General solution to Laplace's equation where V depends only on r

[jrc5135]

Homework Statement

Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends on on r. Do the same for cylindrical coordinnates assuming V depends only on r.

Homework Equations

Laplace's Eq (spherical): 1/r^2 (d/dr)(r^2(dV/dr)) + 1/(r^2sin(theta))(d/dtheta)(sin(theta)(dV/dtheta)) + 1/(r^2sin^2(theta))(d^2V/dphi^2))

The Attempt at a Solution

Having it only depend on r should I just use the first term of the eq. Everything before the first plus.

and, should I use this EQ for V:

V = kq/r?

 

[gabbagabbahey]

Just use V(r,\theta,\phi)=V(r) (that is V depends only on r)

What are \frac{\partial V}{\partial \theta} and \frac{\partial V}{\partial \phi} then?

What does Laplace's equation look like now?

 

[jrc5135]

wouldnt it just be

1/r^2 (d/dr)(r^2(dV/dr)) dV/dr = (-1/4*pi*e0)(q/r^2) and the two r^2 cancel and you get

1/r^2(d/dr)((-1/4*pi*e0)(q)) and that goes to 0 because there are not any r's inside the partial.

 

[gabbagabbahey]

No, Laplace's equation is \nabla^2V=0 NOT \nabla^2V=V_{pointcharge}.

\Rightarrow \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dV(r)}{dr} \right) = 0

Which gives you a second order, ordinary differential equation to solve for V(r).

Why would you think that the Laplacian of V was equal to the potential of a point charge?

 

[jrc5135]

I was saying if you use the point charge formula for V(r) when you do the partial the r^2 that is being multiplied by the partial cancels the r^2 from V'(r) and you get a constant, so when you take the second partial with respect to r, you get 0 and you have that (dell^2)V = 0

 

[gabbagabbahey]

The potential due to a point charge at the origin will satisfy Laplace's equation (except at r=0) but it is not the general solution to Laplace's equation! Can you solve the above ODE I posted?

 

[jrc5135]

not really sure how

 

[gabbagabbahey]

Okay, then I think you need to review the basics of ODE's.