Finding electric potential of an infinite line charge at z axis

[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.