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Electric field

  1. Jan 29, 2004 #1
    The problem is:

    Prove that the gradient of an electric potential V(r) which depends only on the distance r=(x^2 + y^2 + z^2)^1/2 from the origin has the vlaue gradV(r) = V'(r)r-hat where r-hat := r/r is a unit vector in the direction of r, and V'(r) := dV(r)/dr. Use this to evaluate the electric field E= -gradV(r) for the Coulomb potential
    V(r)= kq/r from a point charge +q, where k=1/(4*pi*E0)

    E0 stands for epsilon.

    My question is:

    I proved that gradV(r) = V'(r)r-hat. How do I evaluate the electric field for the Coloumb potential? Would I take the gradient of V(r)=kq/r? I don't get what I need to do.
     
  2. jcsd
  3. Jan 29, 2004 #2

    Doc Al

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    Staff: Mentor

    Yes.
     
  4. Jan 29, 2004 #3
    does this look right?

    This is the answer i got:

    kq*(x^2+y^2+z^2)^1/2 /(xi+yj+zk)
     
  5. Jan 29, 2004 #4

    Doc Al

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    Staff: Mentor

    Re: does this look right?

    Check your work. Your answer must be equivalent to:
    [tex]\frac{kq}{r^2}\hat{r}[/tex]
     
  6. Jan 29, 2004 #5
    i tried it again

    this time i got, -kq(xi+yj+zk)/(r^5)
     
  7. Jan 30, 2004 #6

    Doc Al

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    Staff: Mentor

    I'm not sure what you're doing, but keep trying. You're probably making some simple error. Here's how I did it:

    If you use spherical coordinates, the gradient is trivial to calculate. By symmetry, the gradient must be along the [itex]\hat{r}[/itex] direction. So E = -grad(V)= -kq(∂/∂r)(1/r), which gives [itex]\frac{kq}{r^2}\hat{r}[/itex].

    If you wish to use cartesian coordinates, no problem:
    [tex]V = \frac{kq}{(x^2 + y^2 + z^2)^\frac{1}{2}}[/tex]
    so, -grad(V) =
    [tex]\frac{kq}{(x^2 + y^2 + z^2)^\frac{3}{2}}(x\hat{i}+y\hat{j}+z\hat{k}) = \frac{kq}{r^2}\hat{r}[/tex]
     
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