# Solving Gauss's Law Problem: Electric Fields in 4 Regions

• awvvu
In summary, the conversation is discussing the electric fields in four different regions near an infinitely wide conductor and an infinitely wide plane of charge with a surface charge density of delta. The electric fields in regions 1 and 3 are equal and opposite, while the electric field in region 4 is half the strength of the fields in regions 1 and 3. The calculation of the electric field in region 4 can be checked using superposition.
awvvu

## Homework Statement

An infinitely wide conductor (shaded thing on top) is parallel to and at a distance d from an infinitely wide plane of charge (line on bottom) with surface charge density delta. What are the electric fields $E_1$ to $E_4$ in the regions 1 to 4?

Code:
1
-------------------------
2  /  /  /  /  /  /  /  /
-------------------------
3
_________________________
4

## The Attempt at a Solution

$E_2 = 0$, as the charges move to the surface.

A Gaussian pillbox that extends a bit out of the conductor has a net flux $E A = \delta A / \epsilon_0$, and so $E = \delta / \epsilon_0$. So, $E_1 = E$ and $E_3 = -E$. I'm getting a little confused about finding E_4. Is it 2E, since there's twice as much flux as there was through a pillbox just above the conductor?

Actually, I'm getting confused about the whole thing. When I have a Gaussian surface that encompasses two different electric fields I want to find, which one am I "getting" when I do the calculations?

Also, is there some way to check this by using superposition or something?

Last edited:
Bump?

E1 and E3 at the surfaces? If that's the case you're right, they should be equal but opposite. What do you mean by a Gaussian pillbox? If the coductor is an infinetely wide plane, then what you need to do is construct a gaussian surface surrounding the plane. To do that for a plane that approaches infinite distances, just surround it by two parallel planes of area A that also approach infinite distances:
___________1 (plane)

-------------
////////////// (Conductor)
-------------
___________2 (plane)

In that sense, set E*2A = q/e... dq = (delta)*dA; therefore, E*2A = [(delta)*A]/e, and E = (delta)/2e, in which delta is the surface charge density

Last edited:
Gear300 said:
In that sense, set E*2A = q/e... dq = (delta)*dA; therefore, E*2A = [(delta)*A]/e, and E = (delta)/2e, in which delta is the surface charge density

What does that E represent? The region 4? So it has half the field strength as regions 1 and 3?

Well...dont know how reliable my answer is, but that's what I'm thinking. I could be wrong if the conductor has two sides with surface charge density...but if its thin enough to be a single plane, it should be alright.

Last edited:

## 1. What is Gauss's Law and why is it important in solving electric field problems?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed within that surface. It is important in solving electric field problems because it provides a quick and efficient way to calculate the electric field in a given region by considering the charge distribution.

## 2. What are the four regions in a Gauss's Law problem and how do they affect the calculation of the electric field?

The four regions in a Gauss's Law problem are: inside a charged conductor, outside a charged conductor, on the surface of a charged conductor, and in the region between two charged conductors. These regions affect the calculation of the electric field differently, as the charge distribution and geometry of each region will vary.

## 3. How do you determine the direction of the electric field in a Gauss's Law problem?

The direction of the electric field can be determined by finding the net charge enclosed within a closed surface and using the principle of superposition to determine the direction of the electric field vectors. The electric field will always point away from positive charges and towards negative charges.

## 4. Can you use Gauss's Law to solve for the electric field in non-uniformly charged regions?

Yes, Gauss's Law can be used to solve for the electric field in non-uniformly charged regions. In these cases, the electric field must be broken down into small sections, and the law must be applied to each section separately to find the total electric field.

## 5. Are there any limitations to using Gauss's Law to solve electric field problems?

Yes, there are limitations to using Gauss's Law. It can only be applied to problems with high levels of symmetry, such as spherical, cylindrical, or planar symmetry. In addition, it can only be used to find the electric field at points outside of the charge distribution. For points inside the charge distribution, other methods such as the superposition principle must be used.

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