Method of images: infinite line of charge above plate

[zweebna]

Homework Statement

An infinite line of charge with charge density λ is parallel to and a distance d above an infinite grounded conducting plate. What is the charge density σ that is induced in the plate? For simplicity, consider the line of charge to lie along the line x = 0.

Homework Equations

$V_{line} = \frac {\lambda}{2 \pi \epsilon_0} ln(\frac{s_1}{s_2})$
$\sigma = -\epsilon_0 \frac {\delta V}{\delta n}$

The Attempt at a Solution

So by the method of images, I know this problem is (for z>0) equivalent to an infinite line of charge density λ a distance d above z=0, and an infinite line of charge density -λ a distance d below z=0. I know the potential for an infinite line charge is $V = \frac {\lambda}{2 \pi \epsilon_0} ln(\frac{s_1}{s_2})$, where I believe $s_2$ would be the distance from the line and $s_1$ is an arbitrary point where $V \rightarrow 0$. I can then add together the potentials for these line charges:
$$V = \frac {\lambda}{2 \pi \epsilon_0} ln(\frac{s_1}{s_2}) - \frac {\lambda}{2 \pi \epsilon_0} ln(\frac{s_1}{s_2})$$
Now I'm confused. I can't just set the distance $s_2$ to be $d$ as this results in a potential of zero at all points. I can't set one to $z+d$ and one to $z-d$ as that doesn't seem to satisfy boundary conditions ($V=0$ when $z=0$, $V \rightarrow 0$ when $z \rightarrow \infty$).

I know that once I can get the potential, then I can get the induced charge by taking the partial derivative with respective to $z$ at $z=0$, but I'm confused about getting the potential.

 

[Orodruin]

$s_2$ is the distance from the line to the point you want to know the potential at. The $s_2$ in your second term is not the same as that in your first so do not call them the same thing. Instead, express the distance from the point you want to compute the potential for to the lines and use those values.

 

[zweebna]

$s_2$ is the distance from the line to the point you want to know the potential at. The $s_2$ in your second term is not the same as that in your first so do not call them the same thing. Instead, express the distance from the point you want to compute the potential for to the lines and use those values.

So if the point where we're computing potential is $0<z<d$, then in the first term it would be $d-z$ and in the second term it would be $z+d$?PS
What forum should this be in? Since posting it's been moved from intro to advanced, back to intro, and now it's back in advanced.

 

[Orodruin]

So if the point where we're computing potential is $0<z<d$, then in the first term it would be $d-z$ and in the second term it would be $z+d$?

Are you not forgetting some directions?

 

[zweebna]

Are you not forgetting some directions?

Ah you're right. So the first time is $\sqrt{x^2+y^2+(d+z)^2}$ and the second term is $\sqrt{x^2+y^2+(d-z)^2}$? Thank you I think I have it.

 

[Orodruin]

Ah you're right. So the first time is $\sqrt{x^2+y^2+(d+z)^2}$ and the second term is $\sqrt{x^2+y^2+(d-z)^2}$? Thank you I think I have it.

Almost, remember that you want the distance to the line charge. What you have given is the distance to a particular point.

 

[zweebna]

Almost, remember that you want the distance to the line charge. What you have given is the distance to a particular point.

Ah so since the line is along $x=0$ would I just ignore the y component? So they are $\sqrt{x^2+(d+z)^2}$ and $\sqrt{x^2+(d-z)^2}$?

 

[Orodruin]

Ah so since the line is along $x=0$ would I just ignore the y component? So they are $\sqrt{x^2+(d+z)^2}$ and $\sqrt{x^2+(d-z)^2}$?

Right.

 

[zweebna]

Right.

Thank you!

 

[vela]

What forum should this be in? Since posting it's been moved from intro to advanced, back to intro, and now it's back in advanced.

I think your thread was used for a forum moderation investigation/experiment which is why it got moved so often. I was the original one who moved it to advanced physics because I felt it belonged here.