Recent content by Blistering Peanut

Well another question: I worked out the potential for a capacitor: \phi = \frac{2 V}{\pi}\sum_{n=0}^{\infty}\arcsin{(2a/(\sqrt{(2nL+|z-L|)^2 +(r+a)^2}+\sqrt{(2nL+|z-L|)^2 +(r-a)^2}))} \\ \\ -\arcsin{(2a/(\sqrt{(2nL+|z+L|)^2 +(r+a)^2}+\sqrt{(2nL+|z+L|)^2 +(r-a)^2}))} The discs are located at...

Yes I have been mainly using pen and paper. I was getting sick of those long expressions (and they double in length for the capacitor) so I started using mathematica. However since I was not used to watching out for singularities etc I started running into problems. I've learned now though which...

I think I've got it now. Thank you very much for your help. I have learned a lot today and hopefully I'll be able to apply this to the rest of the problem. nrged sorry, I was still confused at that stage and I don't know why I posted that.

I solved the laplace equation in cyclindrical coordinates with the appropriate boundry condition. Weber's discontinuous integrals really helped. Are you using the Series function in mathematica? Why does the sign matter so much when taking the square root of the square in this problem? This...

Yes mine is the same as yours. What do you mean by a few taylor series for the big denominator? Should I expand it around z=0? I haven't had any experience with this so thanks for bearing with me.

I have this so far E_z = \frac{\frac{2 a z}{\sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)}}}{\sqrt{2(r^2 + z^2 - a^2 + \sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)})}} I'm unsure how to proceed taking the limit. My first attempt gave E_z = \frac{\frac{2 a z}{\sqrt{((r-a)^2 +z^2)((r+a)^2 +z^2)}}}{\sqrt{2 z^2}}}...

Here are my contour plots for the electric fields. You can imagine where the disc of radius 1 is. Ez: http://www.maths.tcd.ie/~dleen/ez.jpg horizontal axis is r and vertical z and Er: http://www.maths.tcd.ie/~dleen/er.jpg horizontal axis is r and vertical zand while I'm at it...

Well I wasn't doing anything wrong with the derivative since looking at some graphs Ez does go to zero as z->0.

Do a taylor expansion of what is left inside the arcsin for r<a and you will end up with 1. arcsin[1]=Pi/2

Dear Michel, Thanks for the prompt reply. I'm still thinking about it

Oh since this got moved: It's not for any course or class so I'm not following any book. I'm not sure if anyone will be able to answer this but I'll ask anyway: I've solved laplaces equation in cyclindrical coordinates for a disc of radius 'a' and constant potential V on the disk (disc in z=0...

I got my results from the Irish equivalent of the A-Levels (although nowdays it is considered far more difficult). Physics - A1 Chemistry - A1 Biology - A1 Maths - A1 Applied Maths - A2 English - B2 German - D3 Irish - C2 I'm extremely pleased and I'm off to theoretical physics in...

There are two ways of doing this but you want to do it the graph way. The most important thing to remember is that the area under a velocity/time graph is equal to the distance travelled. So isolate the area under the graph from said interval and divide it up into simple shapes -...

Microwaves have wavelengths measured in centimeters I think so they cannot be diffracted or pass through the holes in the door.

Photoelectric emission occurs when e.m radiation of a certain frequency falls on a metal, each photon gives a single electron energy which causes it to be ejected from the surface of the metal. Now think what happens when the reverse occurs? Electrons, using a high voltage are slammed into a...