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In summary: Due to the symmetry the field has to be the same everywhere on the far side of the plane. What would a non-zero field imply for the total charge on the plate?Not sure I understand what you mean by "the field below the plane is 0". If there is a net charge on the plane, then the field below the plane would be nonzero.

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Silviu said:

This would be a grounded, infinite plane.

So try this analogous situation. You have a charge at the center of a grounded, conducting spherical shell. What is the field outside of the shell?

Zz.

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An arm waving description of the situation:Silviu said:How do we know this?

The Field at the surface of the plate has to be normal to the surface If it were not then charge would flow across the surface to make it so. This is the steady state situation which took a little time to establish itself as the charge was originally moved into its position. There will be a resulting distribution of imbalanced charges over the surface and these charges have the same field as would a single charged, placed in the mirror image position as the 'real charge'.

The plate will have no net charge so there must be polarisation, with a same sign charge distributed along the back face of the plate (to a minimum energy configuration).

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Not sure I understand what you mean. Far from the center the field goes to zero, but I am not sure how does this imply it to be zero below the plane everywhere.mfb said:

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Yes, grounded, sorry for missing that. Well in this case I assume that if you have q at the center, you get -q to gather on the sphere. By Gauss law, the field outside is zero and hence the potential too for ##r>R##, with ##R## the radius of the sphere. Inside, as the potential is continuous, the potential would be ##\frac{1}{4 \pi \epsilon_0}\frac{q}{r}-\frac{1}{4 \pi \epsilon_0}\frac{q}{R}##. However in my case I am not sure how to take the gaussian surface to proceed the same wayZapperZ said:This would be a grounded, infinite plane.

So try this analogous situation. You have a charge at the center of a grounded, conducting spherical shell. What is the field outside of the shell?

Zz.

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Silviu said:Yes, grounded, sorry for missing that. Well in this case I assume that if you have q at the center, you get -q to gather on the sphere. By Gauss law, the field outside is zero and hence the potential too for ##r>R##, with ##R## the radius of the sphere. Inside, as the potential is continuous, the potential would be ##\frac{1}{4 \pi \epsilon_0}\frac{q}{r}-\frac{1}{4 \pi \epsilon_0}\frac{q}{R}##. However in my case I am not sure how to take the gaussian surface to proceed the same way

But look at what the grounded spherical shell did! It

So what do you think your grounded, infinite conducting plane might do?

Zz.

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I am not sure I understand what you mean by "The plate will have no net charge". If the point charge is q, the surface charge density on the plane will be such that the total charge on the plane will be -1, so the plane will have a net charge, won't it? But this still doesn't tell my the field is zero below.sophiecentaur said:An arm waving description of the situation:

The Field at the surface of the plate has to be normal to the surface If it were not then charge would flow across the surface to make it so. This is the steady state situation which took a little time to establish itself as the charge was originally moved into its position. There will be a resulting distribution of imbalanced charges over the surface and these charges have the same field as would a single charged, placed in the mirror image position as the 'real charge'.

The plate will have no net charge so there must be polarisation, with a same sign charge distributed along the back face of the plate (to a minimum energy configuration).

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Oh ok, I see your point. So the plane would simply shield the charge and no field would go below. However, how do I proceed mathematically to show it (similarly to the way I did with Gauss law for the sphere)?ZapperZ said:But look at what the grounded spherical shell did! Itshieldedthe charge's electric field, so that none of them escaped and is detected outside of the shell!

So what do you think your grounded, infinite conducting plane might do?

Zz.

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Silviu said:Oh ok, I see your point. So the plane would simply shield the charge and no field would go below. However, how do I proceed mathematically to show it (similarly to the way I did with Gauss law for the sphere)?

You can't do gauss's law easily in this case, because you can't construct a gaussian surface that will get you either a constant electric flux, or zero flux.

What you need to do is find the induced surface charge density on the conducting plane, and then integrate this over the entire surface, i.e. infinite 2D plane. This will give you the total charge, and this total charge should be equal to the original charge, but opposite in sign. This implies that all the field lines from the original charge terminates on the conducting surface, but due to the fact that the plane is grounded, no net charge forms on the other side of the plane. Thus, no electric field on the other side.

Zz.

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In the plane (or plane plus 'Earth') there can be no net charge so the charge at the surface of interest will have an equal and opposite charge' somewhere else' . It will be spread over the 'other side' of the plate, uniformly.Silviu said:I am not sure I understand what you mean by "The plate will have no net charge". If the point charge is q, the surface charge density on the plane will be such that the total charge on the plane will be -1, so the plane will have a net charge, won't it? But this still doesn't tell my the field is zero below.

If the plane is a perfect conductor then there can be no internal field. If there were some perturbation of charge, this would result in a flow of charge until the internal fields were all zero.

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But in the derivation from Griffiths, he shows that there is a total charge of -q on the plane.sophiecentaur said:In the plane (or plane plus 'Earth') there can be no net charge so the charge at the surface of interest will have an equal and opposite charge' somewhere else' . It will be spread over the 'other side' of the plate, uniformly.

If the plane is a perfect conductor then there can be no internal field. If there were some perturbation of charge, this would result in a flow of charge until the internal fields were all zero.

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Griffiths would not argue with Charge Conservation as a concept. Who could? If there is a charge on the surface, there must be an equal and opposite charge that has been repelled by the charge that's been brought next to the plane. As far as Griffiths is concerned, he is only dealing with the effect on the surface and that's all. The concept of an Image Charge is just for convenience (and it happens to work!). How could there possibly actually be a charge in the image position? What field could keep it there if there are, in fact, no internal fields. The only charges that are present are actually on the surface and their density varies over the plane, with a maximum density right under the test charge and a decreasing density as you go further away.Silviu said:But in the derivation from Griffiths, he shows that there is a total charge of -q on the plane.

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Wait, so you have the charge q and on the plane your have a surface charge density ##\sigma## such that its integral over the whole plane is -q. This is what I understand from Griffiths derivation. This still doesn't help me understand why is it obvious that the field just below the plane is 0.sophiecentaur said:Griffiths would not argue with Charge Conservation as a concept. Who could? If there is a charge on the surface, there must be an equal and opposite charge that has been repelled by the charge that's been brought next to the plane. As far as Griffiths is concerned, he is only dealing with the effect on the surface and that's all. The concept of an Image Charge is just for convenience (and it happens to work!). How could there possibly actually be a charge in the image position? What field could keep it there if there are, in fact, no internal fields. The only charges that are present are actually on the surface and their density varies over the plane, with a maximum density right under the test charge and a decreasing density as you go further away.

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I understand the second part. But it is the first part that I am not sure about. He shows that, as you said, the charge on the plane is -q. However, the equation he uses to show this is: ##\frac{\partial V}{\partial n} = - \frac{\sigma}{\epsilon}##. However in general the equation for a plane is: ##\frac{\partial V_{above}}{\partial n} - \frac{\partial V_{below}}{\partial n} = - \frac{\sigma}{\epsilon}##. So he discards the ##\frac{\partial V_{below}}{\partial n}## part, without saying why and uses this to show that the total charge is -q. As far as I understand it, it seems a bit circular: he uses ##\frac{\partial V_{below}}{\partial n} = 0## to show that the charge is -q which is used to show that the field below the plane is zero, but we just used that in the first place. What am I missing here?ZapperZ said:You can't do gauss's law easily in this case, because you can't construct a gaussian surface that will get you either a constant electric flux, or zero flux.

What you need to do is find the induced surface charge density on the conducting plane, and then integrate this over the entire surface, i.e. infinite 2D plane. This will give you the total charge, and this total charge should be equal to the original charge, but opposite in sign. This implies that all the field lines from the original charge terminates on the conducting surface, but due to the fact that the plane is grounded, no net charge forms on the other side of the plane. Thus, no electric field on the other side.

Zz.

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Take a short circuit and introduce a charge at one end. Where does it go and what happens to the field across it?Silviu said:Wait, so you have the charge q and on the plane your have a surface charge density ##\sigma## such that its integral over the whole plane is -q. This is what I understand from Griffiths derivation. This still doesn't help me understand why is it obvious that the field just below the plane is 0.

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Silviu said:I understand the second part. But it is the first part that I am not sure about. He shows that, as you said, the charge on the plane is -q. However, the equation he uses to show this is: ##\frac{\partial V}{\partial n} = - \frac{\sigma}{\epsilon}##. However in general the equation for a plane is: ##\frac{\partial V_{above}}{\partial n} - \frac{\partial V_{below}}{\partial n} = - \frac{\sigma}{\epsilon}##. So he discards the ##\frac{\partial V_{below}}{\partial n}## part, without saying why and uses this to show that the total charge is -q. As far as I understand it, it seems a bit circular: he uses ##\frac{\partial V_{below}}{\partial n} = 0## to show that the charge is -q which is used to show that the field below the plane is zero, but we just used that in the first place. What am I missing here?

He discards the "below" solution because he has already assumed that the reader realizes that the E-field below the plane is zero. Thus, potential gradient is zero.

Again, go back to the metal shell of outer radius R with a charge +Q at the center. If it is NOT grounded, then there will be a -Q induced on the inside surface, while a +Q induced on the outer surface. If you do a Gaussian sphere at r>R, you will enclose a total net charge of +Q. You will have a E-field for r>R.

However, if you GROUND the sphere, the +Q on the outer surface goes away. A gaussian sphere at r>R will enclose no net charge. Thus, no net E-field.

This is an analogous situation with your infinite plane. If it is NOT grounded, then you will have -Q induced on the upper surface and +Q induced on the bottom surface. There will be E-field in the region below the plane. But ground the plane, and only the -Q is left on the upper surface, and the E-field in the region below the plane goes away.

Zz.

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I understand the analogy. As I said before, I am not sure how to do the math. I also understood the explanation with field lines starting on q and ending on the plane. But I feel that this is not mathematically rigorous. In the case of the sphere you can easily apply Gauss law and show that the field outside is zero and hence the potential, too. It is obvious that in the case of the plane the derivation is not that easy, but I am sure there is a derivation. My question is how you show this mathematically i.e. more than an intuitive description of the situation. (I am fine also if you can point me towards an article with the proof)ZapperZ said:He discards the "below" solution because he has already assumed that the reader realizes that the E-field below the plane is zero. Thus, potential gradient is zero.

Again, go back to the metal shell of outer radius R with a charge +Q at the center. If it is NOT grounded, then there will be a -Q induced on the inside surface, while a +Q induced on the outer surface. If you do a Gaussian sphere at r>R, you will enclose a total net charge of +Q. You will have a E-field for r>R.

However, if you GROUND the sphere, the +Q on the outer surface goes away. A gaussian sphere at r>R will enclose no net charge. Thus, no net E-field.

This is an analogous situation with your infinite plane. If it is NOT grounded, then you will have -Q induced on the upper surface and +Q induced on the bottom surface. There will be E-field in the region below the plane. But ground the plane, and only the -Q is left on the upper surface, and the E-field in the region below the plane goes away.

Zz.

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Silviu said:I understand the analogy. As I said before, I am not sure how to do the math. I also understood the explanation with field lines starting on q and ending on the plane. But I feel that this is not mathematically rigorous. In the case of the sphere you can easily apply Gauss law and show that the field outside is zero and hence the potential, too. It is obvious that in the case of the plane the derivation is not that easy, but I am sure there is a derivation. My question is how you show this mathematically i.e. more than an intuitive description of the situation. (I am fine also if you can point me towards an article with the proof)

Instead of a point charge above the plane, replace it with an infinite plane of charge with some constant surface charge density. Here, the E-field will now be uniform, and you may apply Gauss's law to your heart's content.

Zz.

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You mean have 2 parallel planes? I am not sure I understand this.ZapperZ said:Instead of a point charge above the plane, replace it with an infinite plane of charge with some constant surface charge density. Here, the E-field will now be uniform, and you may apply Gauss's law to your heart's content.

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Silviu said:You mean have 2 parallel planes? I am not sure I understand this.

Yes... instead of the charge source being a point charge, replace it with an infinite plane of charge that is parallel to the grounded, conducting plane. Is this not something you have done when you did Gauss's law?

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I remember that if you have 2 infinite parallel planes with opposite surface charge densities, the field outside is zero on both sides (I hope I said it right). But I am not sure how this is equivalent to my situationZapperZ said:Yes... instead of the charge source being a point charge, replace it with an infinite plane of charge that is parallel to the grounded, conducting plane. Is this not something you have done when you did Gauss's law?

Zz.

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It is not "equivalent" but it is something to help you think round the problem and gain some understanding. A linear approach can often hamper your learning.Silviu said:But I am not sure how this is equivalent to my situation

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Silviu said:I remember that if you have 2 infinite parallel planes with opposite surface charge densities, the field outside is zero on both sides (I hope I said it right). But I am not sure how this is equivalent to my situation

I thought you wanted "mathematical derivation" of this. If you can show mathematically that the E-field is zero with an infinite plane of charge, then shouldn't replacing it with just one point charge result in a similar situation below the conducting plane?

Zz.

The electric field below an infinite plane is equal to 0 because of the symmetry of the system. The infinite plane has an equal and opposite charge distribution on both sides, resulting in a cancellation of electric field below the plane.

Yes, this result holds true for any distance below the plane. As long as the plane is truly infinite and has a uniform charge distribution, the electric field below the plane will always be 0.

The electric field above the plane is not 0, but rather has a constant value determined by the surface charge density of the plane. This is because the electric field lines above the plane do not cancel out due to the plane's symmetry.

No, a finite plane cannot have an electric field of 0 below it. This is because a finite plane does not have the same symmetry as an infinite plane, and the electric field lines will not cancel out below the plane.

This result was first discovered by physicist David J. Griffiths in his book "Introduction to Electrodynamics". He used the method of images and symmetry arguments to derive the electric field below an infinite plane as 0.

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