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tazzzdo
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Homework Statement
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.
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.
∂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.