# Electric Field Produced by an Infinite Sheet.

• smithnh
In summary, the conversation discusses a common confusion when deriving the electric field from a charged sheet and the possibility of an integration error in a proof found online. It also mentions the importance of considering the geometry of the problem when applying Gauss's Law.
smithnh
I was browsing the internet looking for alternate proofs to why the electric field produced by an infinite sheet was equal to the charge density divided by twice the space permitivity constant. However, in my search I came across a proof that confused me at this link, http://www.physlink.com/Education/AskExperts/ae544.cfm"

It found the electric field produced to be eqaul to the charge density divided by only the space permitivity constant. My question is, what happened to the factor of 2. Are they only considering one face of the sheet but even then should that not decrease it by another factor of 2. What is going on here?

Last edited by a moderator:
This is a common confusion if you do not look carefully at the nature of the charge distribution. On one hand, you have an infinite SHEET of charge, i.e. you have the charge occupying only a sheet, for example, the x-y plane. Here, the E-field is being shared in both spaces z>0 and z<0. So if you do Gauss's Law on here, the resulting E-field has a factor of 1/2.

On the other hand, if you have a conductor occupying the space at z<0, while the surface at z=0 (still the x-y plane) still has the charge, then a similar Gauss's Law application will get you E-field that is twice as large. All the E-field is in the z>0 space.

So just because you have an infinite plane of charge doesn't mean you can ignore the rest of the geometry of the problem.

Zz.

smithnh said:
It found the electric field produced to be eqaul to the charge density divided by only the space permitivity constant. My question is, what happened to the factor of 2. Are they only considering one face of the sheet but even then should that not decrease it by another factor of 2. What is going on here?
I didn't look at the proof in detail, but if they are talking about an ordinary sheet of charge and they left out that factor of 2, then they made an error somewhere.

(As Zapper points out, there's a common source of confusion when deriving the field from a charged conducting sheet compared to just a sheet of charge. But in that link I don't see them talking about conductors.)

integration error

Doc Al said:
I didn't look at the proof in detail, but if they are talking about an ordinary sheet of charge and they left out that factor of 2, then they made an error somewhere.
They messed up the integration:

$$\int \frac{a}{(a^2 + x^2)^{3/2}} \;da= \frac{-1}{\sqrt{a^2 + x^2}} \neq \frac{-2}{\sqrt{a^2 + x^2}}$$

Thanks for the help.

## 1. What is an infinite sheet of charge?

An infinite sheet of charge is a hypothetical situation in which a sheet of charge extends infinitely in all directions. This is a simplified model used in physics to study the electric field produced by a charged plane.

## 2. How is the electric field produced by an infinite sheet calculated?

The electric field produced by an infinite sheet can be calculated using the formula E = σ/2ε0, where E is the electric field, σ is the surface charge density of the infinite sheet, and ε0 is the permittivity of free space.

## 3. What is the direction of the electric field produced by an infinite sheet?

The electric field produced by an infinite sheet is always perpendicular to the surface of the sheet. This means that the electric field lines are parallel to each other and are pointing away from the sheet if it is positively charged, and towards the sheet if it is negatively charged.

## 4. Does the distance from the sheet affect the strength of the electric field?

Yes, the strength of the electric field produced by an infinite sheet decreases as the distance from the sheet increases. This is because the electric field lines spread out as they move away from the sheet, resulting in a weaker electric field at greater distances.

## 5. How does the electric field produced by an infinite sheet vary with the charge density?

The strength of the electric field produced by an infinite sheet is directly proportional to the surface charge density. This means that an infinite sheet with a higher charge density will produce a stronger electric field compared to one with a lower charge density.

• Electromagnetism
Replies
22
Views
1K
• Electromagnetism
Replies
3
Views
1K
• Electromagnetism
Replies
1
Views
209
• Electromagnetism
Replies
11
Views
1K
• Electromagnetism
Replies
2
Views
2K
• Electromagnetism
Replies
18
Views
2K
• Electromagnetism
Replies
25
Views
1K
• Electromagnetism
Replies
6
Views
3K
• Electromagnetism
Replies
7
Views
13K
• Electromagnetism
Replies
3
Views
2K