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.

 

[paranormal]

Yes, in this case, computing the gradient would be the first step in finding the electric field inside the sphere. However, since the potential function is given in terms of spherical coordinates, it would be more efficient to use the Laplacian operator in spherical coordinates to find the electric field. The Laplacian in spherical coordinates is given by \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}. Using this operator, you can solve for the electric field inside the sphere by setting the Laplacian of the potential function equal to the negative of the charge density \rho inside the sphere. This will give you the necessary equations to find the electric field in terms of spherical coordinates.