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Homework Help: Point Charge Work

  1. Jan 23, 2009 #1
    A point charge, Q=+1mC, is fixed at the origin. How much work is required to move a charge, q=+8(micro)C, from the point (0,4meters) to the point (3meters,0)?



    I was unable to derive anything.



    I guessed at around 40J. Correct? If so HOW! lol
     
  2. jcsd
  3. Jan 23, 2009 #2
    How are electric potential and work related?
     
  4. Jan 23, 2009 #3
    Potential=E*d
     
  5. Jan 24, 2009 #4
    You have there an equation relating the electric potential to the electric field and the separation between the change in potential. While it's useful you need to find a way to relate V to W...
     
  6. Jan 24, 2009 #5
    I don't know.... V=Eq*d ? because F=Eq ?
     
  7. Jan 24, 2009 #6
    Ok so it V = E * d then it seems logical to say that

    V = (F*d)/q, right? Now the nice thing about electric potential is that it is a scalar.
     
    Last edited: Jan 24, 2009
  8. Jan 24, 2009 #7
    Since the electic field created by Q is not uniform E(x) must be used

    [tex]E(x)=K\frac{Q}{x^2}[/tex]

    where K is

    [tex]\frac{1}{4\pi\varepsilon_0}[/tex]

    The force on q is

    [tex]F(x)=qE(x)[/tex]

    So, the work done on q is

    [tex]W=-\int\mbox{F(x)dx}[/tex]
     
  9. Jan 24, 2009 #8
    This problem doesn't require integration.

    and.... how did you derive V = (F*d)/q? if F=Eq, then by your derivation V=(Eq)*d/q, which is reduced to V=Ed? how does V=Eqd=Ed?
     
  10. Jan 24, 2009 #9
    Typing error... V=E*d => V = F*d / q
     
  11. Jan 24, 2009 #10
    Integration is a way to solve the problem. Simply choose a convenient path starting at (0,4) and ending at (3,0). Any path will work because E is a conservative field. Using potential is another way. Finding the potential difference between the two points then multiplying this difference by q will give the desired result. Because E is not uniform use integration.

    [tex]V=-\int_{\infty}^{r}\mbox{E(r)dr}=\frac{Q}{4\pi \varepsilon_0\mbox{r}}[/tex]

    [tex]V_1=\int_{4}^{\infty}\mbox{E(y)dy}=\frac{Q}{4\pi \varepsilon_0\mbox{4}}[/tex]

    [tex]V_2=\int_{3}^{\infty}\mbox{E(x)dx}}=\frac{Q}{4\pi \varepsilon_0\mbox{3}}[/tex]

    [tex]W=(V_2-V_1)q[/tex]
     
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