Finding the maximum value of the electric field

[Mike400]

Homework Statement

Suppose you have a surface of finite area with a fixed surface charge distribution. Does a maximum electric field magnitude $|\vec{E}|_{max}$ exist? If yes, how shall we find $|\vec{E}|_{max}$ or any other value greater than $|\vec{E}|_{max}$?

Relevant Equations

The electric field due to an arbitrary surface charge is:

$\displaystyle\vec{E}=k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA$

I tried to find it the following way but to no avail:

Let maximum value of $\sigma$ be $S$

Now unfortunately, we do not have a maximum value for $\dfrac{1}{r^2}$ because the field point can be as close as we want to the arbitrary surface charge. (The field at a point on the surface is undefined.) This is where I can't proceed further.

But we know even though the integrand blows up at points near surface charge, there in no blowing up of the integral at points near surface charge and it approximately equals $2 \pi k\ \sigma (\hat{n})$. Therefore there must be a maximum value for $|\vec{E}|$.

Another try of mine:

\begin{align}
\vec{E} &= k \iint_A \dfrac{\sigma}{r^2}(\hat{r})dA\\
&= k \iint_A \dfrac{\sigma}{r^2}(\hat{r}) \cos{\alpha} \sec{\alpha}\ dA\\
&= k \iint_A \sigma\ (\hat{r})\ \sec{\alpha}\ d\omega\\
\end{align}

where

$\alpha$ is the angle between $\vec{r}$ and unit normal vector to $dA$

$d\omega$ is element solid angle

Here again, unfortunately the maximum value for $\sec{\alpha}$ is infinity. And I cannot proceed further.

 

[kuruman]

Your expression for the electric field due to a finite area is inappropriate. You should be using
$$\vec E(\vec r) = k \iint_A \dfrac{\sigma(\vec r')(\vec r-\vec r')}{|\vec r-\vec r'|^3}dA'$$where $\vec r$ and $\vec r'$ are, respectively, field and source vectors relative to an arbitrary origin.
Having said that, what exactly do you mean by "maximum" electric field? Are you fixing the shape of the surface and the charge distribution and look for a point in space where the field has its largest magnitude?