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Calculating the electric field from the potential

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the magnitude of the electric field at the point (3.00[tex]\hat{i}[/tex] - 2.00[tex]\hat{j}[/tex] + 4.00[tex]\hat{k}[/tex])m if the electric potential is given by V = [tex]2.00xyz^2[/tex], where V is in volts and x, y, and z are in meters?

    2. Relevant equations

    To calculate the field from the potential, [tex]E_{s}=-\delta V/ \delta s[/tex] (ie [tex]E=-dV/ds[/tex] for each component).

    3. The attempt at a solution


    I don't know what they mean by [tex]2.00xyz^2[/tex].
    Last edited: Mar 25, 2008
  2. jcsd
  3. Mar 25, 2008 #2


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    It is a function describing V. What is actually confusing you about it? Is it that it involves more than one variable?
  4. Mar 26, 2008 #3
    Haha...yes. It's probably something really simple. I just don't know what to do to find each component
  5. Mar 28, 2008 #4


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    O ok. Maybe it is the concept of the partial derivative that is confusing you?

    For instance when you want to find the x component of the field you need to find:

    [tex]\frac{\partial V}{\partial x}[/tex].

    In this case, just think of y and z as constants. Similarly, think of x as constant when taking the derivatives with respect to the other variables.

    Does this help?
  6. Mar 30, 2008 #5
    [tex]2.00yz^2 \frac{\partial x}{\partial x}

    [tex]2.00xz^2 \frac{\partial y}{\partial y}

    [tex]2.00xy \frac{\partial z^2}{\partial z}

    [tex]\sqrt{64^2 + 96^2 + 48^2}

    The answer in the back of the book is 150 N/C.
  7. Mar 30, 2008 #6


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    Hi paralian! :smile:

    It should be [tex]2.00xy \frac{\partial z^2}{\partial z}

    That should give you 150.09. :smile:
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