Solving Possion's equation: electrostatic potential

[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.

http://sites.google.com/site/physicsinc/phy

 

[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.

 

[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.:

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

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

 

[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?

 

[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?

 

[LocationX]

i've changed the constant scheme over to what was suggested, it's much more reasonable then what I was using. Here are the results, how does it look?

http://sites.google.com/site/physicsinc/phy

 

[gabbagabbahey]

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

 

[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