Potential, field, Laplacian and Spherical Coordinates

[AdkinsJr]

Homework Statement

Say I am given a spherically symmetric potential function V(r), written in terms of r and a bunch of other constants, and say it is just a polynomial of some type with r as the variable, \frac{q}{4\pi\varepsilon_o}P(r), and we are inside the sphere of radius R, so r<R…

Homework Equations

\vec E =-\vec\nabla V

The operator should reduce since there is are no components for phi or theta, so in spherical
\vec\nabla =\frac{\partial}{\partial r}\hat r

So is it that simple? Just compute the gradient?

 

[vanhees71]

Your expression doesn't make much sense, at least I don't understand it. The gradient in spherical coordinates reads
\vec{\nabla} V=\vec{e}_r \partial_r V + \vec{e}_{\vartheta} \frac{1}{r} \partial_{\vartheta} V + \vec{e}_{\varphi} \frac{1}{r \sin \vartheta} \partial_{\varphi} V.
Now you can calculate the gradient of the potential, which is the electric field.

 

[AdkinsJr]

The expression I gave was just meant to be the operator, not the actual gradient, but I omitted the other terms of the operator because my function is of the form V(r)=\frac{q}{4\pi \varepsilon_o}P(r) where P(r) is just a polynomial with r as the only variable. So infact the operator would be an ordinary derivative, so can I just apply \frac{d}{dr}\hat r to V(r) to get my field?

The question is arising from a homework problem although I don't really want to post the exact problem on the net, it's not from a textbook or anything. He has just given us this function V(r) and I was just skeptical about the approach I was taking to find the field.

 

[vanhees71]

As I said, I don't understand your expression. If \hat{r}=\vec{e}_r is the unit vector in r direction then its derivative wrt. r is obviously 0. The gradient in spherical coordinates is as given in my previous posting.