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Difference of potential between 2 points

  1. Sep 6, 2009 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data

    Point [tex]A[/tex] is situated at a distance [tex]d[/tex] from a charge [tex]q[/tex]. Point [tex]B[/tex] is situated at a distance [tex]2d[/tex] from the charge [tex]q[/tex], in the same straight line than the one of [tex]A[/tex] and [tex]q[/tex].
    Calculate the difference of potential [tex]V_A-V_b[/tex].
    1.1 The attempt at a solution

    [tex]V_a-V_B=k \int_ A^B E ds[/tex].
    But [tex]E=\frac{F}{q}=\frac{k}{r^2}[/tex] and [tex]ds=dr[/tex], hence [tex]V_a-V_B=k \int_ A ^B =-k \left [ \frac{1}{r} \right ]_A ^B =-k \left [ \frac{1}{B} - \frac{1}{A} \right ] = k \left ( \frac{1}{A} - \frac{1}{B} \right )[/tex].

    Is that right?
     
  2. jcsd
  3. Sep 6, 2009 #2

    acr

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    I think that

    [tex]
    F= \frac {kq_1 q_2} {r^2}
    [/tex]

    so

    [tex]
    E=\frac{F}{q} = \frac{kq}{r^2}
    [/tex]

    not

    [tex]
    \frac {k}{r^2}
    [/tex]

    Right?
     
  4. Sep 6, 2009 #3

    fluidistic

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    Gold Member

    You're right. So my result is the one I gave multiplied by [tex]q[/tex]. I hope I'm right this time.
     
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