Solving Possion's equation: electrostatic potential

1. Sep 28, 2008

LocationX

I have the problem and my work uploaded to this webpage. I believe I have the general idea down, but I'm just afraid that I've done a careless calculation and missed a minus sign somewhere. I feel that my results have a lot of negative signs, making the potential constantly negative. I'm not sure if this is what is expected. Any help is appreciated.

2. Sep 28, 2008

gabbagabbahey

You've made an error when applying the boundary conditions:

$${\varphi}_{I}(R)={\varphi}_{II}(R), \quad {\varphi}_{II}(R+d)={\varphi}_{III}(R+d)$$

You've for some reason interchanged your constants $C_0$ and $C_1$ with $C_3$ and $C_4$; you can't do this.

You have 4 unknown constants to solve for, so you will need 4 boundary conditions. Luckily, $\varphi$ isn't the only thing that is continuous; $\vec{E}(r)=\vec{\nabla} \varphi$ must be continuous as well. This should give you your other two boundary conditions.

3. Sep 28, 2008

LocationX

$${\varphi}_{I}(r)=C_1^I$$
$${\varphi}_{II}(r)=-\frac{4 \pi r^2 \rho }{6} + \frac{C_0^II}{r}+C_1^II$$
$${\varphi}_{III}(r)=\frac{C_0^{III}}{r}$$

$${\varphi}_{I}(R)=C_1^I$$
$${\varphi}_{II}(R)=-\frac{4 \pi R^2 \rho }{6} + \frac{C_0^II}{R}+C_1^II$$
$${\varphi}_{II}(R+d)=-\frac{4 \pi (R+d)^2 \rho}{6} + \frac{C_0^II}{R+d}+C_1^II$$
$${\varphi}_{III}(R+d)=\frac{C_0^{III}}{R+d}$$
$$\frac{d {\varphi}_{I}{dr} |_{r=R} = 0$$
$$\frac{d {\varphi}_{II}{dr} |_{r=R} = \frac{8 \pi R \rho }{6} - \frac{C_0^{II}}{R^2}$$
$$\frac{d {\varphi}_{II}{dr} |_{r=R+d} = \frac{8 \pi (R+d) \rho }{6} - \frac{C_0^{II}}{(R+d)^2}$$
$$\frac{d {\varphi}_{III}{dr} |_{r=R+d} = - \frac{C_0^{III}}{(R+d)^2}$$

We use the 4 B.C.:

$${\varphi}_{I}(R) = [tex]{\varphi}_{II}(R)$$
$${\varphi}_{II}(R+d) = {\varphi}_{III}(R+d)$$
$$\frac{d {\varphi}_{I}{dr} |_{r=R} = \frac{d {\varphi}_II}{dr} |_{r=R}$$
$$\frac{d {\varphi}_{II}{dr} |_{r=R+d} = \frac{d {\varphi}_III}{dr} |_{r=R+d}$$

I've already caught 2 minus sign errors... but this is how i'm setting up everything to solve for the 4 constants.

4. Sep 28, 2008

gabbagabbahey

Why not write it as:

$${\varphi}_{I}(r)=C_0$$

$${\varphi}_{III}(r)=\frac{C_1}{r}$$

$${\varphi}_{II}(r)=-\frac{4 \pi r^2 \rho }{6} + \frac{C_2}{r}+C_3$$

to get rid of those cumbersome superscripts and make it a little easier to work with?

5. Sep 28, 2008

gabbagabbahey

You are still missing a couple of negative signs:

$$\frac{d {\varphi}_{II}}{dr} |_{r=R} = -\frac{8 \pi R \rho }{6} - \frac{C_0^{II}}{R^2}$$

$$\frac{d {\varphi}_{II}}{dr} |_{r=R+d} = -\frac{8 \pi (R+d) \rho }{6} - \frac{C_0^{II}}{(R+d)^2}$$

What do you get for your constants now?

6. Sep 29, 2008

LocationX

7. Sep 29, 2008

gabbagabbahey

You are missing a negative sign for C_2, which is affecting your other constants.

8. Sep 29, 2008

LocationX

thanks for the help, i'm confident that i do not have any more errors after solving for the e-field and checking the BC and limiting cases