Calculating the Potential using Laplaces Equation

[zerakith]

1. The question asks to prove that for a sphere (radius R) of constant charge density, $$\rho$$, the potential inside the sphere is $$\Phi=\frac{\rho}{6\varepsilon_0}(3R^2-r^2)$$

Homework Equations

$$\nabla^2\Phi=0$$
$$\nabla^2\Phi=\frac{-\rho}{\varepsilon_0}$$

The Attempt at a Solution

This should be relatively simple. Use the expression for the Laplacian in Spherical Polar Coodinates:
$$\nabla^2\Phi=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \Phi}{\partial r}\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial \Phi}{\partial \theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{ \partial^2 \Phi}{\partial \phi^2}$$
Make symmetry arguments to say that only the first term is non-zero. Thus:
$$\frac{\partial}{\partial r}\left(r^2\frac{\partial \Phi}{\partial r}\right)=0$$
Here comes my issue, which can be broken down into two parts. Firstly, is
$$\int f(x)dx\equiv\int_a^x f(x')dx'$$
If not when would you use each?
Secondly, no matter which method i choose i just can not get it to come out, i know the potential is right and can, weirdly, get there by finding the Electric field first but the question asks you explicitly to solve it using the laplacian. Here is the working:
$$\Phi_\text{outside}=\frac{\rho a^3}{3\varepsilon_0r}$$
Which arises from the solution to the laplacian and the condition that the sphere must appear as a point charge as r tends to infinity.
$$\frac{d\left(r^2\frac{d\Phi_\text{Inside}}{dr} \right)}{dr}=-\frac{\rho r^2}{\varepsilon_0}$$
$$\left(r^2\frac{d\Phi_\text{Inside}}{dr}\right)=-\int\frac{\rho r^2}{\varepsilon_0}dr$$
$$\left(r^2\frac{d\Phi_\text{Inside}}{dr}\right)= \frac{ \rho r^3}{3\varepsilon_0}+D$$
$$\frac{d\Phi_\text{Inside}}{dr}=\frac{ \rho r}{3\varepsilon_0}+\frac{D}{r^2}$$
$$\Phi_\text{Inside}= \frac{ \rho r^2}{6\varepsilon_0}-\frac{D}{r}+E$$
Then we know we need phi to be continuous so the potential inside must be equivalent to potential outside at the radius of the sphere. This leads to $$D=\frac{-\rho a^3}{12\varepsilon_0}$$,$$E=\frac{\rho a^2}{12\varepsilon_0}$$. Which is not correct. I cannot see my mistake and have tried this in several way and not managed it either.
Thanks in advance,
Zerakith

 

[capandbells]

You can't have a 1/r term for the potential inside the sphere because it will go to infinity at the center of the sphere. If you integrate from 0 to r instead of just taking an indefinite integral you will see the D/r term goes away.

 

[zerakith]

Am I right in saying that i could simply say D is 0 for physical reasons (i.e the one you described). Also I tried putting the limits as 0 and r if I apply those limits on the two integrations i get simply that:
$$\Phi_\text{inside}=\frac{-\rho r^2}{6 \varepsilon_0}$$
If I only apply it to the first integration, i do indeed get the right answer, I am confused as to why you would use one integration with limits and one without. Also, I was under the vague impression that since the integration with limits in not a definite integral it was equivalent to the indefinite integral (when boundary conditions are applied), am I wrong in this?

Thankyou for the Help,
Zerakith

 

[capandbells]

You should to apply limits to both integrals. When you integrate a second time, you will get a $$\phi(0)$$ term. This is your arbitrary constant and you can find it from the boundary condition $$\phi_{in} = \phi_{out}$$. If you do this you will recover the answer given.

You can discard the D/r term on physical grounds, yes. In fact, you can consider this as one of your boundary conditions. If you apply this boundary condition it is equivalent to doing the first integral as a definite integral. Yes, simply doing the second integral as an indefinite integral and applying the boundary conditions will give you the right answer, but I think you lose a little physical insight.

 

[zerakith]

Ah I understand my error now. Thankyou very much!