- #1

NasuSama

- 326

- 3

## Homework Statement

*Uniform plane of charge*. Charge is distributed uniformly over a large square plane of side ##\ell##, as shown in the figure from the uploaded file (see file). The charge per unit area (C/m[itex]^2[/itex]) is [itex]\sigma[/itex]. Determine the electric field at a point ##P## a distance ##z## above the center of the plane, in the limit ##\ell \rightarrow \infty##.

The hint is as followed:

Divide the plane into long narrow strips of width ##dy## and determine the electric field due to each of the strips. Then, sum up the fields due to each strip to get the total field at ##P##.

## Homework Equations

##dE = \dfrac{1}{4\pi\varepsilon_0} \dfrac{dQ}{r^2}##

## The Attempt at a Solution

Found ##r##, which is ##\sqrt{z^2 + y^2}##.

With trig, I found that

##dE_z = dE\cos(\theta)##

##dE_y = dE\sin(\theta)##

By integration,

##\int dE_y = \int dE\sin(\theta)##

##E_y = 0##

##\int dE_z = \int dE\cos(\theta)##

##E = E_z = \dfrac{\lambda}{4\pi\varepsilon_0} \int \dfrac{\cos(\theta)}{z^2 + y^2} dy##

Since ##y = z\tan(\theta)##, then ##dy = z(\sec(\theta))^2 d\theta##. Since ##\cos(\theta) = \dfrac{z}{\sqrt{z^2 + y^2}}##, then the integrand becomes ##\dfrac{\cos(\theta) d\theta}{z}##. Thus,

##E = \dfrac{\lambda}{4z\pi\varepsilon_0} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta) d\theta = \dfrac{\lambda}{2z\pi\varepsilon_0}##

That is for one of the strips. I don't know how to sum up the fields.