vanhees71
Gold Member
2019 Award
That's the electric potential for the described situation. The sphere is among the few problems which can be solved analytically, because you can apply the method of image charges. It's not too difficult but a bit of algebraic work. Have fun!

Gold Member
Well, I see that they used equation (2.8) on page 61 by neglecting the term with q/|x-y| since it's infinite.

vanhees71
Gold Member
2019 Award
Obviously the solution uses the usual spherical coordinates, where ##\theta## is the angle between ##\vec{r}## and the ##z## axis, i.e., the ##xy##-plane is at ##\theta=\pi/2##, and there the boundary conditions change given by the potential values on the conductors in the upper and the lower hemisphere.

MathematicalPhysicist
Gold Member
Ah, right.

It's three dimensional spherical coordinates...

Gold Member
I am reading now the solution of problem 3.19 from Jackson (if you know of another more complete solution let me know);

here:
http://homerreid.dyndns.org/physics/jackson/jack3c.pdf

And it seems he's stuck at calculating the integral ##\int_0^\infty \frac{\sinh(kz_0)}{\sinh(kL)}kdk##, can you help both of us?

Cheers!

Gold Member
I read someone's remark on solving Jackson's problems:
"Getting any single Jackson problem completely correct could be a life’s work!"

LOL!!! wish I had the time to solve completely correctly the 350 problems in the third edition, but I have exams... :-D

vanhees71
Gold Member
2019 Award
Mathematica gives as a result (of course only valid for ##L>z_0##)
$$\frac{1}{4L^2} \left [\psi'\left (\frac{L-z_0}{2L} \right )-\psi' \left(\frac{L+z_0}{2L} \right ) \right ],$$
where
$$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$$
is the socalled "digamma" function.

MathematicalPhysicist and dextercioby
Gold Member
Mathematica gives as a result (of course only valid for ##L>z_0##)
$$\frac{1}{4L^2} \left [\psi'\left (\frac{L-z_0}{2L} \right )-\psi' \left(\frac{L+z_0}{2L} \right ) \right ],$$
where
$$\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$$
is the socalled "digamma" function.
Does this expression simplifies to the expression in problem 3.19 in Jackson's? i.e with the ##\sec^2##?

Gold Member
I have a question regarding problem 5.5 from Jackson's third edition.

Do I actually need to post it (I mean I assume that you have a legal or illegal copy of the book).

I've found a solution to problem 5.5 from here:
http://pages.uoregon.edu/gbarello/Resources/Papers/Homework/Electrodynamics/HW4.pdf
on page 8 the solution to assignment 5.5b, the last solution has the wrong coefficients, so I wonder where did he go wrong there?

Thanks, jesus and out... :-)

Gold Member
Does someone know if some of the problems in Jackson's third edition can be found in reference books?

There are some problems about superconducting material which I thought may be found in Tinkham's reference.