# Method of images: infinite line of charge above plate

• zweebna
In summary: The forum staff moved it back to the intro physics forum and it looks like it's been moved back to advanced physics now.
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.

##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
Orodruin said:
##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.

Last edited:
zweebna said:
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
Orodruin said:
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.

zweebna said:
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
Orodruin said:
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}##?

zweebna said:
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
Orodruin said:
Right.
Thank you!

zweebna said:
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.

## 1. What is the method of images in electrostatics?

The method of images is a mathematical tool used in electrostatics to find the electric field and potential of a charge distribution by introducing imaginary or "image" charges in locations where the boundary conditions are known.

## 2. How does the method of images work?

The method of images works by using the principle of superposition to add the contributions of the actual and image charges to obtain the total electric field and potential at any point in space.

## 3. What is the application of the method of images?

The method of images is commonly used in solving problems involving conductors and dielectrics, such as finding the electric field and potential of a point charge near a conducting plane or a dielectric sphere.

## 4. What is the "infinite line of charge above plate" configuration in the method of images?

The "infinite line of charge above plate" configuration refers to a scenario where an infinite line of charge is placed above a grounded conducting plate. This is a common problem used to illustrate the application of the method of images.

## 5. What are the advantages of using the method of images?

One advantage of using the method of images is that it provides an elegant and efficient way to solve electrostatic problems involving conductors and dielectrics. It also allows for the visualization of the electric field and potential in complex configurations.

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