Solving Griffiths 4.10: Calculating V(r) Using Equation 4.13

[ehrenfest]

[SOLVED] Griffiths 4.10

Homework Statement

This question refers to Griffths E and M book.

Homework Equations

The Attempt at a Solution

part a was easy.
For part b, I am trying to calculate V(r) using equation 4.13 and I am facing an absurdly difficult integration which means I must be doing something wrong.
V(\vec{r}) = \frac{1}{4 \pi \epsilon_0}\left(\int_0^{2\pi} \int_p^{\pi}\frac{\sigma_b}{|\vec{r}-\vec{r&#039;}|}R^2 \sin \theta&#039; d\theta&#039; d\phi&#039; + \int_0^{2\pi} \int_0^{\pi} \int_0^R \frac{\rho_b}{|\vec{r}-\vec{r&#039;}|}r&#039;^2 \sin \theta&#039; dr&#039; d\theta&#039; d\phi&#039; \right)<br />
It is those denominators that are going to kill me!

 

[kdv]

Homework Statement

This question refers to Griffths E and M book.

Homework Equations

The Attempt at a Solution

part a was easy.
For part b, I am trying to calculate V(r) using equation 4.13 and I am facing an absurdly difficult integration which means I must be doing something wrong.
V(\vec{r}) = \frac{1}{4 \pi \epsilon_0}\left(\int_0^{2\pi} \int_p^{\pi}\frac{\sigma_b}{|\vec{r}-\vec{r&#039;}|}R^2 \sin \theta&#039; d\theta&#039; d\phi&#039; + \int_0^{2\pi} \int_0^{\pi} \int_0^R \frac{\rho_b}{|\vec{r}-\vec{r&#039;}|}r&#039;^2 \sin \theta&#039; dr&#039; d\theta&#039; d\phi&#039; \right)<br />
It is those denominators that are going to kill me!

Have you tried using eq 3.94?
The sigma_b and rho_b are not complicated right? So it should not be too hard.

 

[ehrenfest]

I don't see how inserting an infinite series into the integral will make it simpler. I don't see any way to make 3.94 not an infinite series in this case.

Yes sigma_b and rho_b are not complicated.

 

[kdv]

I don't see how inserting an infinite series into the integral will make it simpler. I don't see any way to make 3.94 not an infinite series in this case.

Yes sigma_b and rho_b are not complicated.

The point of using the infinite series is that you can then use the orthonormality of the polynomials to do all the angular integrations trivially! Have you used 3.94 in an actual calculation? That's very powerful. What are sigma_b and rho_b?

 

[ehrenfest]

\rho_b = -2k/r
and
\sigma_b = Rk

In equation 3.94 \theta&#039; is the angle between r and r', not the zenith angle.

 

[ehrenfest]

Never mind. I got it. kdv, there is a much easier way to do it than to use eqn 3.94. BTW the bound charges I gave are wrong.