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Homework Help: Integrating a charge along a line

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    total charge Q = 200[nC] is spread along a line from x=−100[mm] to x=+100[mm] .
    a) write the dV caused by each dQ=λ dx .
    b) integrate along the line of charge ; write V as a function of y .
    c) take the derivative with y , to obtain that component of the Electric (vector) field .
    d) take the derivative with y , again , to find where (along y-axis) the Ey is maximum


    2. Relevant equations
    We've used potential energy U=1/2QV...V=kQ/r...that's about it.


    3. The attempt at a solution
    What I can't get started with is the writing dV caused by each dQ. I'm not even sure what that means. And how can I write V as a function of y and take its derivative if there are no y's in any of these equations?

    Any guidance here would be awesome, thanks
     
  2. jcsd
  3. Feb 9, 2010 #2

    rl.bhat

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    hi burns12, welcome to PF.
    Take a small element dx at a distance x from the origin.
    Consider a point P at a distance y from the origin along y-axis.
    Charge dq in the dx element is λ*dx, where λ = Q/L.
    As you have mentioned
    dV = k*dq/r. Put the value of dq and r.
    To find V at P, take the integration between the limits x = -0.1 m to 0.1 m.
     
  4. Feb 9, 2010 #3
    Thanks!

    Ok so, the y comes into play as a distance on the y axis that this dq charge is going to affect?
    And do I integrate the dV = k*dq/r ? Or do I sub in λdx for dq and then integrate with respect to x to find dV?
    I'm still not sure where this distance y would come into this.
     
  5. Feb 9, 2010 #4

    rl.bhat

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    dq = λ*dx and r = sqrt( x^2 + y^2)
     
  6. Feb 9, 2010 #5
    Alright awesome I think I got it. Is the r = to sqrt(x^2 +y^2) because it's like the components of r in a sense?
     
  7. Feb 9, 2010 #6

    rl.bhat

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    Yes.
     
  8. Feb 9, 2010 #7
    Awesome, thanks for your help, that had me stumped for 2 days now ha
     
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