Since upper and lower plate are earthed and electric field within plate (or any other conductor) is $E = 0$, what you really get is that on the bottom surface of the upper plate and on the top surface of the lower plate you get charge density $$\sigma' = -\frac{\sigma}{2}.$$

Electric fields *are* independent of the separations between plates and plane. Electric field between upper plate and plane is $E = +\frac{\sigma}{2\epsilon_0}$ and electric field between lower plate and plane is $E = -\frac{\sigma}{2\epsilon_0}$ (opposite direction). This comes from the well-known expression for the electric field of the infinitive plane with constant charge density. Electric field within both plates, above upper plate and below lower plate is $E = 0$.

The idea of using Gauss's law in your case is to use a cylinder, enclosing e.g. the bottom surface of the upper plate with top and bottom of the cylinder parallel to surface. Since upper side of the cylinder is within plate, $E = 0$, so $\oint \vec{E} \cdot \text{d}\vec{s} = 0$, while for the bottom side of the cylinder $\oint \vec{E} \cdot \text{d}\vec{s} = - E A = - \frac{\sigma}{2\epsilon_0} A$. Since the total charge enclosed in the cylinder is $\sigma' A$, you get the result above.

I'm assuming here that the cylinder is "infinitely long", or at least very long so that $h >> r$. Otherwise, there are complicated non-integrable "end effects", but it doesn't look like you're interested in those.

As you said, you need Gauss's law. You're losing something important if you drop the vector notation, though, so I've added it back in.
$$
\phi_E=\int_S \vec{E} \cdot d\vec{A}= \frac Q{\epsilon_0}
$$
The critical part, which you've already done, is to choose a surface on which $E$ is constant, so the integral is easy to evaluate.

You should pick a cylindrical surface of radius $R$ and height $L$, centered on the axis of the charged cylinder. An illustration of that surface (in green) is shown here:

With that choice of surface, simply by symmetry, the $E$ field must have the same value, and the same direction relative to the normal vector $d\vec{A}$, everywhere on the curved part of the surface!

On the flat ends of your cylindrical surface, the $E$ field is not constant, but it is parallel to the surface, and therefore perpendicular to the surface vector $d\vec{A}$, so the dot product $\vec{E} \cdot d\vec{A}$ is zero on those surfaces, and we can ignore them in the integral.

Here I've split the integral into the two parts - the integral over the curved part of the surface, and the flat ends, and I've evaluated both parts of the integral.

$$\int_{curved} \vec{E} \cdot d\vec{A} + \int_{ends} \vec{E} \cdot d\vec{A}= \frac Q{\epsilon_0}$$
$$\int_{curved} E dA + 0= \frac Q{\epsilon_0}$$
$$E \int_{curved} dA= \frac Q{\epsilon_0}$$
$$E ~A_{curved} = \frac Q{\epsilon_0}$$
$$E ~2 \pi R L = \frac Q{\epsilon_0}$$

Notice that on the curved part, since $\vec{E}$ and $d\vec{A}$ are in the same direction, their dot product is simply $E dA$, and since the magnitude $E$ is the same everywhere, we can remove it from the integral as a constant. Then we're simply integrating $dA$, which simply gives us the area of that part of the surface.

The last job we have is to find how much charge ($Q$) is inside our surface. The charge density is $\sigma = \frac{q}{2 \pi r h}$

There are two cases.

- If $R > r$, then the $Q = 2 \pi r L \sigma = q \frac{L}{h}$.
- If $R < r$, then no charge is inside our surface, so $Q = 0$.

So, using our final version of Gauss's law:

$$E ~2 \pi R L = \frac Q{\epsilon_0}$$
$$E = \frac{Q}{2 \pi \epsilon_0 R L}$$

And our final answer is:

$R<r$;$$E(R) = 0$$
$R>r$;$$E(R) = \frac{q}{2 \pi \epsilon_0 R h}$$

## Best Answer

$K=\frac1{4\pi\epsilon_0}$

Using elemental rings in disc we can get the Electric field at point P.

$dE = \frac{Kdq}{r^2}$ where $dq = \sigma.2\pi xdx $

Integrating from 0 to R we get

$E = \frac{Kqr}{(R^2+r^2)^{3/2}}$ where $q = \sigma\pi R^2$ $ = $

Using this in the cylinder

$dE = \frac{Kqx}{(x^2+R^2)^{3/2}}$ where $q = \rho\pi R^2dx$

Integrating this field from 0 to $\ell$ gives the required result.

$E = \frac{\rho R^2}{4\epsilon_0}\{\frac 1R - \frac 1{(R^2+l^2)^{3/2}}\}$ $ = \frac{\rho R}{4\epsilon_0}\{1 - \frac 1{R(1+\frac{l^2}{R^2})^{3/2}}\} $

and for $l\to\infty$

$E = \frac{\rho R}{4\epsilon_0}$