] (6)

    Q.e.d.

    Daniel.
     
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