Recent content by physics1000

The fact that I have to find a function such that the potential will be zero and the general thingy at the end. But I just realized it is general, so I dont really have to understand that general integral, only to understand what to do. And I just understood I think, I will try tomorrow this way...

Oh I see, it is confusing, but I will try this out of curiosity :) Thanks! And thanks again for explaining it to me, and elaborating on it.

Oh sorry, I know how to latex, I did it before at this post. I will write it now: ##\vec{r\:\:}=\left(x,\:y,\:z\right)\:## ##\:\vec{r'\:\:}=\left(a,\:0,\:z'\right)## ##\vec{R_{\:\:}}\:=\:\left(x-a,\:y,\:-z'\right)##...

LOL Sadly my English is bad, the sentence I say in my language is hard to translate to English :) Another thing learned :cool:

Oh sorry, I will try to upload in good resolution, I will edit. ( My camera has problem at boundarys, the focus it bad, It is the best I could do ) But anyway, the solution is good. About potential approach not being good, I had lot of problems there as you saw, it was impossible to me without...

it worked. Huge thanks :) Now I learned a new trick on my sleeve, how to solve such questions.

Ohh, now I see what you say. I will do the same thing, but with electic field, and then just integrate it. ( because with ##(K)^{3/2}## it is integrable easily ) That I actually accepts. I went out of my home for a few days, be back tommorow or today. I will try it. Huge thanks!!

But it will never be zero, the denominator, it is a absolute value that has ##()^2## on each one. If you talk about numerator, you have 1 only, it can never be 0, and I can not add from x and y to there, only to the denominator. Even if I put ##z'=0##, I will still have ##a^2## which is a...

Okay, that I 100% do not understand, that I can say for sure. ##\frac{1}{\left(\left(x-a\right)^2+y^2+\left(z'\right)^2\right)^{\frac{1}{2}}}## Why should I seperate the ##X## and ##Y## functions here and how do I do it exactly? it is combined.

Ahh, my English really downs me here, but still, I don't understand. ##r-r'## is not 0 if ##r=0##, that you are right. But still, I really do not understand it :\ I can not understand about that constant ##V_0##, or where do I put that constant \ other constant, where exactly, I know I sound...

Oh sorry, it is an immediate integral from my formula of the test ##A=\sqrt{\left(x-a\right)^2+y^2}\:\:\:\:A^2=\left(x-a\right)^2+y^2## ##\int \:\frac{1}{A^2+z^2}dz=ln\left(z+\sqrt{A^2+z^2}\right)+C## Hope it is clear now!!

The original exercise ( as you will see in the picture has two infinite line of charge, but don't mind it, one at ##x=a## the other at ##x=-a##, the problem is in the integral ). Ahh, I seriously do not understand what do you mean by that, nothing... I tried ( what I managed to understand from...

Yea, but about the constant thingy. Where do I add it? I could not understand it. You did not say where I need to say that my potential is the integral + constant? and if so, it still wont help. I dont know how do I use the ##Phi(0,0)## thing. I never encountered such a situtation where it is...

yea, but you use Gauss, that is my problem. There is 100% no way to solve what I want using my way? Usually at test I will not have the electric field or I will have some harder questions. Which I usually do good, but when it is cartesian, I am having hard time.

That actually would help me at Quasi-static exercises, bookmarked it for if I will need. Here sadly it wont, but for Quasi-static it is good.