Griffith's E&M page 124: Explanation for Nonzero Field

• ehrenfest
In summary: Spherical conductor.An infinite plane can be considered a sphere of infinite radius.Ok. So if I have a sphere and I want to divide it into two parts, the part that has the charge and the part that doesn't, how do I do that?You could just take a cube of infinite side length and then cut it in half. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.And as you well know, the electric field inside a conductor is zero.Here is a crude drawing:
ehrenfest
[SOLVED] Griffiths page 124

Homework Statement

Stop reading if you do not have Griffith's E and M book.

Can someone explain why he says "But in the second case only the upper region contains a nonzero field, ... " ?

I don't understand why the field is necessarily zero in the lower region?

ehrenfest said:

Homework Statement

Stop reading if you do not have Griffith's E and M book.

Can someone explain why he says "But in the second case only the upper region contains a nonzero field, ... " ?

I don't understand why the field is necessarily zero in the lower region?

The Attempt at a Solution

Because the lower region is inside a conductor, no?

The conductor is just the x-y plane I think.

Take a spherical conductor. Remember that the potential is constant on the inside. Now simply take the limit as the radius goes to infinity. Now, you have an infinite conducting plane. One side of it (half-space) is "outside", the other side is "inside". And, the region inside a conductor has zero electric field (and therefore zero energy density).

Thus, the conducting plane divides infinite space in half; one half is occupied by charges, but the other half is empty. The field within the latter half is zero.

When you solve the problem via images, you are introducing a fictitious charge on the "other side" that replicates the boundary conditions. When you calculate the total energy, you have to ignore the energy due to this fictitious charge.

Alternatively, you can say that the field on the "other side" of the conductor due to the charge q is exactly canceled by the field due to the surface charge distribution on the conductor. Again, since the field is zero, the energy density in that half of space is zero.

Ben Niehoff said:
Take a spherical conductor.

Do you mean circular? And do you mean filled or hollow?

No, spherical. An infinite plane can be considered a sphere of infinite radius.

What?!? In the problem, the x-y plane is the conductor. There is now way you can blow up a sphere large enough so that the it will coincide with the x-y plane since it will only be locally flat. You're example might make sense if it were a circular disk. I have no idea what you are saying and why you are using a sphere. Please clarify.

Yes, a sphere is only locally flat. But if you blow it up to infinite radius, while simultaneously taking the center infinitely far away, you will end up with a plane. Essentially, that "locally flat" part expands until it fills the entire xy plane. It's the same reason the Earth looks flat, even though it isn't. The difference is that the Earth has only a finite radius--in the case of an infinite radius, the "looks flat" part becomes all that matters.

At any rate, you can just take a cube of infinite side length, if you want. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.

And as you well know, the electric field inside a conductor is zero.

Here is a crude drawing:

Code:
       * Q (point charge)

(outside)
-------------------------------------- (infinite conducting plane)
(inside)

The electric field is zero on the "inside".

Ben Niehoff said:
Yes, a sphere is only locally flat. But if you blow it up to infinite radius, while simultaneously taking the center infinitely far away, you will end up with a plane. Essentially, that "locally flat" part expands until it fills the entire xy plane. It's the same reason the Earth looks flat, even though it isn't. The difference is that the Earth has only a finite radius--in the case of an infinite radius, the "looks flat" part becomes all that matters.

At any rate, you can just take a cube of infinite side length, if you want. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.

And as you well know, the electric field inside a conductor is zero.

Here is a crude drawing:

Code:
       * Q (point charge)

(outside)
-------------------------------------- (infinite conducting plane)
(inside)

The electric field is zero on the "inside".

But its only inside the "meat" of the conductor that the electric field is zero. See the first sentence on page 90. That's why I was asking whether your conductor was "hollow" or not. Apparently it is hollow, and thus I don't think we can say just a priori that the field inside is zero. But maybe I am mistaken.

It doesn't matter whether the conductor is hollow or not. Often, it doesn't even matter whether the shell of the conductor has holes in it or not! This is why a Faraday cage works.

If a (solid) conductor has a net charge, then all of the charge resides on the surface. From this, it should be clear that the conductor can be hollowed out without changing anything.

It is, perhaps, a curious mathematical fact that the field within a region R can always be canceled by some appropriate distribution of charges on the boundary of R. This follows from the Laplace equation.

1. What is Griffith's E&M page 124?

Griffith's E&M page 124 refers to a specific page in the textbook "Introduction to Electrodynamics" by David J. Griffiths. It is a widely used textbook in the field of electromagnetism.

2. What is the significance of page 124 in Griffith's E&M?

Page 124 is significant because it presents an explanation for why the electric and magnetic fields in a vacuum are not always zero, even in the absence of any charges or currents. This is known as the non-zero field phenomenon.

3. How does Griffith's E&M page 124 explain the non-zero field?

The explanation given on page 124 is based on the concept of virtual particles. According to quantum mechanics, even in vacuum, there is a constant creation and annihilation of particle-antiparticle pairs. These virtual particles can briefly interact with the electric and magnetic fields, causing them to have non-zero values.

4. Is the explanation on Griffith's E&M page 124 widely accepted?

Yes, the explanation for the non-zero field on page 124 is widely accepted in the scientific community. It is supported by experimental evidence and is consistent with the principles of quantum mechanics.

5. Are there any practical applications of the non-zero field phenomenon?

Yes, the non-zero field phenomenon has practical applications in various fields such as quantum mechanics, cosmology, and particle physics. It helps to understand the behavior of vacuum and plays a crucial role in modern theories of the universe.

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The content structure of the provided text is a discussion from a physics forum about a specific concept in electromagnetism, specifically from Griffith's "Introduction to Electrodynamics" textbook. The discussion revolves around the electric field inside and outside a conductor, particularly in the context of an infinite conducting plane.

Analysis:
no_question

the text does not contain a mathematical question or a calculation problem. it is a discussion and explanation of a physics concept, specifically the electric field inside and outside a conductor. while there are questions in the text, they are conceptual and not mathematical in nature.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Griffith's E&M page 124: Explanation for Nonzero Field

• ehrenfest
In summary: Spherical conductor.An infinite plane can be considered a sphere of infinite radius.Ok. So if I have a sphere and I want to divide it into two parts, the part that has the charge and the part that doesn't, how do I do that?You could just take a cube of infinite side length and then cut it in half. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.And as you well know, the electric field inside a conductor is zero.Here is a crude drawing:
ehrenfest
[SOLVED] Griffiths page 124

Homework Statement

Stop reading if you do not have Griffith's E and M book.

Can someone explain why he says "But in the second case only the upper region contains a nonzero field, ... " ?

I don't understand why the field is necessarily zero in the lower region?

ehrenfest said:

Homework Statement

Stop reading if you do not have Griffith's E and M book.

Can someone explain why he says "But in the second case only the upper region contains a nonzero field, ... " ?

I don't understand why the field is necessarily zero in the lower region?

The Attempt at a Solution

Because the lower region is inside a conductor, no?

The conductor is just the x-y plane I think.

Take a spherical conductor. Remember that the potential is constant on the inside. Now simply take the limit as the radius goes to infinity. Now, you have an infinite conducting plane. One side of it (half-space) is "outside", the other side is "inside". And, the region inside a conductor has zero electric field (and therefore zero energy density).

Thus, the conducting plane divides infinite space in half; one half is occupied by charges, but the other half is empty. The field within the latter half is zero.

When you solve the problem via images, you are introducing a fictitious charge on the "other side" that replicates the boundary conditions. When you calculate the total energy, you have to ignore the energy due to this fictitious charge.

Alternatively, you can say that the field on the "other side" of the conductor due to the charge q is exactly canceled by the field due to the surface charge distribution on the conductor. Again, since the field is zero, the energy density in that half of space is zero.

Ben Niehoff said:
Take a spherical conductor.

Do you mean circular? And do you mean filled or hollow?

No, spherical. An infinite plane can be considered a sphere of infinite radius.

What?!? In the problem, the x-y plane is the conductor. There is now way you can blow up a sphere large enough so that the it will coincide with the x-y plane since it will only be locally flat. You're example might make sense if it were a circular disk. I have no idea what you are saying and why you are using a sphere. Please clarify.

Yes, a sphere is only locally flat. But if you blow it up to infinite radius, while simultaneously taking the center infinitely far away, you will end up with a plane. Essentially, that "locally flat" part expands until it fills the entire xy plane. It's the same reason the Earth looks flat, even though it isn't. The difference is that the Earth has only a finite radius--in the case of an infinite radius, the "looks flat" part becomes all that matters.

At any rate, you can just take a cube of infinite side length, if you want. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.

And as you well know, the electric field inside a conductor is zero.

Here is a crude drawing:

Code:
       * Q (point charge)

(outside)
-------------------------------------- (infinite conducting plane)
(inside)

The electric field is zero on the "inside".

Ben Niehoff said:
Yes, a sphere is only locally flat. But if you blow it up to infinite radius, while simultaneously taking the center infinitely far away, you will end up with a plane. Essentially, that "locally flat" part expands until it fills the entire xy plane. It's the same reason the Earth looks flat, even though it isn't. The difference is that the Earth has only a finite radius--in the case of an infinite radius, the "looks flat" part becomes all that matters.

At any rate, you can just take a cube of infinite side length, if you want. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.

And as you well know, the electric field inside a conductor is zero.

Here is a crude drawing:

Code:
       * Q (point charge)

(outside)
-------------------------------------- (infinite conducting plane)
(inside)

The electric field is zero on the "inside".

But its only inside the "meat" of the conductor that the electric field is zero. See the first sentence on page 90. That's why I was asking whether your conductor was "hollow" or not. Apparently it is hollow, and thus I don't think we can say just a priori that the field inside is zero. But maybe I am mistaken.

It doesn't matter whether the conductor is hollow or not. Often, it doesn't even matter whether the shell of the conductor has holes in it or not! This is why a Faraday cage works.

If a (solid) conductor has a net charge, then all of the charge resides on the surface. From this, it should be clear that the conductor can be hollowed out without changing anything.

It is, perhaps, a curious mathematical fact that the field within a region R can always be canceled by some appropriate distribution of charges on the boundary of R. This follows from the Laplace equation.

1. What is Griffith's E&M page 124?

Griffith's E&M page 124 refers to a specific page in the textbook "Introduction to Electrodynamics" by David J. Griffiths. It is a widely used textbook in the field of electromagnetism.

2. What is the significance of page 124 in Griffith's E&M?

Page 124 is significant because it presents an explanation for why the electric and magnetic fields in a vacuum are not always zero, even in the absence of any charges or currents. This is known as the non-zero field phenomenon.

3. How does Griffith's E&M page 124 explain the non-zero field?

The explanation given on page 124 is based on the concept of virtual particles. According to quantum mechanics, even in vacuum, there is a constant creation and annihilation of particle-antiparticle pairs. These virtual particles can briefly interact with the electric and magnetic fields, causing them to have non-zero values.

4. Is the explanation on Griffith's E&M page 124 widely accepted?

Yes, the explanation for the non-zero field on page 124 is widely accepted in the scientific community. It is supported by experimental evidence and is consistent with the principles of quantum mechanics.

5. Are there any practical applications of the non-zero field phenomenon?

Yes, the non-zero field phenomenon has practical applications in various fields such as quantum mechanics, cosmology, and particle physics. It helps to understand the behavior of vacuum and plays a crucial role in modern theories of the universe.

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user
The content structure of the provided text is a discussion from a physics forum about a specific concept in electromagnetism, specifically from Griffith's "Introduction to Electrodynamics" textbook. The discussion revolves around the electric field inside and outside a conductor, particularly in the context of an infinite conducting plane.

Analysis:
TYPE: NO_QUESTION

The text does not contain a mathematical question or a calculation problem. It is a discussion and explanation of a physics concept, specifically the electric field inside and outside a conductor. While there are questions in the text, they are conceptual and not mathematical in nature.