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Electric Field

  1. Jul 22, 2009 #1
    Can someone remind me why the electric field is defined as the negative gradient times the electric potential, rather than the gradient times the electric potential?

    Thanks,


    JL
     
  2. jcsd
  3. Jul 22, 2009 #2

    jix

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    If you move up an electric field, the potential increases.

    Suppose that for points A and B, the potential at B is higher. Then moving from A to B, the distance is, say, positive, and so the potential difference is also positive. But then the eletric field is directed from B to A, and so is negative, which is accounted for by using the negative gradient.

    I hope that helps.
     
  4. Jul 22, 2009 #3

    rock.freak667

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    Homework Helper

    [tex]V = \frac{kQ}{r}[/tex]

    [tex]\frac{dV}{dr}= \frac{-kQ}{r^2} \times \frac{q}{q}[/tex]

    [tex]\frac{dV}{dr} = \frac{ \frac{-kQq}{r^2}}{q} = - \frac{F}{q}[/tex]

    [tex]\frac{dV}{dr}= -E \Rightarrow E= - \frac{dV}{dr}[/tex]
     
  5. Jul 22, 2009 #4
    That helps a lot, I was looking for a conceptual explanation.

    I will try to understand this later, I'm not sure about that cross or is it cross product.


    Thanks guys,


    Jeffrey
     
  6. Jul 22, 2009 #5

    rock.freak667

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    Homework Helper

    No cross-product, just multiplication.
     
  7. Jul 23, 2009 #6
    Here's another way they derived the formula:

    Work-Energy Theorem relates the potential difference to work as:

    [tex]\Delta U=U_{f}-U_{i}=-W[/tex]

    If the "electrical potential energy", [tex]U[/tex], at infinity is defined to be 0, then the electrical potential energy of a single point charge is defined as:

    [tex]U=-W[/tex]

    To define the electric potential of a charge, we divide it's energy by it's charge.

    [tex]V=\frac{-U}{q}[/tex]

    The electrical potential difference from infinity to a point is therefore:

    [tex]\Delta V=V_{f}-V_{i}=-\frac{W}{q}[/tex]

    Subbing the definition of W from mechanics:

    [tex]V=- \frac{ \int \textbf{F} \cdot d\textbf{r}}{q}[/tex]

    [tex]V=-\frac{1}{q}\int\textbf{F} \cdot d\textbf{r}[/tex]

    [tex]V=-\frac{1}{q}\int\frac{kqq}{r^{2}}\cdot d\textbf{r}[/tex]

    [tex]V=-\int\frac{kq}{r^{2}}\cdot d\textbf{r}[/tex]

    [tex]V=-\int E \cdot d\textbf{r}[/tex]

    Taking the gradient of both sides, you get:

    [tex]\nabla V=-\nabla\int \textbf{E} \cdot d\textbf{r}[/tex]

    [tex]\nabla V=-\textbf{E}[/tex]

    [tex]\textbf{E}=-\nabla V[/tex]
     
    Last edited: Jul 23, 2009
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