Finding the charge density of an infinite plate

[Eitan Levy]

Homework Statement

First of all let me apologise for asking so many questions today, this is just my collection of unsolved problems and they have been driving me crazy.



An infinite plate with has a charge density σ. We move a conductor under it as described in the figure below. A is the surface area of the table.



The bottom of the conductor is grounded.



It is also known that A>>9d^2, and the system is stable.



Find the charge density at the top of the conductor (the part marked with an A on it)

Relevant Equations

Gauss law

The answer is that the charge density would be -σ, I cannot for the life of me understand why would that be the case. Of course it makes sense but I can't convince myself that it would be the only possible answer.

I have tried to apply Gauss law a few times, but it doesn't yield anything.

 

[kuruman]

Is the infinite plate with charge density σ conducting? If so, you have a parallel plate capacitor with equal and opposite surface charge densities. More formally, you can construct the standard Gaussian pillbox with its flat ends completely inside the conductors. The electric flux through this box is zero. What is the enclosed charge that you already know is there?

 

[haruspex]

Is the infinite plate with charge density σ conducting?

Wouldn't that create a runaway migration of the charge to the region above the earthed block?

you can construct the standard Gaussian pillbox with its flat ends completely inside the conductors.

For the plate not being a conductor, allow the pillbox to extend above it?

 

[kuruman]

Wouldn't that create a runaway migration of the charge to the region above the earthed block?

Good question. The figure provided by OP requires some interpretation. The plate that carries the charge density is said to be "infinite", yet it is shown as having finite area A or maybe what is shown is only the area in the infinite plane that is directly above the area of the bottom conductor. My thought when I posted #2 was that, assuming that the top charge distribution is on a conductor and given that A >>9d², the system forms a parallel plate capacitor which is easy to handle. I think the author's intention is to consider the field to be uniform in the in-between space.

For the plate not being a conductor, allow the pillbox to extend above it?

Yes. The key question to be answered is "When the earthed conductor is put in place, does charge migrate to the lower surface or not?"

 

[Eitan Levy]

Is the infinite plate with charge density σ conducting? If so, you have a parallel plate capacitor with equal and opposite surface charge densities. More formally, you can construct the standard Gaussian pillbox with its flat ends completely inside the conductors. The electric flux through this box is zero. What is the enclosed charge that you already know is there?

I didn't realize I could choose the pillbox to be completely inside the plate.
One question about this - if I try to calculate the field of an infinite plate with the said pillbox, I get the field to be twice as large. Is the charge density inside the pillbox (looking at the part inside the infinite plate) only half of the density of the plate? Otherwise I get:
EA=4*pi*K*(density*A)
And I know that the field should be 2*pi*K*density overall.

Edit: Another thing which bothers me is that following what you said, if there are two conducting infinite plates next to each other they must have opposite densities. Is this always the case? Is it impossible for other situations to happen?

 

[rude man]

Edit: Another thing which bothers me is that following what you said, if there are two conducting infinite plates next to each other they must have opposite densities. Is this always the case? Is it impossible for other situations to happen?

It is 3am here so I need to get some sleep.
I haven't looked at this thread at all but I can hopefully answer this question, as follows: if charges were not equal & opposite, what would your pillbox indicate running from inside one plate to anywhere in the gap between the plates? Would it make a difference where in the gap you run the gap end? (Remember that we are neglecting edge effects so think infinite plates and E field perpendicular to the plates everywhere).

BTW this is a vitally important issue to settle for you).

 

[Eitan Levy]

It is 3am here so I need to get some sleep.
I haven't looked at this thread at all but I can hopefully answer this question, as follows: if charges were not equal & opposite, what would your pillbox indicate running from inside one plate to anywhere in the gap between the plates? Would it make a difference where in the gap you run the gap end? (Remember that we are neglecting edge effects so think infinite plates and E field perpendicular to the plates everywhere).

BTW this is a vitally important issue to settle for you).

I am having troubles understanding the following: What exactly would be Q_in in that case? If we take a pillbox that intersects with the plate in an area with size A, is that charge going to be σA? Is it going to be σA/2?

If it is σA then there is a problem with the case of a single infinite plate. The field outside should be 2πKσ, but if we take a pillbox running from inside one plate to any other point, we get a field with size 4πKσ.

If it σA/2 then there is a problem with the case of two infinite plates with opposite densities. Taking the same pillbox, I will get that the field size is 2πKσ instead of 4πKσ.

I added my work for the first case in order to explain better, the second case is similar.

 

[kuruman]

Perhaps this will clarify the situation for you. We consider two cases of a very thin rectangular sheet of area $A$ that has total charge $Q$. We neglect edge effects and assume uniform charge distribution.
Case I - The charges are "pasted" on the sheet and are not free to move.
The surface charge density is $\sigma=\frac{Q}{A}$. Application of Gauss's law using the standard Gaussian pillbox with two faces outside either side of the sheet gives$$E_{left}\Delta A+E_{right}\Delta A =\frac{q_{enc}}{\epsilon_0}=\frac{\sigma\Delta A}{\epsilon_0}$$ By symmetry, $E_{left}=E_{right}=E_{I}$ which gives $$2E_{I}\Delta A=\frac{\sigma\Delta A}{\epsilon_0}=\frac{Q\Delta A}{\epsilon_0A}~\Rightarrow~E_I=\frac{Q}{\epsilon_0A}.$$
Case II - The sheet is conducting and the charges are free to move.
In this case, if there are no other charges nearby, the surface charge density on each face is $\sigma=\frac{Q}{2A}$. Application of Gauss's law using the standard Gaussian pillbox with two faces outside either side of the sheet gives$$E_{left}\Delta A+E_{right}\Delta A =\frac{q_{enc}}{\epsilon_0}=2\times\frac{\sigma\Delta A}{\epsilon_0}$$ By symmetry, $E_{left}=E_{right}=E_{II}$ which gives $$2E_{II}\Delta A=2\times\frac{\sigma\Delta A}{\epsilon_0}=2\times\frac{(Q/2)\Delta A}{\epsilon_0A}~\Rightarrow~E_{II}=\frac{Q}{2\epsilon_0 A}.$$ What if we chose a pillbox that has one face inside the conductor and the other face sticking out on the left? Then, $$E_{left}\Delta A+0\times \Delta A =\frac{q_{enc}}{\epsilon_0}=\frac{\sigma\Delta A}{\epsilon_0}=\frac{(Q/2)\Delta A}{\epsilon_0A}~\Rightarrow~E_{left}=\frac{Q}{2\epsilon_0 A}=E_{II}=E_I.$$See how it works?

 

[Eitan Levy]

Perhaps this will clarify the situation for you. We consider two cases of a very thin rectangular sheet of area $A$ that has total charge $Q$. We neglect edge effects and assume uniform charge distribution.
Case I - The charges are "pasted" on the sheet and are not free to move.
The surface charge density is $\sigma=\frac{Q}{A}$. Application of Gauss's law using the standard Gaussian pillbox with two faces outside either side of the sheet gives$$E_{left}\Delta A+E_{right}\Delta A =\frac{q_{enc}}{\epsilon_0}=\frac{\sigma\Delta A}{\epsilon_0}$$ By symmetry, $E_{left}=E_{right}=E_{I}$ which gives $$2E_{I}\Delta A=\frac{\sigma\Delta A}{\epsilon_0}=\frac{Q\Delta A}{\epsilon_0A}~\Rightarrow~E_I=\frac{Q}{\epsilon_0A}.$$
Case II - The sheet is conducting and the charges are free to move.
In this case, if there are no other charges nearby, the surface charge density on each face is $\sigma=\frac{Q}{2A}$. Application of Gauss's law using the standard Gaussian pillbox with two faces outside either side of the sheet gives$$E_{left}\Delta A+E_{right}\Delta A =\frac{q_{enc}}{\epsilon_0}=2\times\frac{\sigma\Delta A}{\epsilon_0}$$ By symmetry, $E_{left}=E_{right}=E_{II}$ which gives $$2E_{II}\Delta A=2\times\frac{\sigma\Delta A}{\epsilon_0}=2\times\frac{(Q/2)\Delta A}{\epsilon_0A}~\Rightarrow~E_{II}=\frac{Q}{2\epsilon_0 A}.$$ What if we chose a pillbox that has one face inside the conductor and the other face sticking out on the left? Then, $$E_{left}\Delta A+0\times \Delta A =\frac{q_{enc}}{\epsilon_0}=\frac{\sigma\Delta A}{\epsilon_0}=\frac{(Q/2)\Delta A}{\epsilon_0A}~\Rightarrow~E_{left}=\frac{Q}{2\epsilon_0 A}=E_{II}=E_I.$$See how it works?

One question: When are the charges free to move and when are they "pasted"?

 

[kuruman]

It depends on the material. Generally speaking, charges are free to move in a conductor. They are not free to move in a dielectric. There are in between cases like semiconductors but for an oversimplified introduction to the subject this distinction suffices.

 

[Eitan Levy]

It depends on the material. Generally speaking, charges are free to move in a conductor. They are not free to move in a dielectric. There are in between cases like semiconductors but for an oversimplified introduction to the subject this distinction suffices.

One more thing regarding the original question. How did you know that on the grounded part the charge is 0?

 

[kuruman]

One more thing regarding the original question. How did you know that on the grounded part the charge is 0?

I didn't know. In fact I knew there will be net negative charge on the grounded conductor regardless of whether the charged plane near it is a conductor or dielectric. What is described here is part of a process known as charging by induction where one stops after step 2 and before the grounding wire is disconnected.

"Grounding" a conductor implies that there is zero charge on it only if there are no other charge distributions nearby. Grounding means that the conductor is at the same potential as the Earth which is also a conductor. With a grounded neutral conductor, the negative electrons that are free to move inside it have no reason to do so. If you bring a positive (negative) charge distribution near it while it remains grounded, you raise (lower) the electrostatic potential of the conductor relative to the Earth. Because there is a potential difference between conductor and Earth. the free electrons will flow from the Earth (conductor) into the conductor (the Earth) until the electrostatic potential difference is zero in which case no more electrons need to move.

Grounding a conductor is analogous to connecting a vessel of water with a tube at its bottom to a calm lake. Water will move into or from the lake depending on whether the initial level in the vessel is higher or lower than the lake level. Here, the water molecules that are free to move do so until the gravitational potential difference between the levels is zero.