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Work done - moving a charge in an electric field

  1. Nov 22, 2013 #1
    work done -- moving a charge in an electric field

    1. The problem statement, all variables and given/known data
    Q=10uC from origin to (3m,pi/4,pi/2) E=10r unit r + 5/(rsin(theta) unit phi V/m

    answer is -475uJ

    I found this problem in a reference book that gave answers, this problem is similar to a homework problem.


    2. Relevant equations

    w = -q ∫ E*dl or w = 1/2 integral (εE^2) dv

    3. The attempt at a solution

    w= -10x10-6 ∫ 10rdr + 5d(phi)
     
  2. jcsd
  3. Nov 22, 2013 #2

    berkeman

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    Staff: Mentor

    Could you please clarify the question a bit? When you give the electric field, is that in spherical or rectangular coordinates? Would it be possible to scan the problem and upload it into this thread?
     
  4. Nov 24, 2013 #3
    The problem is in spherical coordinates.

    My work is w = -10x10^-6 ( 20( sqrt(exp(1) -1) +10 pi/2 = 5.736 uJ , which is incorrect from the answer in the book.
     
  5. Nov 24, 2013 #4
    I am not sure what I am doing wrong, dL = dr (unit r) + r dtheta (unit theta) + r sin theta d phi (unit phi) .
    E=10r (unit r) + 5/[(rsin(theta)] (unit phi) V/m

    so integral of work = q * (10r dr + r phi dphi )

    from origin 0,0,0 to 3, pi/4, pi /2 so r goes 0 to 2 theta goes 0 to pi/4 and phi goes 0 to pi/2
     
  6. Nov 25, 2013 #5

    rude man

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    Are you sure about the second term in the E field expression? Because that implies an infinite E field at the origin.

    Anyway, two hints:

    1. the E field is conservative so what does that tell you about the path of integration?

    2. You can split the E field into two fields, corresponding to the two terms, and use superposition. Then go back to hint #1.

    I assume theta is the angle with the z axis and phi is the azimuth angle, and that you're giving the final position as (r, theta, phi).

    (My answer is close to -475 uJ but not quite.)
     
    Last edited: Nov 25, 2013
  7. Nov 25, 2013 #6
    Since E is conservative, the path of integration does not matter.
     
  8. Nov 25, 2013 #7

    rude man

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    Right. So what path would you take for E dot dl?
     
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