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Conservative Vector Field Potential

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]\vec{E}[/itex]([itex]\vec{r}[/itex]) = [itex]\vec{r}[/itex]/r2, r = |[itex]\vec{r}[/itex]|, [itex]\vec{r}[/itex] = x[itex]\hat{i}[/itex] + y[itex]\hat{j}[/itex] + z[itex]\hat{k}[/itex] be a vector field in ℝ3. Show that [itex]\vec{E}[/itex] is conservative and find its scalar potential.

    2. Relevant equations

    All of the above.

    3. The attempt at a solution

    [itex]\vec{\nabla}[/itex] [itex]\times[/itex] [itex]\vec{E}[/itex] = [itex]\vec{0}[/itex] [itex]\Rightarrow[/itex] [itex]\vec{E}[/itex] is conservative [itex]\Rightarrow[/itex] [itex]\vec{E}[/itex] = [itex]\vec{\nabla}[/itex]f, where f is a scalar function.

    xf = x/r2
    yf = y/r2
    zf = z/r2

    r2 = x2 + y2 + z2

    By implicit differentiation:

    2r[itex]\frac{∂r}{∂x}[/itex] = 2x [itex]\Rightarrow[/itex] [itex]\frac{∂r}{∂x}[/itex] = x/r

    And as follows:

    [itex]\frac{∂r}{∂y}[/itex] = y/r, [itex]\frac{∂r}{∂z}[/itex] = z/r

    [itex]\frac{x}{r}[/itex] [itex]\cdot[/itex] [itex]\frac{1}{r}[/itex] = r-1[itex]\frac{x}{r}[/itex] = r-1[itex]\frac{∂r}{∂x}[/itex] = [itex]\frac{∂}{∂x}[/itex](log r)

    [itex]\Rightarrow[/itex] f = log r


    This is an extra credit question in my Vector Calculus class. I think this is the correct solution, but I'm not positive.
     
  2. jcsd
  3. Mar 28, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It looks fine to me.
     
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