Recent content by ApexOfDE

Griffith's Classical electrodynamics p.147 has a detailed explanation for multipole expansion. You may take a look on the that book.

Homework Statement Consider a 3-level laser. Give the occupation numbers N1(R), N2(R), the inversion \DeltaN(R)/N = [N2(R) - N1(R)]/N and the flux below and above the threshold as a function of pump rate R. F - the photon flux Homework Equations The Attempt at a Solution I have...

my current bg: [PLAIN]http://img293.imageshack.us/img293/8058/vistawallpaperaerowoods.jpg

I don't understand. When we calculate the electrostatic energy, we must integrate in all space (from zero to infinity) and the formula depends only in electric field and doesn't mention the material. In region (3), there is a field here. Therefore, I think W_3 must not be zero.

I think you have no mistake with your calculation and there is perhaps typo in the book.

After step 3, the potential at point P is: V = \frac{q}{2 \pi \epsilon_0 z} [\ln \frac{d-x}{x} - \ln \frac{d-a}{a}] Now plug V = 12 and x = d-a into this formula, and you will have: 12 = -2 \frac{q}{2 \pi \epsilon_0 z} \ln \frac{d-a}{a} This will yield constant "6" in your solution.

use your eyes.

Sry for this inconvenience. I will try again. 1/ You should write down the form of electric field caused by a cylinder conductor at point x outside the conductor. 2/ Find total field at point P (beware the sign of RHS field). 3/ Find potential (integrate from a to d-a) 4/ After getting step...

Firstly, you find the total electric field at P. From this result, you calculate the potential at point P (from a to d-a). When you arrive this point, consider conditions given in problem: VL = 0 (x = a) and VR = 12 (x = d-a). Plug these into the potential you just found to determine some constants.

In normal condition, how can you have a free proton? Electron, proton, neutron exist in our body in the form of molecular. Therefore, I think the only dipole we may have is induced dipole, which is created only by outer field. Moreover, spark is something like current, and current is the flow...

in sphere, integral intervals [r inf] ==> [r R] + [R inf]. Because there is no field inside sphere, only the 2nd integrals survives.

in conductors, all charges will concentrate on the surface. That's why there is no field inside a conducting sphere. However, there is field outside the sphere. Calculate it, then by taking its integrals (from r to inf.) to find potential inside sphere. p/s: I am confused with cylinder. Can...

iirc, Tesla coil was used some experiments in transmitting radio in late-19th and early-20th century. If you are interested in Tesla coil, read "Tesla - man out of time" (first 10 chapters covers the period Tesla built the coil). You can also take a look on "The ultimate Tesla coil design...

Its true and we can consider that function as some kind of constant. But it is useless and just makes our problem more complicated. from CE book (3rd edition - Jackson): "The expression for energy density is intuitively reasonable, since regions of high fields 'must' contain considerable...

General formula for E is: dE = 1/4πεo (dq/r^2)\hat{r} However, r = r.\hat{r} (r_hat is unit vector) So you can write" dE = 1/4πεo (dq/r^3).r -- Electric field of a point above distance Z from a line is a cumulative Electric field caused by every charge in straight line. Lets...