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Homework Help: Quick Potential Query

  1. Apr 14, 2009 #1
    In my electrostatics notes, when calculating the work done in moving a point charge from r1 to r2 we use

    [itex]W=\int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot \mathbf{dr}[/itex]
    which ends up giving W=qV where V is the potential difference.

    However in previous years we used [itex]\mathbf{F}=-\nabla W \Rightarrow W=-\int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot \mathbf{dr}[/itex]

    (this works if you consider W as potential energy (mgz) and then you get a force due to gravity of -mg in the z direction.

    so which is correct and why?
     
  2. jcsd
  3. Apr 14, 2009 #2

    dx

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    Work done by what? The work done by you in moving a particle from r1 to r2 in a force field F is always

    [tex] - \int_{{r_1}}^{{r_2}} \mathbf{F} \cdot \mathbf{dr} [/tex]
     
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