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Electric Field Due to a disk

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A disk is located at 0<r<1, z = 1. It has a uniform charge distribution of 200pC/m^2. A 30 [tex]\mu[/tex]C point charge is located at the origin. Determine the force acting on the point charge due to the electric dield produced by the disk.

    2. Relevant equations

    [tex]E = \frac{F}{Q}[/tex]

    [tex]F = \frac{kQ_{1}Q_{2}}{R^{2}}[/tex]

    [tex]dQ = \rho_{s}ds[/tex]

    [tex]Q = \int \rho _{s}ds[/tex]

    3. The attempt at a solution

    I know the idea behind finding the solution but I can't get the right answer. Thinking back to my calculus classes I need to take very very small peices to make the integral. One of the problems that I keep running into is that the book I'm using seems to keep everything in Cartesian and at last minute without telling you they switch to polar. I think that this problem should be easier if it was in polar so I am attempting that.

    The first thing that I need to do is find the E-Field. If I take a very small peice of disk the area would be

    da = r * dr * d(theta)

    then I use this to find the charge

    [tex]dq = \rho_{s} * d\rho * d\Theta[/tex]

    Then I plug that into the e-field equation.

    [tex]dE = \frac{\rho_{s}*\rho d\Theta d\rho}{4\pi\epsilon_{0}R^{2}}[/tex]

    Now I need to find R wich is the distance from the point of observation from the charge. This is what I think is causing most of my problem. The book says to ignore all vectors other then the Z vector but, it never works out. Here is what I'm doing.

    R is going to be the point of observation minus whereever the charge is. So

    ([tex]R= (0-\rho a_{\rho}-1a_{z}[/tex]

    I need the magnitude of this


    Now plug that into the E-field equation along with multiplying by the unit vector and integrating to find the actual E-Field I get the following Equation

    [tex]E = \int\int\frac{\rho_{s}*\rho d\Theta d\rho(-\rho a_{\rho}-1a_{z})}{4\pi\epsilon_{0}\sqrt{(-\rho^{2}-1^{2}})^{3/2}}[/tex]

    This gives me basically two sets of integrals one for each seperate componet of the unit vector. This is what my final equation looked like before I plugged in the numbers. I already taken out the integral of theata since the equation didn't depend on it and so it equalled 2pi.

    [tex]\frac{-\rho_{s}}{2\epsilon_{0}}(\int\frac{\rho^{2}}{\sqrt{\rho^{2}+1}^{3/2}}d\rho a_{\rho}+\int\frac{\rho}{\sqrt{\rho^{2}+1}^{3/2}}d\rho a_{z})[/tex]

    I end up with .233887 for the first integral and .3784 for the second integral. Then multiply by the constants out front and my final answer is 2.701 in the A_p and 4.279 in the A_z. Wich is not the correct answer.

    The Answer is supposed to be 99.24 in the Az
  2. jcsd
  3. Sep 17, 2009 #2
    Can someone look at my math and tell me if I did the integrations correctly? I'm still stumped on this problem. I've been looking over the steps and it all seems right to me except for the E-field in the rowe direction. I know that this should be zero. I'm wondering if the answer in the back of the book is incorrect.

    Thanks for the help
  4. Sep 18, 2009 #3
    I think I've figured something out. I need R to be the distance from the point of observation to the charge. I'm making the mistake of trying to find the distance using polor wich dosn't work. I need to convert to cartesien to find the distance.
  5. Sep 18, 2009 #4
    First of all it makes sense that the electric field at any point on the axis of symmetry have only a component along the axis of symmetry!!! Even if you don't know the result you have to become suspicious about your results.

    You have to perform the integration more attentive, there are several mistakes in your text!

    (I have suspend the disk to z'=0, so the point charge is located at (0,0,-1)!)

    In general the electric field of a know charge distribution is (with the volume element of cylinder coordinates)

    [tex]\vec E(\vec r) = \frac{1}{4\pi \epsilon_0} \int \limits_{R_0}^{R_1} d\rho^{\prime} \, \rho^{\prime} \, \int \limits_{\varphi_0}^{\varphi_1} d\varphi^{\prime} \, \int \limits_{z_0}^{z_1} dz^{\prime} \, \rho_V(\vec r^{\, \prime}) \, \frac{\vec r - \vec r^{\, \prime}}{|\vec r - \vec r^{\, \prime}|^3} [/tex]​

    Now we are searching for
    [tex]\vec E(0,0,-1) = ?[/tex]​

    Let us now inspect
    [tex]\vec r - \vec r^{\, \prime} = ?[/tex]​

    You have written sth. like this

    [tex]\vec r - \vec r^{\, \prime} = (\rho - \rho^{\prime}) \vec e_{\rho} + (z - z^{\prime} ) \vec e_z ~~~~~~~~~~~~ \mbox{with} ~~~ \vec e_z = \vec e_3[/tex]​

    Ok, e_\rho is a unit vector of the cylindrical orthogonal system, but remember that it also depends on \varphi!!! So, if we want to work exactly (otherwise we get the wrong result) we have to write the above like

    [tex]\vec r - \vec r^{\, \prime} = \rho \, \vec e_{\rho}(\varphi) - \rho^{\prime} \, \vec e_{\rho}(\varphi^{\prime}) + (z - z^{\prime} ) \vec e_z[/tex]​

    because in the integration we vary \varphi^{\prime} (in our problem from 0 to 2\pi) but retain \varphi.
    So even we take the volume element of cylinder coordinates we must write the unit vectors of the cartesian orthogonal system, unless we are geniusses in mental arithmetics.

    Ok then, let's write it in cartesians
    [tex]\vec r - \vec r^{\, \prime} = \left( \rho \cos(\varphi) - \rho^{\prime} \cos(\varphi^{\prime}) \right) \vec e_1 ~ + ~ \left( \rho \sin(\varphi) - \rho^{\prime} \sin(\varphi^{\prime}) \right) \vec e_2 ~ + ~ \left( z - z^\prime) \, \vec e_3 [/tex]​

    The point charge is at \vec r = (0,0,-1). This yields

    [tex]\vec r - \vec r^{\, \prime} = - \rho^{\prime} \cos(\varphi^{\prime}) \, \vec e_1 ~ - ~ \rho^{\prime} \sin(\varphi^{\prime}) \, \vec e_2 ~ - ~ \left(1 + z^\prime \right) \, \vec e_3 [/tex]​


    [tex]|\vec r - \vec r^{\, \prime}| = \sqrt{ \rho^{\prime}^2 + \left(1 + z^\prime\right)^2}[/tex]​

    With the assumption that the disk is of infinitesimal height the integral must be

    [tex]\vec E(0,0,-1) = - \frac{\rho_S}{4\pi \epsilon_0} \int \limits_{0}^{R} d\rho^{\prime} \, \rho^{\prime} \, \int \limits_{0}^{2\pi} d\varphi^{\prime} \,
    ~ \frac{\rho^{\prime} \cos(\varphi^{\prime}) \, \vec e_1 ~ + ~ \rho^{\prime} \sin(\varphi^{\prime}) \, \vec e_2 ~ + ~ \vec e_3 }{\sqrt{ \rho^{\, \prime \, 2} + 1}^{~3}}[/tex]​

    Now we can guess, that the terms with \cos(\varphi) and \sin(\varphi) vanishing, because we integrate over a whole period!

    Hope i could help you...

    with best regards
    Last edited: Sep 18, 2009
  6. Sep 18, 2009 #5
    You took words out of my mouth. I can't beleive I made such a small mistake. Well actually it was a rather large mistake. I just figured it out in my last period. You can't find a distance in cylindrical cordinates very efficently so you have to convert to cartesian. x = rho cos(phi) y = rho sin(phi). Then you can take the integrals wich the x and y vectors go to zero like they are supposed to and your left with z. Which will give you the right E-field. Muliply by the charge at the observed point and viola. Correct answer. Thanks for your help. It's like being hit by lighting when you realize that you made a silly mistake and created a ton of extra work.
  7. Sep 18, 2009 #6
    That's not quite true! We are using cylindrical coordinates but not the unit vectors of cylindrical symmetry, that's a big difference.

    If we are using cartesian coordinates our independent variables are


    if we are using cylindrical coordinates we use


    as independent! So we are still using cylindrical coordinates but do not use the unit vectors of a cylindrical orthogonal coordinate system.

    If you have remaining questions please ask...

    with best regards
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