Recent content by MrCoffee2004

You're lucky your professor uses Griffiths. Mine uses Reitz, its terrible by comparison. I brought Griffiths to supplement where my book is weak. They do compliment each other though... Good luck with your course

I don't think that is right. If it were, the Br2 would have dissolved into the hexane or toluene. The Br was in an ionic state in the aqueous solution and that is why it didn't dissolve.

Aqueous bromine is bromine dissolved in a water based solution. Usually HBr and H20. Your bromine preparation did not mix with either hexane or toluene because of their non-polar nature. They are non-miscible with water. Gaseous bromine, either Br2(g) or HBr(g), would dissolve however.

Never mind guys, solution came as a superposition of the binomial expansion of the above terms...fell apart and was easy to show to be proportional to legendre polynomial of degree n=2. May post the rest of the work if it will help anyone here.

Ok so yea, someone give me a hand? I am basically screwed here.

Forgot to ask, with respect to which variable will I be doing this expansion?..I believe that r and r' are both in the eqn.

Yea I also started doing a superposition of potentials... \varphi = \frac{1}{4\pi\epsilon_{0}} [ \frac{q}{\sqrt{x^{2} + y^{2} + (z+L)^{2}}} + \frac{q}{\sqrt{x^{2} + y^{2} + (z-L)^{2}}} + \frac{-2q}{\sqrt{x^{2} + y^{2} + z^{2}}} ] Noting that r^{2} = {\sqrt{x^{2} + y^{2} + z^{2}}}...

Homework Statement a) Find the potential of an axial quadrupole: point charges q, -2q, and q placed on the z-axis at distances L, 0, and -L from the origin. b) Find the potential only at distances r>>L. c) Show that this potential is proportional to one of the zonal harmonics.Homework...