# Three infinite sheets creating two regions - find the potential difference

• bluelensflares
In summary: I'm not sure about this.Thanks!In summary, the problem involves three infinite sheets with different surface charge densities and the task is to calculate the electric fields and potential difference between two regions. The electric field for an infinite sheet is given by n/(2e) and the net electric field is the sum of the individual fields. The potential is Es, where s is the distance from the negative sheet. The solution for the first part involves adding the electric fields and considering direction, while for the second part, the potential function must be calculated for a single sheet and then added for all three sheets.
bluelensflares

## Homework Statement

Hi all, I have this problem that was on a recent exam but I did not know how to make sense of it.

So, suppose you have three infinite sheets, each layered on top of each other each separated by a distance d. So the first is d above the second, and the second is d above the third. The first has a surface charge density of n, the second sheet n, and the third sheet -n. The first part, which I believe I got correct, was to calculate the electric fields of the two inner regions. The second part, which tripped me up and made me panic, was to calculate the difference in potential between the two regions.

## Homework Equations

E(for an infinite sheet) = n/(2e), where e is the permittivity constant, 8.85E-12.

E(net) = E1 + E2 +E3...

V = Es, for two infinite sheets, where s is the distance from the negative sheet.

## The Attempt at a Solution

So for the first part, calculating the fields, I just added them up and watched the directions: so for the top region, E = (n/2e) + (-n/2e) + (n/2e), since they don't depend on distance, and found the top region to be E = n/2e.

The bottom region was just E = 3n/2e, since they all have fields that point downward at the bottom sheet.

The second part I was stuck - since V = Es, where s is the distance from the negative sheet, I tried to calculate ΔV = Vf- Vi by subtracting the potentials. The problem is, we're talking about regions, not well-defined points... So I was at a loss about what to use for s. Also, instantaneously, it goes from the top voltage to the bottom voltage as you cross the sheet... So I was wondering how to do this. I ended up putting the difference in the fields, and getting ΔV = n/e... but I'm not sure about this.

Thanks!

What is the relationship between the electric field and the potential - in general?Notes:
Technically there's four regions - but you don't care about the regions outside the outermost sheets?

Were you careful to take into account that the electric field is a vector, when you did the superposition?

bluelensflares said:

## Homework Statement

Hi all, I have this problem that was on a recent exam but I did not know how to make sense of it.

So, suppose you have three infinite sheets, each layered on top of each other each separated by a distance d. So the first is d above the second, and the second is d above the third. The first has a surface charge density of n, the second sheet n, and the third sheet -n. The first part, which I believe I got correct, was to calculate the electric fields of the two inner regions. The second part, which tripped me up and made me panic, was to calculate the difference in potential between the two regions.

## Homework Equations

E(for an infinite sheet) = n/(2e), where e is the permittivity constant, 8.85E-12.

E(net) = E1 + E2 +E3...

V = Es, for two infinite sheets, where s is the distance from the negative sheet.

## The Attempt at a Solution

So for the first part, calculating the fields, I just added them up and watched the directions: so for the top region, E = (n/2e) + (-n/2e) + (n/2e), since they don't depend on distance, and found the top region to be E = n/2e.

The bottom region was just E = 3n/2e, since they all have fields that point downward at the bottom sheet.

The second part I was stuck - since V = Es, where s is the distance from the negative sheet, I tried to calculate ΔV = Vf- Vi by subtracting the potentials. The problem is, we're talking about regions, not well-defined points... So I was at a loss about what to use for s. Also, instantaneously, it goes from the top voltage to the bottom voltage as you cross the sheet... So I was wondering how to do this. I ended up putting the difference in the fields, and getting ΔV = n/e... but I'm not sure about this.

Thanks!

E=n/2e is just giving you the magnitude of the E field. The direction is different depending on which side of the sheet you are on, right? Suppose z is the coordinate defining the separation between the sheet and the sheet is located at z=b. Pick ##\hat z## to be a vector pointing in the positive z direction. Then above the sheet the E field is ##\frac{{\hat z}n}{2e}## and below it's ##\frac{{-\hat z}n}{2e}##. Can you work out the potential function for that sheet? Hint: it will involve |z-b|. Once you have the potential function for a single sheet you can just add the potential functions for all of your three sheets.

bluelensflares said:

## Homework Statement

Hi all, I have this problem that was on a recent exam but I did not know how to make sense of it.

So, suppose you have three infinite sheets, each layered on top of each other each separated by a distance d. So the first is d above the second, and the second is d above the third. The first has a surface charge density of n, the second sheet n, and the third sheet -n. The first part, which I believe I got correct, was to calculate the electric fields of the two inner regions. The second part, which tripped me up and made me panic, was to calculate the difference in potential between the two regions.

## Homework Equations

E(for an infinite sheet) = n/(2e), where e is the permittivity constant, 8.85E-12.

E(net) = E1 + E2 +E3...

V = Es, for two infinite sheets, where s is the distance from the negative sheet.

## The Attempt at a Solution

So for the first part, calculating the fields, I just added them up and watched the directions: so for the top region, E = (n/2e) + (-n/2e) + (n/2e), since they don't depend on distance, and found the top region to be E = n/2e.

The bottom region was just E = 3n/2e, since they all have fields that point downward at the bottom sheet.

The second part I was stuck - since V = Es, where s is the distance from the negative sheet, I tried to calculate ΔV = Vf- Vi by subtracting the potentials. The problem is, we're talking about regions, not well-defined points... So I was at a loss about what to use for s. Also, instantaneously, it goes from the top voltage to the bottom voltage as you cross the sheet... So I was wondering how to do this. I ended up putting the difference in the fields, and getting ΔV = n/e... but I'm not sure about this.

Thanks!

You got the electric field correctly, and you have the proper formula for the potential difference ΔV=Es where s is the distance between the two plates. Here s=d . The potential decreases in the direction of the electric field. Is plate 1 at higher or lower potential than plate 2? What is the electric field between them?

Remember, the potential difference was defined by the work done on unit positive charge. All points of one plate are at the same potential. No work is done when a charge is moved along a plate. Also the work does not depend on the actual path taken between two points.

So the work is the same if you move that unit charge from a point of one plate to any point of the other one - equal to the work between opposite points.

ehild

Hello, the potential difference between the two regions can be calculated by integrating the electric field over the distance between the two regions. Since the electric field for an infinite sheet is constant, the integral becomes a simple multiplication of the field strength (n/2e) and the distance between the two regions. So, the potential difference would be (n/2e) x d.

In this case, since the electric field for the top region is pointing upward and the electric field for the bottom region is pointing downward, the potential difference would be negative, indicating a decrease in potential as you move from the top region to the bottom region.

As for your attempt at using the difference in potential between the two regions, it is not incorrect, but it would only give you the potential difference at a specific point on the sheets. In order to calculate the potential difference between the two regions as a whole, you would need to integrate the electric field over the entire distance between the two regions.

I hope this helps clarify the problem for you. Let me know if you have any other questions or concerns.

## 1. What is the concept of "infinite sheets" in this scenario?

In this scenario, "infinite sheets" refer to theoretical planes that extend infinitely in all directions. These sheets have a constant surface charge density that creates an electric field.

## 2. How are the two regions created by the three infinite sheets defined?

The two regions are defined by the position of the sheets relative to each other. The sheets create a boundary between the two regions, with the electric field being different on either side.

## 3. What is the potential difference and how is it related to this scenario?

Potential difference, also known as voltage, is the difference in electric potential between two points in an electric field. In this scenario, the potential difference is calculated by finding the difference in electric potential between the two regions created by the three infinite sheets.

## 4. How can the potential difference be calculated in this scenario?

The potential difference can be calculated using the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the two regions. In this scenario, the electric field can be calculated from the surface charge density of the infinite sheets.

## 5. What are the practical applications of studying the potential difference in this scenario?

Studying the potential difference in this scenario can help us understand and predict the behavior of electric fields in more complex systems. This knowledge is essential in fields such as electronics, power systems, and telecommunications.

Replies
7
Views
1K
Replies
26
Views
1K
Replies
1
Views
350
Replies
1
Views
1K
Replies
2
Views
4K
Replies
8
Views
758
Replies
6
Views
781
Replies
9
Views
3K
Replies
1
Views
1K
Replies
7
Views
1K