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Potential of a circle boat

  1. Jun 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A line charge has the total charge Q evenly distributed over a circle boat with radius a and sector 2β, placed according to the figure

    figure.jpg
    [​IMG]

    Find the Electric field E and the potential V in the origin.

    2. Relevant equations
    I know for this case that E(r) = (1/4πε) ∫ (λ(r')/R2)R dl' , R is unit a vector

    λ(r') = Q/(2βa) R = - a *(cos(φ)x + sin(φ)y) dl'=a*dφ R2= a^2

    V(r) = (1/4πε) ∫ (λ(r')/R) dl'
    3. The attempt at a solution
    The electric field is not so hard to calculate I know
    E(r) =Q/(2βa)(1/4πε) ∫ ((-a *(cos(φ)x + sin(φ)y))/a3) adφ
    which is : (Q/8πεa2β) * [ -sinφx + cosφy]

    Now if I want to calculate the potential I know :
    V(r) = ∫ E*dl where dl = a*dφ*R = -a*dφ*(cos(φ)x + sin(φ)y)
    But I see (Q/8πεa2β) * [ -sinφx + cosφy] * (-a*dφ*(cos(φ)x + sin(φ)y)) = 0!

    And the real solution is :

    V(r) = (1/4πε) ∫ (λ(r')/R) dl' = (1/4πε) * (Q/2βa) ∫(adφ/a) and for a boat circle from π-β to π+β

    Or :
    sol-jpg.jpg

    What is actually wrong ?! If I do the same algorithm for a Sphere, it gives me a correct answer but why can't I do the same thing here through ∫ E*dl ?
     
    Last edited: Jun 16, 2017
  2. jcsd
  3. Jun 16, 2017 #2

    mfb

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    There is a step missing to evaluate the electric field.

    For the potential, there is an easier way, without using the electric field. Otherwise you would have to calculate the electric field everywhere along your integration path and that gets much more complicated.
     
  4. Jun 16, 2017 #3
    Ok the whole solution is :

    sol.jpg


    Why can't we just integrate E*dl from this solution for E ?!
     
  5. Jun 16, 2017 #4

    mfb

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    That looks right.
     
  6. Jun 16, 2017 #5
    But is this wrong to think that E =(Q/8πεa2β) * [ -sinφx + cosφy] ?! If we just want to find V through E *dl ?!
    I know that there is an easier way. Just using formula. But when I will find potential for a solid sphere it seems much more easier to use ∫E*dl. First we find E and then we integrate over that E and find V. But here I see we can not do exact the same thing .... why ?!
     
  7. Jun 16, 2017 #6

    mfb

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    What is φ in that formula? How can the electric field depend on it?
    The electric field has a zero y component due to the symmetry of the setup.
    You can do it, but it is much more work because there is no simple expression for E at other places.
     
  8. Jun 16, 2017 #7
    It's for polar coordinate.
     
  9. Jun 16, 2017 #8
    OK and because of that the easiest way is : V(r) = (1/4πε) ∫ (λ(r')/R) dl' ....
     
  10. Jun 16, 2017 #9

    vela

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    Yes, it's wrong. You're integrating over ##\phi## with definite limits, so that variable shouldn't appear in the final result.

    You're calculating the electric field at one point, the origin. To find the potential by integration, you need to find ##\vec{E}## at all points along whatever path you choose from infinity to the origin. In other words, what you're proposing to do is calculate
    $$-\int_C \vec{E}(0)\cdot d\vec{r}$$ instead of
    $$-\int_C \vec{E}(r)\cdot d\vec{r}$$ where ##C## is some path from infinity to the origin.
     
  11. Jun 16, 2017 #10

    mfb

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    The polar coordinate of what?
    The electric field can only depend on parameters of the setup (##\beta##, a and Q). It cannot depend on intermediate parameters introduced in the calculation. If someone gives you values for these three things you have to be able to calculate the electric field in V/m. What would you plug in for ##\phi##?
    Right.
     
  12. Jun 17, 2017 #11
    OK than you !

    What do I see; it's not easy to find E(r) for this case....
     
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