# Potential, field, Laplacian and Spherical Coordinates

## 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?

## Answers and Replies

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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.

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?
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.