Finding electric potential of an infinite line charge at z axis

[haruspex]

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 problem, I can not do a minus.
Only if I demand that $z'=0$ and $a=0$ so It will be true. but it seems weird.

Asking for the potential at (x,y) given that it is zero at (0,0) is the same as asking for the potential difference, $\phi(x,y)-\phi(0,0)$. You can write that as one big integral. Each by itself will diverge, but you can manipulate the combined expression so that the two divergent components cancel.

 

[Orodruin]

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 problem, I can not do a minus.
Only if I demand that $z'=0$ and $a=0$ so It will be true. but it seems weird.

What do you mean it will never be zero? Whatever your function $f(x)$ you can create a function with the same derivative that is zero in $x_0$ by creating $g(x) = f(x) - f(x_0)$

 

[kuruman]

Except this does not work here because the potential derived from assuming the potential to be zero at infinity diverges everywhere.

And that's because this is a 2D problem algebraically and is independent of $z$. The equipotential surfaces in this case are concentric cylinders with their common axis coincident with the line of charge. Lines parallel to the line of charge are also equipotentials, which means that a calculation for $\Phi(x,y)$ can be done in the $xy$-plane without loss of generality.

In the $xy$-plane, point with Cartesian coordinates $\{x,y\}$ lies on an equipotential circle of radius $r=\sqrt{(x-a)^2+y^2}$ centered at $\{a,0\}$. Likewise, the origin of coordinates lies on an equipotential circle of radius $r_0=a$ also centered at $\{a,0\}.$

Asking for the potential at (x,y) given that it is zero at (0,0) is the same as asking for the potential difference, ϕ(x,y)−ϕ(0,0).

Precisely, but why not use the electric field to find said potential difference? If the use of Gauss's law to find the electric field is not an option as stipulated by the OP, then one can always do the integral $$E_r=\frac{\lambda r}{4\pi\epsilon_0}\int_{-\infty}^{\infty }\frac{dz}{(r^2+z^2)^{3/2}}$$ to find the radial electric field and finally integrate to find the potential function, $$\Phi(x,y)=-\int_a^{\sqrt{(x-a)^2+y^2}}E_r~dr.$$ This problem is equivalent to finding the electrostatic potential of a grounded conducting very long cylinder with a line of charge along its axis relative to an origin that is on the cylinder's surface with the x-axis crossing the cylinder's axis.

 

[physics1000]

And that's because this is a 2D problem algebraically and is independent of $z$. The equipotential surfaces in this case are concentric cylinders with their common axis coincident with the line of charge. Lines parallel to the line of charge are also equipotentials, which means that a calculation for $\Phi(x,y)$ can be done in the $xy$-plane without loss of generality.

In the $xy$-plane, point with Cartesian coordinates $\{x,y\}$ lies on an equipotential circle of radius $r=\sqrt{(x-a)^2+y^2}$ centered at $\{a,0\}$. Likewise, the origin of coordinates lies on an equipotential circle of radius $r_0=a$ also centered at $\{a,0\}.$

Precisely, but why not use the electric field to find said potential difference? If the use of Gauss's law to find the electric field is not an option as stipulated by the OP, then one can always do the integral $$E_r=\frac{\lambda r}{4\pi\epsilon_0}\int_{-\infty}^{\infty }\frac{dz}{(r^2+z^2)^{3/2}}$$ to find the radial electric field and finally integrate to find the potential function, $$\Phi(x,y)=-\int_a^{\sqrt{(x-a)^2+y^2}}E_r~dr.$$ This problem is equivalent to finding the electrostatic potential of a grounded conducting very long cylinder with a line of charge along its axis relative to an origin that is on the cylinder's surface with the x-axis crossing the cylinder's axis.

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!!

 

[physics1000]

it worked.
Huge thanks :)

Now I learned a new trick on my sleeve, how to solve such questions.

 

[Orodruin]

it worked.
Huge thanks :)
View attachment 337077

Now I learned a new trick on my sleeve, how to solve such questions.

I can't really see what you have written here, the resolution is too low. I will say that it is highly probable that you are making incorrect statements about the potential approach not being good.

 

[nasu]

Now I learned a new trick on my sleeve, how to solve such questions.

Aren't the tricks usually up your sleeve? If they are on the sleeve they'll be quite obvious.

 

[physics1000]

I can't really see what you have written here, the resolution is too low. I will say that it is highly probable that you are making incorrect statements about the potential approach not being good.

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 doing electric field as kuruman said sadly. since I had $+-infinity$ at boundarys.
But anyway, now I know how to do the solution, so if I will see it in the future, it is now easy :)

Thanks to all of you, sorry for being "block head" as not wanting to use Gauss and such.

 

[physics1000]

Aren't the tricks usually up your sleeve? If they are on the sleeve they'll be quite obvious.

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

 

[Orodruin]

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 doing electric field as kuruman said sadly. since I had $+-infinity$ at boundarys.
But anyway, now I know how to do the solution, so if I will see it in the future, it is now easy :)View attachment 337102

Thanks to all of you, sorry for being "block head" as not wanting to use Gauss and such.

Still impossible to see what is written in text. This is how it looks:

For the future, please learn to use the LaTeX features of the forum.

About potential approach not being good, I had lot of problems there as you saw, it was impossible to me without doing electric field as kuruman said sadly. since I had +−infinity at boundarys.

As I told you several times, this was never a problem. Your problem was not shifting the potential to keep the zero level at the correct place. The point-particle potential you should be using was
$$
G(\vec r, \vec r') = \frac{1}{4\pi \epsilon_0} \left( \frac{1}{|\vec r - \vec r'|} - \frac{1}{|\vec r'|}\right)
$$
This has the same derivatives with respect to the unprimed coordinates as the typical point-particle potential (the first term) but is constructed to have the zero-level at the origin $\vec r = 0$. The integral will converge without issues.

Alternatively you can just take the integral from $-Z$ to $Z$, subtract the potential at the origin after, and then let $Z \to \infty$.

 

[physics1000]

Still impossible to see what is written in text. This is how it looks:
View attachment 337103
For the future, please learn to use the LaTeX features of the forum.As I told you several times, this was never a problem. Your problem was not shifting the potential to keep the zero level at the correct place. The point-particle potential you should be using was
$$
G(\vec r, \vec r') = \frac{1}{4\pi \epsilon_0} \left( \frac{1}{|\vec r - \vec r'|} - \frac{1}{|\vec r'|}\right)
$$
This has the same derivatives with respect to the unprimed coordinates as the typical point-particle potential (the first term) but is constructed to have the zero-level at the origin $\vec r = 0$. The integral will converge without issues.

Alternatively you can just take the integral from $-Z$ to $Z$, subtract the potential at the origin after, and then let $Z \to \infty$.

Oh sorry, I know how to latex, I did it before at this post.
I will write it now:
$\vec{r\:\:}=\left(x,\:y,\:z\right)\:$
$\:\vec{r'\:\:}=\left(a,\:0,\:z'\right)$
$\vec{R_{\:\:}}\:=\:\left(x-a,\:y,\:-z'\right)$
$\:\left|\vec{R\:}\right|=\:\left(\left(x-a\right)^2+y^2+\left(z'\right)^2\right)^{\frac{1}{2}}$
$\lambda \left(\vec{r'\:}\right)=\lambda _0$
$\Phi \left(\vec{r'\:}\right)=\frac{1}{4\pi \epsilon _0}\int _{-\infty }^{+\infty }\frac{\lambda _0}{\left(\left(x-a\right)^2+y^2+\left(z'\right)^2\right)^{\frac{1}{2}}}\:dz'$
Not Good, so we find Electric field
$\:\:\:\:\:\vec{E}\left(\vec{r\:}\right)=\frac{\lambda _0}{4\pi \epsilon _0}\int _{-\infty \:}^{+\infty \:}\frac{\lambda \:_0}{\left(\left(x-a\right)^2+y^2+\left(z'\right)^2\right)^{\frac{3}{2}}}\:dz'\:=\frac{\lambda _0}{2\pi \epsilon _0}\left[\frac{1}{\left(x-a\right)^2+y^2}\right]\hat{r\:}$
And then
$\Phi \left(\vec{r'\:}\right)=-\frac{\lambda _0}{2\pi \epsilon _0}\int _a^{\sqrt{\left(x-a\right)^2+y^2}}\frac{1}{r}dr\:=\frac{-\lambda _0}{2\pi \epsilon _0}ln\left(\frac{\sqrt{\left(x-a\right)^2+y^2}}{a}\right)$

And about your idea, I dont really understand how it shifted it to zero? and what it means by that.

 

[Orodruin]

And about your idea, I dont really understand how it shifted it to zero? and what it means by that.

The (unit charge) point particle potential for a particle in $\vec r'$, which is zero at infinity is on the form:
$$
G_0(\vec r, \vec r') = \frac{1}{4\pi\epsilon_0} \frac{1}{|\vec r - \vec r'|}.
$$
If you just integrate this with the line charge, you will get something divergent. Even if you did not, you would get something that is typically not zero in $\vec r = 0$. In order to get something that is zero in $\vec r = 0$, you need to add a constant (in $\vec r$) to this potential. This is what we mean by a shifted potential. (scalar) Potentials are only defined up to a constant so this is perfectly fine. Integrating a potential that is always zero in $\vec r = 0$ will of course result in a potential that is also zero at that point - satisfying that part of your problem. So, the only question is which constant to shift the potential by. This is also not hard to find out because you want your $G(\vec r, \vec r')$ to be zero in $\vec r = 0$, so you put $G(\vec r, \vec r') = G_0(\vec r, \vec r') + g(\vec r')$ where $g(\vec r')$ can possibly be a function of $\vec r'$ because the only requirement is that it does not depend on $\vec r$. By inserting the requirement that $G(0, \vec r') = 0$, we easily find that
$$
G(0,\vec r') = G_0(0, \vec r') + g(\vec r') = 0 \qquad \Longrightarrow \qquad
g(\vec r') = - G_0(0,\vec r') = - \frac{1}{4\pi\epsilon_0 |\vec r'|}.
$$
Integrating over the entire line charge, in general we will have
$$
\Phi(\vec r) = \int_{-\infty}^\infty G(\vec r, z' \vec e_z + a \vec e_x) \lambda\, dz'
$$
which is a perfectly well converging integral.

 

[physics1000]

The (unit charge) point particle potential for a particle in $\vec r'$, which is zero at infinity is on the form:
$$
G_0(\vec r, \vec r') = \frac{1}{4\pi\epsilon_0} \frac{1}{|\vec r - \vec r'|}.
$$
If you just integrate this with the line charge, you will get something divergent. Even if you did not, you would get something that is typically not zero in $\vec r = 0$. In order to get something that is zero in $\vec r = 0$, you need to add a constant (in $\vec r$) to this potential. This is what we mean by a shifted potential. (scalar) Potentials are only defined up to a constant so this is perfectly fine. Integrating a potential that is always zero in $\vec r = 0$ will of course result in a potential that is also zero at that point - satisfying that part of your problem. So, the only question is which constant to shift the potential by. This is also not hard to find out because you want your $G(\vec r, \vec r')$ to be zero in $\vec r = 0$, so you put $G(\vec r, \vec r') = G_0(\vec r, \vec r') + g(\vec r')$ where $g(\vec r')$ can possibly be a function of $\vec r'$ because the only requirement is that it does not depend on $\vec r$. By inserting the requirement that $G(0, \vec r') = 0$, we easily find that
$$
G(0,\vec r') = G_0(0, \vec r') + g(\vec r') = 0 \qquad \Longrightarrow \qquad
g(\vec r') = - G_0(0,\vec r') = - \frac{1}{4\pi\epsilon_0 |\vec r'|}.
$$
Integrating over the entire line charge, in general we will have
$$
\Phi(\vec r) = \int_{-\infty}^\infty G(\vec r, z' \vec e_z + a \vec e_x) \lambda\, dz'
$$
which is a perfectly well converging integral.

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.

 

[Orodruin]

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.

Which part do you find confusing?

 

[physics1000]

Which part do you find confusing?

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 and update if needed.