Conservative Vector Field Potential

tazzzdo
Messages
47
Reaction score
0

Homework Statement



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

Homework Equations



All of the above.

The Attempt at a Solution



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

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

r2 = x2 + y2 + z2

By implicit differentiation:

2r\frac{∂r}{∂x} = 2x \Rightarrow \frac{∂r}{∂x} = x/r

And as follows:

\frac{∂r}{∂y} = y/r, \frac{∂r}{∂z} = z/r

\frac{x}{r} \cdot \frac{1}{r} = r-1\frac{x}{r} = r-1\frac{∂r}{∂x} = \frac{∂}{∂x}(log r)

\Rightarrow 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.
 
Physics news on Phys.org
It looks fine to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top