# How would one approach such problem?

1. Sep 8, 2013

### NasuSama

1. The problem statement, all variables and given/known data

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$^2$) is $\sigma$. 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 $P$.

2. Relevant equations

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

3. 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.

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2. Sep 9, 2013

### rude man

If the sheet is infinite in its two dimensons then the E field is easily determined by using Gauss' theorem. You don't need to integrate.

If the sheet dimensions are finite that's more difficult.

The problem states that the dimensions are infinite.

3. Sep 9, 2013

### Bryson

Since your point of interest is in the center of the plane, just use symmetry. Even for the finite case, it's not difficult. It's difficult when the point is off axis.

4. Sep 9, 2013

### NasuSama

I haven't learned Gauss' Law 100% yet. I need to use the electric field formula to work out the problem.

5. Sep 9, 2013

### rude man

Oh, OK. I'll keep tabs on the thread & see how it goes.

6. Sep 10, 2013

### Bryson

Note again, this is a fairly straight forward problem; do not over complicate it (thinking too hard only makes matters worse). Use symmetry and your golden.