Finding electric potential of an infinite line charge at z axis

[physics1000]

I mean, the solution for the field is literally given in the article. All one needs to do to find the potential is to integrate …

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.

 

[Orodruin]

There is 100% no way to solve what I want using my way?

I have told you already that there is. You just need to execute it.

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.

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.

Not sure I can be more specific without actually solving the problem …

 

[physics1000]

I have told you already that there is. You just need to execute it.Not sure I can be more specific without actually solving the problem …

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 not zero at infinity but at origin.
Sadly I can not understand what you want me to do...
I dont need you to do the solution, just based on my answer, what do I have to add? I really can not understand, what I did is basically like most of what is needed, from what you say, I miss small detail thingy that I can not answer without it.

 

[haruspex]

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 not zero at infinity but at origin.
Sadly I can not understand what you want me to do...
I dont need you to do the solution, just based on my answer, what do I have to add? I really can not understand, what I did is basically like most of what is needed, from what you say, I miss small detail thingy that I can not answer without it.

Suppose you have a solution to the integral for the potential, $\phi(x,y,z)$, which is zero at infinity. For any constant potential, $V_0$, $\phi(x,y,z)+V_0$ is also a solution, but will be $V_0$ at infinity.
What will it be at $(0,0,0)$?
What value of $V_0$ will make it zero at $(0,0,0)$?

 

[physics1000]

Suppose you have a solution to the integral for the potential, $\phi(x,y,z)$, which is zero at infinity. For any constant potential, $V_0$, $\phi(x,y,z)+V_0$ is also a solution, but will be $V_0$ at infinity.
What will it be at $(0,0,0)$?
What value of $V_0$ will make it zero at $(0,0,0)$?

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 your thing )

Tried to do $X=Y=Z=0$ and not $X'=Y'=Z'=0$ of course.
In The case of the post ( not the two infinite line charge, cus it does not matter ), I will just receive
##V_0 =# #minus of all the first integral. Or again, I do not understand again...

 

[haruspex]

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 your thing )
View attachment 336885
Tried to do $X=Y=Z=0$ and not $X'=Y'=Z'=0$ of course.
In The case of the post ( not the two infinite line charge, cus it does not matter ), I will just receive
##V_0 =# #minus of all the first integral. Or again, I do not understand again...

You have an integral of the form $\int_{-\infty}^{\infty}\frac{dz}{(A^2+z^2)^\frac 12}$. I don't understand the next line. Please post your working of the integration.

 

[Orodruin]

You did not say where

I did:

add a constant (…) to the point charge potential that you are using

(added emphasis)

Your point charge potential is the $\frac{1}{4\pi\epsilon_0 |r - r’|}$ in the integral. This is clearly not zero at r=0.

 

[Orodruin]

Suppose you have a solution to the integral for the potential, $\phi(x,y,z)$, which is zero at infinity. For any constant potential, $V_0$, $\phi(x,y,z)+V_0$ is also a solution, but will be $V_0$ at infinity.
What will it be at $(0,0,0)$?
What value of $V_0$ will make it zero at $(0,0,0)$?

Except this does not work here because the potential derived from assuming the potential to be zero at infinity diverges everywhere. The way to solve it is using a different point particle potential in the integral. One that has the correct zero level. Once you do that, the linear combination of any such potentials will have the correct zero level.

 

[Orodruin]

Alternatively you can separate out the x and y dependence so your potential takes the form $\phi = C + f(x,y)$ where $C$ is a formally infinite constant you can remove. That also works.

 

[physics1000]

You have an integral of the form $\int_{-\infty}^{\infty}\frac{dz}{(A^2+z^2)^\frac 12}$. I don't understand the next line. Please post your working of the integration.

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!!

 

[physics1000]

I did:

(added emphasis)

Your point charge potential is the $\frac{1}{4\pi\epsilon_0 |r - r’|}$ in the integral. This is clearly not zero at r=0.

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 like an idiot already, but I really can not apprehend this.

 

[physics1000]

Alternatively you can separate out the x and y dependence so your potential takes the form $\phi = C + f(x,y)$ where $C$ is a formally infinite constant you can remove. That also works.

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.

 

[Orodruin]

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.

That is not the integrated expression…

 

[Orodruin]

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 like an idiot already, but I really can not apprehend this.

You add a constant (in x and y) to that expression such that the expression becomes zero at r = 0. This you can do without changing the field as it doesn’t change any derivative wrt x or y.

 

[physics1000]

You add a constant (in x and y) to that expression such that the expression becomes zero at r = 0. This you can do without changing the field as it doesn’t change any derivative wrt x or y.

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.

 

[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.