[PhD Qualifier] Charged rod above metal plate

[confuted]

Hey ... I'm trying to study for my PhD qualifying exam, and I have a bunch of questions from previous years, but no answer keys.

Homework Statement

A think uniformly charged rod of length L is positioned vertically above a large uncharged horizontal thick metal plate. The distance between the lower end of the rod and the metal plate is S.

If the total charge of the rod is q, find the charge density \sigma on the upper surface of the metal plate directly below the rod. (Hint: First, consider only the rod and find the electric field due to he rod at a distance S directly below the rod.)

Homework Equations

The Attempt at a Solution

I found the electric field due to the rod to be
E_{rod}=\int_{S}^{S+L}{\frac{kQ}{L^2z^2}dz}
E_{rod}=-\left.\frac{kQ}{Lz}\right|^{S+L}_{S}
E_{rod}=\frac{kQ}{S(S+L)}

I'm not sure where to go from here. If the plate was grounded, we could use an image charge and
\sigma=\epsilon_0E_n; is this also the right approach for the present problem, given that we're only considering the point directly below the rod?

 

[badphysicist]

I think that the method of images should work here. You shouldn't need the plate to be grounded. Electrons will move to one surface leaving the other surface positively charged.

 

[nicksauce]

I seem to recall a formula,
\sigma = -\epsilon_0\frac{\partial V}{\partial S}
But then you would have to calculate the potential, which might not be that easy.Edit: Another thought: The field immediately outside the conductor should be E=\frac{\sigma}{\epsilon_0}, right? So can you equate this with the other electric field expression and solve for \sigma ?

 

[kreil]

I would use the formula nicksauce posted:

\sigma = -\epsilon_0\frac{\partial V}{\partial S}

And since you know the electric field from the rod, finding the potential shouldn't be a problem..

 

[confuted]

As long as the method of image charges is appropriate, the electric field should just be the superposition of the E field I posted earlier and the E field of a negative image charge with its top at -S, correct? In which case I can use the formula I posted, \sigma=\epsilon_0E_n (where the n subscripts denotes the normal component).