- #1

zweebna

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## 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.