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

1. Sep 30, 2008

### jrc5135

1. The problem statement, all variables and given/known data
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.

2. Relevant 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))

3. 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?

2. Sep 30, 2008

### 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?

3. Oct 1, 2008

### 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.

4. Oct 1, 2008

### 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?

5. Oct 1, 2008

### 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

6. Oct 1, 2008

### 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?

7. Oct 1, 2008

### jrc5135

not really sure how

8. Oct 1, 2008

### gabbagabbahey

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