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Radial Electric Field

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Carefully study the following plot of electric field E in N/C versus distance r in m and answer the questions that follow. The electric field is directed radially outward, and the variation of E with r is independent of direction.

    Refering to the figure, determine the work done by the electric field on a 7.75 C charge moved from A to B.



    2. Relevant equations
    [tex] W=q \int\vec{E}\dot\vec{dr}[/tex]

    3. The attempt at a solution
    First I found the charge caused by the electric field( I used E=25 and r=2 since they intersect:
    [tex] q=\frac{E r^2}{\epsilon_{o}}\longrightarrow q= q=\frac{25* 2^2}{\epsilon_{o}}= 1.11E^-8
    C[/tex]
    -----
    I am finding Work done from a to b:
    [tex] W= q\int\vec{E}\dot\vec{dr}[/tex] I am pretty sure I am integrating with respect to r.
    [tex] W= q\int\frac{kq_{1}}{r^2}dr[/tex] a=10 b=4
    So what do you think?
     

    Attached Files:

  2. jcsd
  3. Sep 15, 2007 #2
    anyone?
     
  4. Sep 15, 2007 #3

    learningphysics

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    Homework Helper

    Yes, looks right to me. Although you wrote [tex] q=\frac{E r^2}{\epsilon_{o}}[/tex] you meant [tex] q=\frac{E r^2}{k}[/tex] right?

    Also here:
    [tex] W= q\int\vec{E}\dot\vec{dr}[/tex]

    [tex] W= q\int\frac{kq_{1}}{r^2}dr[/tex]

    q is the 7.75C and q1 is the 1.11E-8C (doesn't matter mathematically... but for your substitution to make sense E = kq1/r^2... hence q1 is the charge creating the field)...

    just making sure because you used q for 1.11E-8 before...

    Everything looks good.
     
  5. Sep 15, 2007 #4
    That is Correct i meant K.
    So: [tex] (9.0e^9* 1.11E^-8* 7.75E^-6 * \int \frac{1}{r^2} dr[/tex] boudries of a=10 to b=4. = -1.161E-4 J, shouldn't work be postive, we are going against the e-field
    Oh and by the way the charge is suppose to be 7.75 uC.
     
  6. Sep 15, 2007 #5

    learningphysics

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    Homework Helper

    It should be negative since we're calculating the work done by the field, not the change in electric potential energy.

    The change in electric potential energy is the negative of the work done by the field... the field does negative work... potential energy increases.
     
  7. Sep 15, 2007 #6
    Ok i get it,
    Thank you
     
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