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Pontential Difference

  1. Feb 22, 2010 #1
    I need to calculate the potential difference between A and B.
    I am having trouble with this problem. I know that change in voltage = ES. I tried finding the distance between the two points and calculaing the difference but that did not seem to work.

    Any suggustions???

    Attached Files:

  2. jcsd
  3. Feb 22, 2010 #2


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    Staff: Mentor

    Welcome to the PF. What are the Relevant Equations involved in this problem? You should know an equation that relates the Voltage, the Electric Field, and the path from one point to the other....
  4. Feb 22, 2010 #3

    the electric potential is defined as

    [tex] \vec E(\vec r) = - \vec \nabla V(\vec r)[/tex]​

    in your sketch the electric field vector can arbitrarily be defined as

    [tex] \vec E(\vec r) = E_0 ~ \vec e_x \qquad \mbox{with} \qquad E_0 = 1000 ~ \mathrm{V}/\mathrm{m}[/tex]​

    because only one component of the electric field is nonzero the equation, which defines the electric potential simplifies to

    [tex] E_0 = - \frac{\partial V(x)}{\partial x}[/tex]​

    That is a very easy to solve differential equation. Integration of both sites yields

    [tex]V(x) = - E_0 x + V_0[/tex]​

    (The electric field vector must points at regions of lower electric potential)
    You see that the electric potential doesn't depend on [tex]y[/tex] and [tex]z[/tex].

    Thus we have the electric potential at two points

    [tex]V_{\mbox{\small A}}(x_A) \qquad \mbox{and} \qquad V_{\mbox{\small B}}(x_A+7cm) [/tex]​

    Know it's your turn to calculate the potential difference between these points ;)

    i hope i could help you!?

    with best regards
  5. Feb 22, 2010 #4
    You have the right equation, but what distance are you using? Remember that is a line integral, meaning E and D are multiplied via dot product.
  6. Feb 22, 2010 #5
    Thank for all the help guys, that cleared it up. I was getting cofused on the y and x directions. I didn't realize that point A is actually at zero.

    But why is the difference negative???
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