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Electric Field of extended mass

  1. Feb 16, 2012 #1
    So here is the scenario (see attachment) - I have a semicircle wire (radius R=15.9cm) which is made of insulator material , the semicircle consist of two combined quartercircle wires parts where one has equally distributed charge +Q and the other has -Q . Required is find the Electric field in direction of x at the origin . Q=5.33nC

    My approach was as follows

    Let E = 1/(4*pi*e)∫1/(R^2).dQ r

    dQ=λ*ds and ds=R*dθ and i also know that unit vector r = cosθ*i+sinθ*j

    therefore for the E in x direction i get this expression

    E = 1/(4*pi*e)*1/(R^2)*λ*R∫cosθ.dθ

    Integrating from 0 to pi ( thus taking only half of the semicircle ) and using λ as 2/(pi*r)

    I get Q/(2*pi^2*e*R^2) .
    Because the other half has opposite charge i can say that the Etot = Eneg +Epos

    Therefore i multiply the equation by two to finaly get


    If i put the values given i get as absolute value 2413 N/C for Electric field at origin of circel in the direction of x

    Unfortunately it is a wrong solution :( !! What is the mistake i hv done ?? Can anyone spot it ? Thanks in advance

    Attached Files:

  2. jcsd
  3. Feb 16, 2012 #2


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    Hi Redoctober! :smile:
    noooo :cry:

    0 to π/2 :redface:
  4. Feb 16, 2012 #3
    Oh !! srrry typo error :S :S , i integrated from 0 to pi/2 .
    The answer 2413 N/C is wrong :/ !
  5. Feb 16, 2012 #4
    Anyone has a solution ?? :/ !
  6. Feb 17, 2012 #5
    The E field for the right part of your semi circle is given by
    [tex]E = \int_0^{\frac{\pi}{2}} \frac{-Q}{4\pi\epsilon_0r^2} d\theta[/tex]
    While the E field for the left part is the same but with +Q charge and integrated from [itex]\frac{\pi}{2}[/itex] to [itex]\pi[/itex]. Just add these two together to get the total field.
  7. Feb 17, 2012 #6


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    Are you sure? An electric field is a vector, and the correct formula is

    [tex]\vec{E}(\vec{x})=\frac{1}{4 \pi \epsilon_0} \int_{\mathbb{R}^3} \rho(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.[/tex]

    Of course, in the here considered case, you have to integrate along the semicircle and use the charge per length instead of the bulk-charge density. However, you should check, whether you have all the geometrical factors for you vector component right.
  8. Feb 17, 2012 #7
    Vanhees is right , Electric field is a vector .
    So regarding the E will be in radial direction .
    For the Y axis we need to consider the j component .

    Vanhees , your mathematical expression is a bit high beyond my math skill xD !

    My question is why the final expression for the problem is wrong :/

    i considered the +Q and -Q E vector addition

    I reached to the conclusion of

    |E| = Q/(4*pi^2*e*r^2) (In the direction of J

    Btw : how do u write the math expressions in that style xD . my way of typing the math is silly :/ !!
  9. Feb 19, 2012 #8
    The final equation i got was actually correct

    i made an error with the calculation thats why my answer was wrong :)
    Last edited: Feb 19, 2012
  10. Feb 19, 2012 #9


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    Yes, I've checked your result too. It's correct. I parametrized the line charge with help of the charge per angle:

    -\frac{2 Q}{\pi} & \text{for} \quad 0 \leq \varphi \leq \pi/2 \\
    +\frac{2Q}{\pi} & \text{for} \quad \pi/2<\varphi\leq \pi \\
    0 & \text{elsewhere}.

    Then you can use my formula for [itex]\vec{x}=0[/itex] to get

    [tex]\vec{E}=\frac{1}{4 \pi \epsilon_0 r^2} \int_0^{2 \pi} \mathrm{d} \varphi' \lambda(\varphi') \begin{pmatrix} -\cos \varphi' \\ -\sin \varphi' \\0 \end{pmatrix}.[/tex]

    Integrating over the two regions gives finally

    [tex]\vec{E}=\frac{Q}{\pi^2 \epsilon_0 r^2} \vec{e}_x.[/tex]
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