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Electric Potential

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data
    An electric field is given by E = 2xî - 3y²j N/C. Find the change in potential from the position RA = î - 2j M to RB = 2î + j +3k m.

    2. Relevant equations
    ∆V = VA - VB = - ∫ E • ds (definite integral from A to B)
    Or in a uniform field, simply:
    ∆V = VA - VB = ±Ed

    3. The attempt at a solution
    I know from the solutions manual that the answer is +6 Volts.

    I began by finding ∆R = RB - RA = i + 3j + 3k m.
    That can easily be converted into ∆R = √19 right?

    My problem comes in when I have to deal with E = 2xî - 3y²j N/C.... what do I do with the square on the y?

    Thank you once again.
  2. jcsd
  3. Feb 7, 2007 #2


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    Have you had much experience with line integrals? Oh well...let me address specific aspects of your solution attempt first:

    Not exactly. Be careful! [itex] \mathbf{r_A} [/itex] and [itex] \mathbf{r_B} [/itex] are vectors! So you were correct in stating that:

    [tex] \mathbf{\Delta r } = \mathbf{r_B - r_A} = \mathbf{\hat{i}} + 3\mathbf{\hat{j}} + 3\mathbf{\hat{k}} [/tex].

    But it doesn't make sense to say that [itex] \mathbf{\Delta r } = \sqrt{19} [/itex], because you're equating a vector to a scalar! Really, what you calculated was the magnitude of [itex] \mathbf{\Delta r } [/itex]:

    [tex] |\mathbf{\Delta r}| = \sqrt{19} [/tex]

    Unfortunately, I don't think that this is going to be of much use to you in solving the problem! I wouldn't start with calculating the difference between the two vectors, because to calculate the potential difference, you need to calculate the line integral of the electric field over some path between [itex] \mathbf{r_A} [/itex] and [itex] \mathbf{r_B} [/itex]. You correctly wrote:

    [tex] \Delta V = V(\mathbf{r_B}) - V(\mathbf{r_A}) = - \int_{\mathbf{r_A}}^{\mathbf{r_B}}{\mathbf{E} \cdot d\mathbf{l}} [/tex]

    But then...

    ...this indicates you're stuck. But if you haven't calculated too many line integrals or taken vector calculus, don't despair. There is a half-assed, "physics" way to interpret it :rofl:

    Think of [itex] d\mathbf{l} [/itex] as an infinitesimal vector that points "along" the direction of the path at any point on the path. Then:

    [tex] d\mathbf{l} = dx\mathbf{\hat{i}} + dy\mathbf{\hat{j}} + dz\mathbf{\hat{k}} [/tex]

    Therefore, you can calculate:

    [tex] \mathbf{E} \cdot d\mathbf{l} = (2x\mathbf{\hat{i}} - 3y^2 \mathbf{\hat{j}}) \cdot (dx\mathbf{\hat{i}} + dy\mathbf{\hat{j}} + dz\mathbf{\hat{k}} ) = ? [/tex]

    your integral will turn into two integrals, one wrt x and the other wrt y, and you're off to the races. :smile:
  4. Feb 7, 2007 #3


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    Uh, looking at the question again, I think I made the further assumption that since the E-field is a conservative vector field, then the answer is path independent, so you can simply use your starting and ending x & y points as limits of integration for each integral (i.e., take the straight line path).
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