How to Determine Scalar Potential Inside and Outside a Charged Sphere?

[dsta]

Homework Statement

Use Poisson's equation and Laplace's equation to determine the scalar potential inside and outside a sphere of constant charge density p_o. Use Coulomb's law to give the limit at very large r, and an argument from symmetry to give the value of E at r=0.

Homework Equations

del²V = - p/epsilon_o
del²V = 0

The Attempt at a Solution

Not sure how to start this just using Poisson's/Laplace's equation.

 

[gabbagabbahey]

What is $\nabla^2V$ in spherical coordinates?...Are there any symmetries that allow you to simplify it?

 

[dsta]

Hmm okay. I'm not sure what symmetries you would need to consider to simplify $\nabla^2V$ in spherical coordinates. Because the charge density is constant, V does not depend on r inside the sphere...I'm not sure about anything else.

 

[gabbagabbahey]

What makes you think $V$ doesn't depend on $r$?

Surely you can say that it doesn't depend on either the azimuthal or polar angles though right?...After all, a test charged placed at any given value of $r$, will "see" the same charge distribution at say $\theta=\phi=\pi/4$ as it would at say $\theta=3\pi$, $\phi=\pi/7$ or any other angle wouldn't it? (Draw a picture to convince yourself of this!)...This type of symmetry is called spherical symmetry.

 

[dsta]

Oops sorry, I meant to say that. Okay so using the simplified form of the equation for $\nabla^2V$ in spherical coordinates, and Poisson's equation, I was able to get the equation for V inside the sphere. For outside the sphere, the charge density is 0 obviously, so you have to use Laplace's equation. I tried to do this but I can't anywhere near the form of the equation I'm meant to be getting

 

[gabbagabbahey]

How about you show me your steps (for both inside and outside) and I'll see if I can spot where you are going wrong...

 

[dsta]

Inside the sphere:
$$\nabla^2V = \frac{1}{r}$$$$\frac{d}{dr}(r^2\frac{dV}{dr}) = \frac{\rho_{o}}{\epsilon_{o}}$$
$$\Rightarrow V = - \frac{\rho_{o} r^2}{6\epsilon_{o}}$$

Outside the sphere:
$$\nabla^2V = \frac{1}{r}$$$$\frac{d}{dr}(r^2\frac{dV}{dr}) = 0$$
$$\Rightarrow r^2\frac{dV}{dr} = constant = a$$
$$\Rightarrow V = \frac{-a}{r} + c$$
I'm assuming that you now have to use some sort of boundary conditions...

 

[gabbagabbahey]

Inside the sphere:
$$\nabla^2V = \frac{1}{r}$$$$\frac{d}{dr}(r^2\frac{dV}{dr}) = \frac{\rho_{o}}{\epsilon_{o}}$$
$$\Rightarrow V = - \frac{\rho_{o} r^2}{6\epsilon_{o}}$$

You're missing a minus sign in your first equation (although it looks like that's just a typo, since your answer has the correct sign)...More importantly, shouldn't you have two constants of integration for the solution to this 2nd order ODE?