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 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)##...
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...
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.