Method of images: infinite line of charge above plate

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1. Dec 3, 2016

zweebna

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Dec 3, 2016

Orodruin

Staff Emeritus
$s_2$ 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.

3. Dec 3, 2016

zweebna

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.

Last edited: Dec 3, 2016
4. Dec 3, 2016

Orodruin

Staff Emeritus
Are you not forgetting some directions?

5. Dec 4, 2016

zweebna

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.

6. Dec 4, 2016

Orodruin

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

7. Dec 4, 2016

zweebna

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 $\sqrt{x^2+(d-z)^2}$?

8. Dec 4, 2016

Orodruin

Staff Emeritus
Right.

9. Dec 4, 2016

zweebna

Thank you!

10. Dec 8, 2016

vela

Staff Emeritus

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.