[Physics] Finding the maximum value of electric field for a given two-dimensional charge distribution

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

The electric field due to an arbitrary surface charge is:

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

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.

If infinite surfaces and arbitrarily high $$\sigma$$ is allowed:

For an infinite flat plane with uniform surface charge $$\sigma$$, the electric field at all points in space is

$$\mathbf{E}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{n}}$$

where $$\hat{\mathbf{n}}$$ is the normal vector to the surface. The magnitude of the electric field at all points in space is

$$E=\frac{\sigma}{2\epsilon_0}$$

If $$\sigma$$ is allowed to be arbitrarily high, then $$E$$ can also be arbitrarily high, so there can be no upper bound on the electric field from an arbitrary surface charge.

As an aside, this also makes it clear that $$E$$ does not always vanish at infinity in this case.

If the surface is constrained to be finite and $$-S\le\sigma\le S$$ for some fixed $$S$$:

Suppose we have a finite flat plate with uniform charge density $$S$$, and parallel to it, we have another finite flat plate with uniform charge density $$-S$$. They both have a finite area $$A$$ and are separated by a distance $$L$$.

The electric field midway between the plates' centers will look like the of two infinite flat plates, plus a finite-size correction that depends on the quantity $$L/A$$ (in other words, the assembly looks more like two infinite parallel plates when the plates get larger or closer together):

$$\mathbf{E}=\frac{\sigma}{\epsilon_0}-f\left(\frac{L}{A}\right)$$

By making the plates big enough (while still having a finite area) or the spacing small enough, we can make the finite-size corrections arbitrarily small.

Now, add a second pair of plates of the same size and with the same surface charge distribution as before (so now we have two plates on one side with surface charge density $$S$$ and two plates on the other side with surface charge density $$-S$$) at $$L/4$$ and $$3L/4$$. Now the field midway between is:

$$\mathbf{E}=\frac{2\sigma}{\epsilon_0}-f\left(\frac{L}{A}\right)-f\left(\frac{L}{2A}\right)$$

You can keep adding more and more pairs of plates, closer and closer to the center charge, and make the electric field midway between arbitrarily large (and you can also make all of the finite-size corrections arbitrarily small). Since you can always add another pair of plates, you will never reach a maximum electric field, even with a constrained charge density.