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The dot product on electric potential

  1. Jul 27, 2008 #1
    there are 2 situations that need explanation. first, the general formula for electric potential as you take potential = 0 at infinity as a reference.

    Second, the general formula for capacitance on a parallel plate.

    In situation one, the negative sign does NOT DISAPPEAR, and in the second situation....the negative sign DISAPPEARS!!!! I'm not too sure why.

    in the first situation.....

    the electric field from a positive charge emanates outward toward infinity. The work of the electric field created by the positive charge is NEGATIVE as you bring a positive test charge from infinity to the positive charge, and V = - integral of E vector dot dL vector. The negative sign tells us that the work of the positive electric field from the positive source charge is negative and the
    -W = U.

    However, as you go from infinity to the positive charge, the dot product between E and dl creates a cos 180 degrees(the dL vector is in the opposite direction of the E vector from the positive source charge). Thus, shouldn't V = + integral of magnitude of E vector times dL vector? then, for a point charge.... V = - (k)(q)/r as you take infinity as a reference point. BUT THIS EQUATION IN NOT CONSISTENT WITH THE IDEA THAT IT CREATES POSITIVE ELECTRIC POTENTIAL AS YOU BRING A TEST CHARGE CLOSER TO THE SOURCE POINT CHARGE. What am I missing?

    in the second situation......

    the electric field between a parallel plate (one with +q and the other with -q) from Gauss's law is (q) (permittivity) / Area

    V = - integral of E vector dot dL vector to figure out the electric potential created when taking a positive test charge from negative plate to positive plate. This makes sense because the negative sign indicates the work done by the positive electric field as one brings the positive test charge from the negative plate to the positive plate.

    However, the negative sign disappears from taking the dot product of the general electric potential formula.
    this gives us...

    V = Ed what gives??????!!!!! why can we not do this in situation 1????
  2. jcsd
  3. Jul 28, 2008 #2
    Please Help Me!!!!

    Any Hw Tutor Or Mentor!!!

    Help Me!!!!
  4. Jul 28, 2008 #3


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    Homework Helper

    The formula is [tex]V = -\int^{r_A}_{\infty} \textbf{E} \cdot d\textbf{r}[/tex] where in this case [tex]d\textbf{r}[/tex] is in the same direction as the field, not pointing in the direction along the path on which you perform the line integral. The path can be described parametrically in spherical coordinates:

    [tex]\textbf{r}(t) = t\hat{r}[/tex]. So, [tex]\textbf{E} = \frac{kq}{t^2}[/tex] and [tex]\textbf{r}'(t) = \hat{r}[/tex]. So plugging this into the line integral:

    [tex]-\int^{r_A}_{\infty} \frac{kq}{t^2} dt[/tex] which gives [tex]\frac{kq}{r_A}[/tex].

    In this case, imagine a simple setup with the negative plate at x=0 and the positive plate at x=a. The electric field is then given in 1D as [tex]\frac{q}{\varepsilon_0 A}[/tex], in the negative x-direction. Invoking the formula for potential, we replace the lower limit of integration with x=0 since we're assuming that the potential at the negative plate is 0V for reference. Hence we have [tex]V = -\int^a_0 -\frac{q}{\varepsilon_0 A} dx[/tex]. This evaluates to [tex]\frac{qa}{\varepsilon_0 A}[/tex].
    Last edited: Jul 28, 2008
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