Finding electric potential of an infinite line charge at z axis

[physics1000]

The question says:
Find the electric potential of the infinite line charge at $\Phi \left(x,y\right)$, when known $\Phi \left(0,0\right)=0$

I am having hard time finding the electric potential of such.
We know that the line charge is infinite at Z axis.
And we know $\Phi \left(0,0\right)=0$
So I tried:
$\vec{r\:\:}=\left(x,\:y\right)\:$
$\vec{r'\:}=\left(a,0\right)$
$\vec{R_1}=\left(x-a,y\right)$
$\:\left|\vec{R_1}\right|=\left(\left[x-a\right]^2+y^2\right)^{\frac{1}{2}}$
$\lambda \left(r'\right)=\lambda _0$
then put at the formula:
$\Phi \left(\vec{r\:}\right)=\frac{1}{4\pi \epsilon _0}\int \:\frac{\lambda \left(\vec{r'\:}\right)dl'}{\left|r-r'\right|}$
But it wont work, the solution should be: ( according to the test )
$\Phi \left(\vec{r\:}\right)=\frac{-\lambda _0}{2\pi \epsilon _0}ln\left(\frac{r}{r_{el}}\right)$

$r_{el}=a$, forgot to mention. ( because its the relative, the $r'$

But the problem is, it wont matter what $dl'$ I put in my formula, I wont be able to reach it.
The problem they said they want $\Phi \left(x,y\right)\:$ and not $\Phi \left(x,y,z\right)$.
Since I can not integrate at z axis from $\int _{-\infty }^{+\infty }\:$

FINAL ANSWER ( forgot to put it before, sorry ):
$\Phi \left(x,y\right)=-\frac{\lambda _0}{2\pi \epsilon _0}ln\left(\frac{\sqrt{\left(x-a\right)^2+y^2}}{a}\right)$

Please help, I am at this question for a long time ( it is a subquestion for the rest of the question and it crazy my mind since I can not answer that basic question ).

 

[nasu]

Have you studied Gauss' law?

 

[kuruman]

But the problem is, it wont matter what $dl'$ I put in my formula, I wont be able to reach it.
The problem they said they want $\Phi \left(x,y\right)\:$ and not $\Phi \left(x,y,z\right)$.
Since I can not integrate at z axis from $\int _{-\infty }^{+\infty }\:$

If you cannot "reach" infinity, why don't you integrate from $-z$ to $+z$, get the result and then let $z$ become very large? What's the difference? The $z$-dependence will drop out. Besides, you can see that the electrostatic potential is independent of $z$ by symmetry: Imagine standing on the wire and looking and looking around. You see the same charge distribution around you at the origin as you would if you walked to any distance $z$ from the origin. How would you know that you are distance $z$ and not at the origin?

 

[physics1000]

Have you studied Gauss' law?

Yea, but the point is to use that technique :\
With the way I wrote.
Gauss I use for different else, I know I can use it here also, since there is symmetry and such, but still.

 

[physics1000]

If you cannot "reach" infinity, why don't you integrate from $-z$ to $+z$, get the result and then let $z$ become very large? What's the difference? The $z$-dependence will drop out. Besides, you can see that the electrostatic potential is independent of $z$ by symmetry: Imagine standing on the wire and looking and looking around. You see the same charge distribution around you at the origin as you would if you walked to any distance $z$ from the origin. How would you know that you are distance $z$ and not at the origin?

But I can not use z... that is the problem.
I was given to use only X and Y.
That is what making me go mad.
and even if I integrate from $-z$ to $+z$, the problem is still, I have to use it somehow in my $R$ and such, and my function again, is $X$ and $Y$ only, I can not use $z$

 

[kuruman]

Is there a diagram that comes with the question? Is the line of charge along the $x$-axis as you show or along the $z$-axis in cylindrical coordinates?

 

[physics1000]

Is there a diagram that comes with the question? Is the line of charge along the $x$-axis as you show or along the $z$-axis in cylindrical coordinates?

hmm, the painting I drew is what was given.
The line of charge at $x=a$ and $y=0$ as said.
It is also infinity at $z$, which basically means $-\infty<z<\infty$
I don't think I have to use cylindrical coordinates, since I was given by $X$ and $Y$.

 

[physics1000]

I will reply the final answer, not just with r, I just saw I forgot to put it in here, sorry!!

$\Phi \left(x,y\right)=-\frac{\lambda _0}{2\pi \epsilon _0}ln\left(\frac{\sqrt{\left(x-a\right)^2+y^2}}{a}\right)$

And also edit at original post.
Again, Very much sorry I forgot it, maybe it made you confuse because of it!!!

 

[kuruman]

You show x and y axes and the charge distribution along the x-axis. OK. It follows that the z-axis passes through the origin and is perpendicular to the screen coming out of it. Still OK.

What confuses me is the information that are you trying to convey with the arrow labeled with $-\infty < \lambda_a < \infty$ and pointing to $\lambda_a.$

 

[Orodruin]

Since I can not integrate at z axis from ∫−∞+∞

Why not?

 

[physics1000]

You show x and y axes and the charge distribution along the x-axis. OK. It follows that the z-axis passes through the origin and is perpendicular to the screen coming out of it. Still OK.

What confuses me is the information that are you trying to convey with the arrow labeled with $-\infty < \lambda_a < \infty$ and pointing to $\lambda_a.$

View attachment 336828

ohh sorry, I will elaborate.
I mean by that arrow, is that the length of the line charge at z axis is infinity.
I will write again
$x=a$
$y=0$
$-\infty<z<+\infty$
That is the placement of the line charge
$\lambda=\lambda_0$

 

[physics1000]

Why not?

my function does not accept $z$. look at the question, do you see $z$ anywhere? I can not use $z$.
It is like you will say
$f(x,y)=5x+3y+6z$.
I dont have $z$ in my function.
That is what making me confused... I dont know what to do :(

 

[nasu]

@physics1000 Can you post the actual, full text of the problem? I belive that the line of charge is parallel to the z axis (or along the z axis). The distance between the charge element ($\lambda dl$) and the point where you calaculate the potential is a function of z. The coordinates of the point (x,y) are fixed. You don't integrate over them.

 

[Orodruin]

my function does not accept $z$. look at the question, do you see $z$ anywhere? I can not use $z$.
It is like you will say
$f(x,y)=5x+3y+6z$.
I dont have $z$ in my function.
That is what making me confused... I dont know what to do :(

r’ should contain z’, it is z’ that changes when you integrate over dl’. The function you search for is still not dependent on z. z’ is not z.

This approach will have other problems, but you’ll discover those along the way.

 

[physics1000]

@physics1000 Can you post the actual, full text of the problem? I belive that the line of charge is parallel to the z axis (or along the z axis). The distance between the charge element ($\lambda dl$) and the point where you calaculate the potential is a function of z. The coordinates of the point (x,y) are fixed. You don't integrate over them.

The question is in other language, I will translate it (hopefully good) and upload it.

 

[physics1000]

r’ should contain z’, it is z’ that changes when you integrate over dl’. The function you search for is still not dependent on z. z’ is not z.

This approach will have other problems, but you’ll discover those along the way.

I thought about it, about $z'$, but how do I contain it?
since I can not use third vector $(x,y)-(a,0,z')$, it is not possible.

 

[Orodruin]

I thought about it, about $z'$, but how do I contain it?
since I can not use third vector $(x,y)-(a,0,z')$, it is not possible.

Wrong. It is not (x,y), it is (x,y,0). Or even (x,y,z) - it is just that the resulting function does not depend on z.

 

[physics1000]

Given infinity line of linear charge density $\lambda_0$ parallel to z axis and placed at point $y=0$ and $x=a$. at $z=constant$ as expressed in paint.
Suppose the space is all dielectric-magnetic linear with $\epsilon$ and $\mu$.
A) Assume that your point potential beginning $\Phi(0,0)=0$ and find an expression of the potential to some point $(x,y)$

I just saw about the parallel, to be honest, I dont think I ever took any notice for it, also any others, we just know it is on z axis, I did not know it was important to say it, that it is parallel.

 

[nasu]

This view is not very useful. You need to have the z axis in your picture. And where does it say that you cannot use Gaus's law?

 

[physics1000]

Wrong. It is not (x,y), it is (x,y,0). Or even (x,y,z) - it is just that the resulting function does not depend on z.

Wait, so $f(x,y,0)=f(x,y)$? I can say such thing? or I did not understand correctly.
Ahh, its confusingg.
So I can start with $(x,y,z)$ and end with $(x,y)$ only? I guess that I kind of understand, I actually receive that alot. just thing subquestion is pretty harsh to me.

 

[physics1000]

This view is not very useful. You need to have the z axis in your picture. And where does it say that you cannot use Gaus's law?

The view and question, is my translation, it is the same as in the question :(
About gauus law, I did not say I cannot use.
But I prefer to use that way, since the later subquestions and all other tests and such, we use this technique.
Actually, its not that I prefer, I will only use it that way, Gauss I will use on other exercises, not here.

 

[physics1000]

Regarding using $(x,y,0)$ and $(a,0,z')$. I will try and update.

 

[physics1000]

$\frac{\lambda _0}{4\pi \epsilon _0}\int _{-\infty }^{+\infty }\frac{dz'}{\left[\left(x-a\right)^2+y^2+\left(z'\right)^2\right]^{\frac{1}{2}}}\:=\frac{\lambda _0}{4\pi \epsilon _0}\left[ln\left(z'+\left[\left(z'\right)^2+\left(x-a\right)^2+y^2\right]^{\frac{1}{2}}\right)\right]_{_{-\infty }}^{^{+\infty }}$
Sadly, did not work also :(
I am stuck with z' as it goes to infinity. I can not seem to find a way to remove $z'$
I guess I will give up and thats it, I see that I can not solve it. I dont want to use gauss law, to solve such questions we do not use gauss law, that I can say 100%, I am using it for "circle" stuff mostly. But again, I do not want to use Gauss.

 

[kuruman]

From your description, you are asked to calculate the potential at point P in the xy-plane as indicated in the figure below. The red line is the linear charge distribution.

 

[Orodruin]

Sadly, did not work also :(
I am stuck with z' as it goes to infinity. I can not seem to find a way to remove z′

Ok, so now you are running into the actual issue with your approach here. The Green’s function you have used (potential for a point charge) presumes that the potential at infinity is zero. However, that is not what your problem specified. It specified that the potential at the origin should be zero.

Luckily, potentials that only differ by a constant are equivalent - they lead to the same field. So instead of adding potentials that have been chosen to be zero at infinity, you need to use potentials that are zero at the origin.

Ultimately, the reason you are getting an infinity is that if you put the zero level of the potential at a finite point, then the potential at infinity diverges. Trying to force the potential at infinity to zero is therefore a bad idea.

 

[Orodruin]

From your description, you are asked to calculate the potential at point P in the xy-plane as indicated in the figure below. The red line is the linear charge distribution.View attachment 336842

Based on the OP, I would say the problem asks for an arbitrary point. However, the problem is invariant under z translation so only the x and y values can possibly affect the potential. Therefore one might as well pick z=0 and infer that the sane x and y values will gove the same result for any other z.

 

[physics1000]

From your description, you are asked to calculate the potential at point P in the xy-plane as indicated in the figure below. The red line is the linear charge distribution.View attachment 336842

Exactly, that is how you should graph it, was not given like that, but that is the point.

 

[physics1000]

Ok, so now you are running into the actual issue with your approach here. The Green’s function you have used (potential for a point charge) presumes that the potential at infinity is zero. However, that is not what your problem specified. It specified that the potential at the origin should be zero.

Luckily, potentials that only differ by a constant are equivalent - they lead to the same field. So instead of adding potentials that have been chosen to be zero at infinity, you need to use potentials that are zero at the origin.

Ultimately, the reason you are getting an infinity is that if you put the zero level of the potential at a finite point, then the potential at infinity diverges. Trying to force the potential at infinity to zero is therefore a bad idea.

First paragraph, yep, my problem sadly :\
Second paragraph, I can not seem to understand it? how do I change it exactly? sorry for my "bad brain".
Third paragraph, Then what should I do? is it impossible to solve it using that technique?
And thanks for elaborating for me.

 

[Orodruin]

Second paragraph, I can not seem to understand it? how do I change it exactly?

You need to add a constant (not depending on x, y, of z - possibly depending on x’, y’, and z’) to the point charge potential that you are using. Since this just adds a constant, it does not affect the fields.

I believe I covered this problem snd its solution in great detail in my book should you have access to it via your university library or similar.

Third paragraph, Then what should I do?

You should do what I just told you. Use a shifted point particle potential.

 

[physics1000]

You need to add a constant (not depending on x, y, of z - possibly depending on x’, y’, and z’) to the point charge potential that you are using. Since this just adds a constant, it does not affect the fields.

I believe I covered this problem snd its solution in great detail in my book should you have access to it via your university library or similar.You should do what I just told you. Use a shifted point particle potential.

Ahh, I will try to understand :(
Thanks anyway for the help :)