# Finding the maximum value of the electric field

• Mike400
In summary, the conversation discusses the maximum value of the electric field in relation to the surface charge and the field point. The first attempt at finding a maximum value for ##\dfrac{1}{r^2}## is unsuccessful due to the undefined field at points on the surface. The second attempt uses an inappropriate expression for the electric field and also leads to an infinite maximum value for ##\sec{\alpha}##. The speaker also asks for clarification on the concept of "maximum" 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.

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?

## 1. What is the maximum value of the electric field?

The maximum value of the electric field is determined by the charge and distance from the source of the field. It is calculated using the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge, and r is the distance.

## 2. How do you find the maximum value of the electric field?

To find the maximum value of the electric field, you need to know the charge and distance from the source of the field. Then, you can use the equation E = kQ/r^2 to calculate the maximum value.

## 3. What factors affect the maximum value of the electric field?

The maximum value of the electric field is affected by the charge and distance from the source of the field. It also depends on the medium in which the field is present, as different materials have different dielectric constants that can alter the strength of the electric field.

## 4. Can the maximum value of the electric field be greater than the speed of light?

No, the maximum value of the electric field cannot be greater than the speed of light. The speed of light, c, is the maximum speed at which any physical object or information can travel in the universe. Therefore, the electric field, being a physical quantity, cannot exceed this speed limit.

## 5. Is there a limit to the maximum value of the electric field?

There is no theoretical limit to the maximum value of the electric field. However, in practical scenarios, the maximum value is limited by the breakdown strength of the medium in which the field is present. If the electric field exceeds this limit, the medium will break down and lose its insulating properties.

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