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E field from potential

  1. Jan 11, 2010 #1
    Say you have a charge distribution in the picture, and you want to find the potential THEN the electric field at P, using [tex]E = -\nabla V[/tex]

    I found the potential to be 0, and yet the field isn't. Why is this?

    And how would we get the actual field from the voltage?

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    Last edited: Jan 11, 2010
  2. jcsd
  3. Jan 11, 2010 #2


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    Science Advisor
    Gold Member

    Potential is arbitrary. It is easy to see, from the relationship that the electric field is the gradient of the potential, that we can give the potential any DC offset and not affect the resulting electric field. What is of true importance for us is the potential difference.

    For point charges, it is often simple to set the potential at infinityto be zero. Under such an assumptions, then yes, the potential at point P, or in fact at any point along the z-axis (the dotted line in the picture) the potential will be zero. However, the electric field is related to the gradient of the potential. So just because a function evaluates to zero at a point does not mean that the derivative of the function evaluates to zero at that point as well.

    If you were to work out the gradient of the general expression for the potential at any point, then you would find that if you solved for the field at point P it would be non-zero.
  4. Jan 11, 2010 #3
    Oh, ok, so by analogy, if you find the potential anywhere, it's like finding

    [tex]f(x)[/tex], from which you can find [tex]f'(x)[/tex]

    However if you find the potential only at 1 point or a set of points, it's like finding

    [tex]f(x_0)[/tex], and from that [tex]\frac{d}{dx}f(x_0) = 0[/tex] since it's a constant?

    Is that kinda right?

  5. Jan 11, 2010 #4


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    Allow me to make an analogy... you know that for a function of a single variable, the derivative at a point is the slope of a graph of the function at that point, right? What you're saying is that if you find the value of a function at only one point, then the slope has to be zero at that point. Does that make sense? :smile:

    In order to find the slope at a point, you need to know the values of the function not only at the point in question, but also at neighboring points. Either you know an explicit value at some particular nearby point, in which case you calculate the slope approximately as [itex]\Delta y / \Delta x[/itex]; or you know a formula for the function, in which case you find the derivative of the formula (which gives you the slope at all points) and evaluate it at the point in question.

    In this case you have to find a general formula for the potential at all points, and then find the negative gradient of that formula (which gives you the field at all points) and evaluate it at the point in question.
  6. Jan 11, 2010 #5
    Ah ok thanks jtbell and Born2bwire
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