- #1

tazzzdo

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## Homework Statement

Let [itex]\vec{E}[/itex]([itex]\vec{r}[/itex]) = [itex]\vec{r}[/itex]/r

^{2}, 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 ℝ

^{3}. Show that [itex]\vec{E}[/itex] is conservative and find its scalar potential.

## Homework Equations

All of the above.

## 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.

∂

_{x}f = x/r

^{2}

∂

_{y}f = y/r

^{2}

∂

_{z}f = z/r

^{2}

r

^{2}= x

^{2}+ y

^{2}+ z

^{2}

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.