Potential, field, Laplacian and Spherical Coordinates

  • Thread starter AdkinsJr
  • Start date
  • #1
150
0

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, [tex]\frac{q}{4\pi\varepsilon_o}P(r)[/tex], and we are inside the sphere of radius R, so r<R…


Homework Equations



[tex]\vec E =-\vec\nabla V[/tex]

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

So is it that simple? Just compute the gradient?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
17,596
8,591
Your expression doesn't make much sense, at least I don't understand it. The gradient in spherical coordinates reads
[tex]\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.[/tex]
Now you can calculate the gradient of the potential, which is the electric field.
 
  • #3
150
0
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 [tex]V(r)=\frac{q}{4\pi \varepsilon_o}P(r)[/tex] 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 [tex]\frac{d}{dr}\hat r[/tex] 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 text book 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.
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
17,596
8,591
As I said, I don't understand your expression. If [itex]\hat{r}=\vec{e}_r[/itex] is the unit vector in [itex]r[/itex] direction then its derivative wrt. [itex]r[/itex] is obviously 0. The gradient in spherical coordinates is as given in my previous posting.
 

Related Threads on Potential, field, Laplacian and Spherical Coordinates

Replies
3
Views
29K
Replies
5
Views
3K
Replies
4
Views
11K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
889
Replies
4
Views
3K
  • Last Post
Replies
3
Views
1K
Top