# Electric field created by a charge seen through an infinite plate (dielectric vs. conducting)

• I
Having come experimentally to an interesting electrostatic effect, I have returned, aged 47, to my old books in physics. It turns out that my books delight in using Gauss theorem etc. in rather ideal geometrical surface charge distribution, but never gave me the tools to answer to this simple question.

Let x,y,z be a system of axes.
In the x,y plane, there is an infinite plate of some width w assumed to be small for the sake of simplicity, made of some dielectric material.
A charge q is placed at (x,y,z) = (1, 0, 0). What is the electrical field for x < 0, and in particular along the x axis.
Same question if the plate is made of some conductive material.
Make any assumption, simplification or approximation you want about all the unspecified parameters like the dielectric constant etc., a partial answer being better than no answer at all. thx.

Delta2
Homework Helper
Gold Member
The charge is placed inside the dielectric slab (in the case of conductive plate, inside the conductive plate)? Or perhaps you mean the charge is placed at (x,y,z)=(0,0,1).

The dielectric slab (or the conductive plate) is the whole x-y plane with some small width ##dw<<1## along the z-axis?

In the case we have a point charge q located at (1,0,0) and the dielectric slab is the whole x-y plane then the electric field along the x-axis that is at a point ##(x,0,0)## will be

$$\vec{E}=\frac{1}{4\pi\epsilon}\frac{q}{(x-1)^2}\hat{x}$$
and with direction along the x-axis. (##\epsilon## is the permittivity of the dielectric)

Generally for a point ##(x,y,z)## inside the dielectric the electric field will be
$$\vec{E}=\frac{1}{4\pi\epsilon}\frac{q}{{((x-1)^2+y^2+z^2)}^{\frac{3}{2}}}[(x-1)\hat{x}+y\hat{y}+z\hat{z}]$$

while for a point ##(x,y,z)## outside the dielectric (assuming there is vacuum outside) it will be
$$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{{((x-1)^2+y^2+z^2)}^{\frac{3}{2}}}[(x-1)\hat{x}+y\hat{y}+z\hat{z}]$$

In the case we have conductive plate and we put a charge q on it, then the charge will distribute evenly along the whole infinite x-y plane and it is a well know result that the electric field at any point ##(x,y,z)## will be constant and equal to:

$$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat{z}$$ but because the charge is finite and the plane is infinite it will be ##\sigma\approx 0## so ##\vec{E}\approx \vec{0}##

Last edited:
Oh my god ! There is a mistake in my question. I meant the dielectric is the whole y-z plane, with some width dw << 1 along the x-axis, and the charge is placed outside the plate at (x,y,z) = (1,0,0) (as I wrote), that is, I'm interested in knowing the field after it has "crossed" the plate.

Note: In the case of a metallic plate, I know that the field cannot cross the plate, nevertheless, it will induce surface charges in the x < 0 side of the plate, and hence an electrical field in the x < 0 side.

Last edited: