1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential, field, Laplacian and Spherical Coordinates

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data



    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…


    2. Relevant 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?
     
  2. jcsd
  3. Oct 17, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
  4. Oct 17, 2013 #3
    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.
     
  5. Oct 17, 2013 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted