# Finding electric field between two conducting plates using Gauss' law

• MatinSAR
In summary, the conversation discusses the use of Gauss's law to calculate the electric field between two plates in a specific charge distribution scenario. The question arises of why the negative charges on the other plate are not considered in the calculation, to which it is explained that Gauss's law considers the total field at a given surface, not just the field due to enclosed charge. The conversation also clarifies the purpose and limitations of Gauss's law in determining electric fields in different scenarios.
MatinSAR
Homework Statement
Finding electric field between two conducting plates using Gauss's law
Relevant Equations
##\epsilon_0 \oint \vec E \cdot d \vec A=q_{net,enc}##

This picture is from Sears and Zemansky's University Physics.
It considers ##S_1## as a gaussian surface then it finds electric field between two plates.
The only thing that I cannot understand is why it doesn't consider the electric field due to negative charges on other plate. Then electric field between plates should be ##\frac {2\sigma}{\epsilon_0}##.Same thing happened in Fundamentals of Physics(Textbook by David Halliday).

MatinSAR said:
The only thing that I cannot understand is why it doesn't consider the electric field due to negative charges on other plate.
Who says it doesn't? It seems you don't understand Gauss's law. The symbol ##E## in Gauss's law stands for the total field at area element ##dA##, not just the field due to the enclosed charge.

At the flat pillbox surface between the plates, the total field is ##E## and at the other surface inside the plate is zero. So $$(E+0)A=\frac{q_{enc.}}{\epsilon_0}=\frac{\sigma A}{\epsilon_0}\implies E=\frac{\sigma}{\epsilon_0}.$$ Another way to see it is this. Each of the planar charge distributions contributes an electric field of magnitude $$E=\frac{\sigma}{2\epsilon_0}.$$ Between the plates, the contributions are in the same direction so the total field is the sum of the magnitudes. Inside the conductor, the contributions are in opposite directions resulting in zero.

MatinSAR
kuruman said:
The symbol ##E## in Gauss's law stands for the total field at area element ##dA##, not just the field due to the enclosed charge.
I've read this statement in my book but I didn't understand it correctly. So I can't use Gauss's law to find the electric field of a charge distribution inside the closed surface.(since this law relates net charge inside the surface and total electric field.)
But I can do that If there's no charge outside of the surface and we have only charge inside.

MatinSAR said:
I've read this statement in my book but I didn't understand it correctly. So I can't use Gauss's law to find the electric field of a charge distribution inside the closed surface.(since this law relates net charge inside the surface and total electric field.)
But I can do that If there's no charge outside of the surface and we have only charge inside.
It looks like you still don't understand what Gauss's law says. Consider it in the form $$\epsilon_0\oint \vec E \cdot d\vec A=q_{enc}$$You can view it as a set of instructions to perform two tasks.

1. Take a closed surface ##S## that forms the boundary of volume ##V##, like the skin of a potato.
2. Subdivide the skin of the potato into small directed area elements ##d\vec A##. This means that each element has magnitude ##dA## and points perpendicular to the surface and outward, away from the skin.
3. At each element find ##\vec E \cdot d\vec A## which is the perpendicular-to-the-surface component of the total electric field multiplied by ##dA## at the location of that particular ##dA## and ##\epsilon_0##.
5. Repeat with the next element and the next and the next until you cover all the elements on the skin.
6. Recall the number in the memory of your calculator which is the sum of all the products. Call it the left-hand-number.
1. Dice the meat of the potato into small element each of volume ##dV##.
2. Find the total charge in element ##dV##.
4. Repeat with the next element and the next and the next until you cover all the elements that make up the meat of the potato.
5. Recall the number in the memory of your calculator which is the sum of all the charges. Call it the right-hand-number.
At this point you have two numbers that you have obtained by performing two completely separate tasks. Gauss's law is the assertion that these two numbers are the same. It is an experimental result and always holds but cannot always be used to calculate electric fields.

Crudely, it says that if you walk on the surface of the potato and find that electric field lines are coming out of it at all locations, you don't need to look inside to figure out that there is a net positive charge inside. Furthermore, you find that as many field lines flow out as flow in for a net of zero flux, again you don't need to look inside to figure out that there is no net charge inside. Note that no net electric flux does not mean no field at the surface. It only means that left-hand-number has just as many positive as negative contributions.

Gauss's law can be used in highly symmetric situations to find the electric field not inside the Gaussian surface, but at the Gaussian surface, i.e. on the skin of the potato. These are situations where (a) the electric field is perpendicular to the surface at each point on the skin and (b) the electric field has the same magnitude at each point on the skin.

MatinSAR
@kuruman Thank you for your valuable reply. I understand much better now.

kuruman

## 1. What is Gauss' law and how is it used to find the electric field between two conducting plates?

Gauss' law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is expressed as ∮E·dA = Q_enclosed/ε₀. To find the electric field between two conducting plates, we consider a Gaussian surface (usually a rectangular box or cylindrical surface) that encloses one of the plates. By symmetry, the electric field is perpendicular to the plates and uniform. Using Gauss' law, we can solve for the electric field by considering the charge density on the plates and the area of the Gaussian surface.

## 2. What assumptions are made when using Gauss' law to find the electric field between two conducting plates?

When using Gauss' law to find the electric field between two conducting plates, we assume that the plates are infinite in extent, which simplifies the problem by making the electric field uniform and perpendicular to the plates. Additionally, we assume that the space between the plates is vacuum or air, which allows us to use the permittivity of free space (ε₀) in our calculations. Lastly, we assume that the plates are perfectly conducting, meaning they can hold charge uniformly on their surfaces.

## 3. How do you set up a Gaussian surface for calculating the electric field between two parallel plates?

To set up a Gaussian surface for calculating the electric field between two parallel plates, we typically use a rectangular box or a cylindrical surface that spans the distance between the plates. The surface should be oriented such that its sides are parallel and perpendicular to the electric field lines. The key is to choose a surface where the electric field is either constant or zero, simplifying the integral in Gauss' law. For a rectangular box, two faces should be parallel to the plates, and for a cylindrical surface, the flat ends should be parallel to the plates.

## 4. How does the distance between the plates affect the electric field calculated using Gauss' law?

The distance between the plates does not directly affect the magnitude of the electric field calculated using Gauss' law. The electric field between two infinite parallel plates with surface charge densities ±σ is given by E = σ/ε₀, and it is independent of the distance between the plates. However, this is an idealization, and in practical scenarios with finite plates, edge effects and the finite size of the plates can cause deviations from this simple relationship.

## 5. What is the significance of the permittivity of free space (ε₀) in the context of Gauss' law and electric fields?

The permittivity of free space (ε₀) is a fundamental constant that quant

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