Electric field of an infinite plane and a conductor

[Niles]

Homework Statement

Hi all.

I have two questions.

1) The electric field of a uniformly charged, infinite plane is given by:

E = \frac{\sigma }{{2\varepsilon _0 }}

The electric field just outside a conductor is given by:

E = \frac{\sigma }{{\varepsilon _0 }}

Why are these expressions not the same? Isn't a plane with a surface charge a conductor?

2)
The electric field just outside a conductor is given by:

E = \frac{\sigma }{{\varepsilon _0 }}

Lets say I have two conducting plates situated a distance d from each other. Is the electric field between the two conducting planes given by

E = \frac{\sigma }{{\varepsilon _0 }}

OR

E = \frac{2\sigma }{{\varepsilon _0 }}

I can't find an answer for these questions in my book nor on the WWW. Help is very much appreciated!

 

[tmc]

The answer to your first question may be near the middle / bottom of this page:
http://farside.ph.utexas.edu/teaching/em/lectures/node58.html; specifically equations (623) onwards.

This might allow you to answer your second question.

 

[rohanprabhu]

Isn't a plane with a surface charge a conductor?

no. a charged plane referred to earlier has a uniform charge on one of it's surface only. The other side of it is uncharged. However, it it is a conductor, the other side HAS TO have a negative charge of the equal magnitude. This is because the field inside a conductor is zero [that is how a conductor is defined].

2)
The electric field just outside a conductor is given by:

E = \frac{\sigma }{{\varepsilon _0 }}

Lets say I have two conducting plates situated a distance d from each other. Is the electric field between the two conducting planes given by

E = \frac{\sigma }{{\varepsilon _0 }}

OR

E = \frac{2\sigma }{{\varepsilon _0 }}

The electric field between them is given by E = \frac{\sigma }{{\varepsilon _0 }}. This follows from a simple application of Gauss Law.

Also, do note here that if two conducting plates are placed closed to each other, the surfaces facing each other will develop opposite and equal charges, no matter what charge they initially had. Same applies to the surfaces facing away.

 

[Niles]

The electric field between them is given by E = \frac{\sigma }{{\varepsilon _0 }}. This follows from a simple application of Gauss Law.

This I cannot see how - don't the electric fields just add?

 

[Dick]

The electric field between two equally charged conductors (or planes) will be zero.

 

[Niles]

The electric field between two equally charged conductors (or planes) will be zero.

Say that I have two conducting planes separated with a vertical distance d. Plate 1 at z=0 has a potential 0, and plate 2 at z=d has a potential V maintained by a battery. How come there is an electric field E = V/z there then and doesn't this equal E = sigma/epsilon_0?

 

[Dick]

Say that I have two conducting planes separated with a vertical distance d. Plate 1 at z=0 has a potential 0, and plate 2 at z=d has a potential V maintained by a battery. How come there is an electric field E = V/z there then?

In that case the charge on the two plates isn't equal. The battery pumps charge from one plate to the other creating the charge difference which in turn creates the field.

 

[Niles]

And this electric field does not equal E = \frac{\sigma }{{\varepsilon _0 }}?

 

[Dick]

It's the difference of two expressions like that with two different charge densities. As you said, it's also V/d.

 

[Niles]

If <br /> E = \frac{\sigma }{{\varepsilon _0 }}<br /> is the difference between the electric field of the upper and lower plate, then must it not equal the total field?

 

[Dick]

In between the two plates the field from the upper plate and the field from the lower plate point in opposite directions. If the charges are equal, they cancel, if they aren't, then it's the difference that determines the net field.

 

[Niles]

Hmm, I think there is a contradiction in what you are telling me and what my book is saying. According to my book, the force that the two plates are attracting each other with is given by

F = sigma^2 / (2*e_0) - which is OK

and they say that E = sigma / e_0 => sigma = V*e_0 / z. So they say that the electric field is <br /> <br /> E = \frac{\sigma }{{\varepsilon _0 }}<br /> <br />

 

[Doc Al]

2)
The electric field just outside a conductor is given by:

E = \frac{\sigma }{{\varepsilon _0 }}

OK.

Lets say I have two conducting plates situated a distance d from each other. Is the electric field between the two conducting planes given by

E = \frac{\sigma }{{\varepsilon _0 }}

OR

E = \frac{2\sigma }{{\varepsilon _0 }}

I suspect you are thinking of two oppositely charged plates close together (like a parallel plate capacitor). The field between them is given by the first expression.

Note that you can consider the field between the plates as the sum of the fields created by each sheet of charge:

E = \frac{\sigma }{2\epsilon _0} + \frac{\sigma }{2\epsilon _0} = \frac{\sigma }{\epsilon _0}

 

[Niles]

I suspect you are thinking of two oppositely charged plates close together (like a parallel plate capacitor). The field between them is given by the first expression.

Note that you can consider the field between the plates as the sum of the fields created by each sheet of charge:

E = \frac{\sigma }{2\epsilon _0} + \frac{\sigma }{2\epsilon _0} = \frac{\sigma }{\epsilon _0}

Thanks!

 

[Niles]

Wait, I haven't understood this thoroughly.

We have two conducting plates. I am not told anything about the charges on the plate, only that the lower plate is grounded and the upper plate has a potential V. So the electric field goes from the upper plate to the lower, and this is given by

E = V/z, where z is the distance between the plates.

For this setup I have the following questions:

1) Can I say anything about the amount of charge on each plate from these informations?

2) If I want to find the force that each plate exerts on each other, according to my book (Griffiths' page 103, eq. 2.51) I should use the expression

<br /> F = PA = A\frac{{\varepsilon _0 }}{2}E^2 <br />
where A is the area of the plate.

How can I justify using this expression, when we are dealing with planes? When we are told they are conducting planes, what expression for the electric field am I supposed to use?

 

[Doc Al]

For this setup I have the following questions:

1) Can I say anything about the amount of charge on each plate from these informations?

Sure. If you have the potential and the distance between the plates, then you have the field between the plates. If you have the field, then you have the charge density.

2) If I want to find the force that each plate exerts on each other, according to my book (Griffiths' page 103, eq. 2.51) I should use the expression

<br /> F = PA = A\frac{{\varepsilon _0 }}{2}E^2 <br />
where A is the area of the plate.

How can I justify using this expression, when we are dealing with planes? When we are told they are conducting planes, what expression for the electric field am I supposed to use?

Realize that in this expression for force, E stands for the total field between the plates, not just the field from one plate. In deriving this expression, use the field from one plate to calculate the force on the other.

 

[Niles]

Sure. If you have the potential and the distance between the plates, then you have the field between the plates. If you have the field, then you have the charge density.


Realize that in this expression for force, E stands for the total field between the plates, not just the field from one plate. In deriving this expression, use the field from one plate to calculate the force on the other.

For 1 and 2) This is what I am confused about. In this setup, am I supposed to use the electric field for two conductors (so the total field is E = 2*sigma / e_0) or the field from two oppositely charged infinitely large planes E = sigma/e_0?

I am just told that it is two conducting planes with a potential difference. Nothing else.

 

[Doc Al]

For 1 and 2) This is what I am confused about. In this setup, am I supposed to use the electric field for two conductors (so the total field is E = 2*sigma / e_0) or the field from two oppositely charged infinitely large planes E = sigma/e_0?

I think I see what's messing you up. Realize that a conducting plate with a total charge Q will have different surface charge densities in these two situations:

(a) The conducting plate is isolated from the world. The charge is evenly divided on both sides of the plate so the surface charge density on each side is Q/(2A). The field from that plate is Q/(2 A e_0).

Note that this is identical to the field from a sheet of charge Q. (Which makes sense, right?)

(b) The conducting plate is parallel to another oppositely charged conducting plate. In this case, the entire charge is on the inner surface, so the surface charge density is Q/A. The field between those plates is Q/(A e_0), which is twice what it would be if there were just one plate.

To answer your question, the field between the two conducting plates is sigma/e_0.

I am just told that it is two conducting planes with a potential difference. Nothing else.

That's all the information you need.

 

[Niles]

I think I see what's messing you up. Realize that a conducting plate with a total charge Q will have different surface charge densities in these two situations:

(a) The conducting plate is isolated from the world. The charge is evenly divided on both sides of the plate so the surface charge density on each side is Q/(2A). The field from that plate is Q/(2 A e_0). That is the field on each side, right?

Note that this is identical to the field from a sheet of charge Q. (Which makes sense, right?) Yes, this does make sense.

(b) The conducting plate is parallel to another oppositely charged conducting plate. In this case, the entire charge is on the inner surface, so the surface charge density is Q/A. So all the charge resides on one surface. From E = sigma / e_0 we get that E = Q/(A*e_0) for this one plate on the side facing towards the other plate.

The field between those plates is Q/(A e_0), which is twice what it would be if there were just one plate. Why not the double?

I hope it is alright that I just comment on your post this way.

 

[Doc Al]

I hope it is alright that I just comment on your post this way.

It makes it a more difficult to respond to your comments.

(b) The conducting plate is parallel to another oppositely charged conducting plate. In this case, the entire charge is on the inner surface, so the surface charge density is Q/A.

So all the charge resides on one surface. From E = sigma / e_0 we get that E = Q/(A*e_0) for this one plate on the side facing towards the other plate.

This is the field due to both conducting plates. (The field from a single sheet of charge would only be half that.) Realize that when deriving the field from a conductor using Gauss's law, you are including the effects of both plates at once.

The field between those plates is Q/(A e_0), which is twice what it would be if there were just one plate.

Why not the double?

It is double! (Double the field that would exist if there were only one plate with the same charge.)

 

[Niles]

I can see it wasn't such a good idea to comment on your post like that. Sorry.

The boundary condition for the electric field tells us that the electric field outside our conductor (in this case at the surface that points toward the other plate) is given by

<br /> E = \frac{\sigma }{{\varepsilon _0 }} = \frac{Q}{A\varepsilon _0 }<br />

since sigma = Q/A for this side. And this is for one of the plates. According to your explanation, this is wrong - why?

 

[Doc Al]

The boundary condition for the electric field tells us that the electric field outside our conductor (in this case at the surface that points toward the other plate) is given by

<br /> E = \frac{\sigma }{{\varepsilon _0 }} = \frac{Q}{A\varepsilon _0 }<br />

since sigma = Q/A for this side. And this is for one of the plates. According to your explanation, this is wrong - why?

That result comes from applying Gauss's law. When you say "this is for one of the plates" what do you mean? That's the field between two plates (where Q is the charge on each plate). (The fact that you are using sigma = Q/A, instead of sigma = Q/2A, means that you are including the effect of both plates.)

 

[Niles]

First, I really appreciate you helping me. Your patience is apparently never-ending.

When the two conducting plates are parallel, the charges reside on the sides facing towards each other. When we look at a conducting plane, where the charge only resides on one side, then the surface-charge of the plane on that side is:

sigma = Q/A.

This I insert into E = sigma / e_0 = Q / (A*e_0). This is the electric field from one of the plates, when they are parallel. There is a similar contribution from the other plate.

What argument is wrong in this?

 

[Doc Al]

When the two conducting plates are parallel, the charges reside on the sides facing towards each other. When we look at a conducting plane, where the charge only resides on one side, then the surface-charge of the plane on that side is:

sigma = Q/A.

This I insert into E = sigma / e_0 = Q / (A*e_0). This is the electric field from one of the plates, when they are parallel. There is a similar contribution from the other plate.

What argument is wrong in this?

What's wrong is your thinking that that's the electric field from one of the plates. Not so. The electric field contribution from one plate is half that.

What's the charge density on one plate? Sigma = Q/A. What's the field from such a charge distribution? E = sigma/2 e_0. (Which makes the total field between the plates = sigma/e_0.)

Don't confuse the field obtained using Gauss's law, which automatically includes the effect of everything--including that second plate, with the field obtained from just looking at the charge distribution of one plate. The electrostatic fields from each plate do add up (superposition), but Gauss's law already includes that effect. (Of course you get the same answer for the field between the plates no matter which method you use, if you use it correctly.)

 

[Niles]

There's two expressions for the electric field in this case:

1) E = sigma / 2*e_0 - this is for the electric field from a charged plane.

2) E = sigma / e_0 - this is for the electric field outside a conductor.

You are saying that I am supposed to use #1, and yes - that will make it work, and #2 will not make it work. I can see how the two expressions are derived in my book, and I understand it fully and I agree - but as I see it, I am "allowed" to use both expressions here. I can use #1 because it's a plane, and #2 because it's a conductor.

 

[Doc Al]

There's two expressions for the electric field in this case:

1) E = sigma / 2*e_0 - this is for the electric field from a charged plane.

2) E = sigma / e_0 - this is for the electric field outside a conductor.

You are saying that I am supposed to use #1, and yes - that will make it work, and #2 will not make it work. I can see how the two expressions are derived in my book, and I understand it fully and I agree - but as I see it, I am "allowed" to use both expressions here. I can use #1 because it's a plane, and #2 because it's a conductor.

No, I'm saying that either method will work--if used correctly. What you can't do is use the result for the field outside a conductor--which assumes electrostatic equilibrium within the conductor--and then double it.

Let's examine that case in a little detail. Imagine a conducting sheet with one side covered with a charge Q and the other side uncharged. The formula E = sigma / e_0 applies for the field outside a conductor because we can assume that the field inside the conductor is zero. But if you just had a single conducting sheet with charge on one side, how can the field inside the conductor be zero? The only way that the field within the conductor can be zero is if some other fields are involved--in particular the field from the other charged plate--to make it zero. Thus that equation already incorporates the field from the second plate--it's not the field from one plate by itself.

Let me know if this is making any sense.

 

[Niles]

That does make sense, and that is a very good explanation to my problem.

My mystery is solved. Thank you!

 

[Doc Al]

You're welcome.