Earthing of two parallel conducting plates

[palaphys]

Homework Statement

There are three plates, X Y and Z, with plate Y given a charge Q. predict what will happen when X and Z are simultaneously earthed, as shown in the figure.

Relevant Equations

E= - gradient V

If both S1 and S2 are switched on, then the charge on the leftmost and rightmost surfaces will be zero. Also, we may assume that a charge q would appear on the left surface of Y, and a charge Q-q on the right surface of the same. due to this, there will be a charge q-Q on the left surface of Z and a charge -q on the right surface of X, as shown in the figure below:

now I feel that X and Z are at the same potentials, so the potential difference between XY and YZ must be the same. however, I am unable to compute the electric field from X to Y and Y to Z, which is necessary to find the potential.
Here is what I tried:
1. I considered the field due to each uniform sheet, which is just $q/2Aε_0$, and summed up the directions vectorially. however, I get zero due to this..
Please help

 

[kuruman]

Perhaps you could see what's gong on if you tried to make a drawing to scale. Labeling "2d" a distance that looks less than a distance labeled "d" cannot help.

You know that the electric field is uniform between plates. I recommend that you make a graph of electrostatic potential between plates as a function of displacement starting at the left plate all the way to the right plate. Note that the end points are at the same potential of zero. So you start at zero potential and end at zero potential because that's what "earthing" means.

What happens in between? In what direction does the electric field point in each region? How do the magnitudes of the electric field in each region compare? What does this imply about the charge density on each surface of the middle plate?

On edit

"Earthing" means providing a conducting connection to the Earth that is an infinite reservoir of electrons. This means that it can provide or accept as many electrons as needed without changing its constant electrostatic potential set at zero.

Consider the schematic on the right. The region of space outlined in red contains charge +Q on the middle plate. When the switch is open, the left and right plates have no charge and all the charge is on the middle plate. What will happen when the switch is closed?

 

[palaphys]

You know that the electric field is uniform between plates. I recommend that you make a graph of electrostatic potential between plates as a function of displacement starting at the left plate all the way to the right plate. Note that the end points are at the same potential of zero. So you start at zero potential and end at zero potential because that's what "earthing" means.

How to do this? I am having difficulty getting the function for electric field between the plates

Also apologies for not making the diagram to scale

 

[kuruman]

How to do this? I am having difficulty getting the function for electric field between the plates

Did you not learn that $$V=Ed$$ where $V$ is the voltage, $d$ is the plate separation and $E$ is the electric field between parallel plates? If the plates are connected to a battery and you double the plate separation, what happens to the electric field?

 

[palaphys]

Did you not learn that $$V=Ed$$ where $V$ is the voltage, $d$ is the plate separation and $E$ is the electric field between parallel plates?

yes, I know this, but do all the plates contribute for the electric field between just X and Y?

 

[kuruman]

yes, I know this, but do all the plates contribute for the electric field between just X and Y?

Perhaps you need to draw electric field lines in the region between the plates. Remember that electric field lines point from regions of high potential to regions of low potential and that they start at positive charges and end at negative charges. There are no lines inside the plates.

 

[Gavran]

If both S1 and S2 are switched on, then the charge on the leftmost and rightmost surfaces will be zero. Also, we may assume that a charge q would appear on the left surface of Y, and a charge Q-q on the right surface of the same. due to this, there will be a charge q-Q on the left surface of Z and a charge -q on the right surface of X, as shown in the figure below:
View attachment 360440

This is not true.
When both $S_1$ and $S_2$ are switched on the charge on the left surface of plate X and the charge on the right surface of plate X will be zero. The same holds for plate Z.

, but do all the plates contribute for the electric field between just X and Y?

The answer is yes.
Calculate the electric field between Y and X and you will see that the ratio of $d_2$ to $d_1+d_2$ contributes to the electric field between Y and X where $d_1$ is the distance between X and Y and $d_2$ is the distance between Y and Z.

 

[palaphys]

This is not true.
When both $S_1$ and $S_2$ are switched on the charge on the left surface of plate X and the charge on the right surface of plate X will be zero. The same holds for plate Z.

not sure how you are coming to this conclusion. In general while earthing we have zero charge on the outermost surfaces of plates.

 

[haruspex]

not sure how you are coming to this conclusion. In general while earthing we have zero charge on the outermost surfaces of plates.

No, earthing produces a zero potential throughout the plate.
If there is an external field acting through the plate then there will be induced charges on the surfaces of the plate opposing it whether or not it is earthed. Earthing means that if external charges would produce a nonzero potential in the plate then additional charge flows on to the plate, equally to both surfaces, to neutralise that potential.

 

[palaphys]

No, earthing produces a zero potential throughout the plate.
If there is an external field acting through the plate then there will be induced charges on the surfaces of the plate opposing it whether or not it is earthed. Earthing means that if external charges would produce a nonzero potential in the plate then additional charge flows on to the plate, equally to both surfaces, to neutralise that potential.

But as a consequence of zero potential throughout the plate, we have zero charge on the outer surfaces. That was what I meant

 

[haruspex]

But as a consequence of zero potential throughout the plate, we have zero charge on the outer surfaces. That was what I meant

No, that does not follow.
The only places charges can be on a conducting plate is on the surfaces. Suppose an external charge creates a potential within the plate. If there were no charges on the surfaces then there would be no charge on the plate at all, so the potential created by the external charge would be the potential at the plate.

 

[Steve4Physics]

now I feel that X and Z are at the same potentials,

It should be more than a feeling - they are both at earth potential!

so the potential difference between XY and YZ must be the same.

Yes.

1. I considered the field due to each uniform sheet, which is just $q/2Aε_0$, and summed up the directions vectorially. however, I get zero due to this.

You may have made a simple mistake.

Consider, for example, the inner face of X and the left face of Y, with charges $q$ and $-q$ respectively, as shown in your 2nd diagram in Post #1.

Using Gauss’s law (and assuming an ideal parallel plate capacitor) you should be able to show that the magnitude of the field between X and Y is given by $E = \frac q {A\epsilon_0}$.

You’ve already correctly noted that, after the switches are closed, $V_{XY} = V_{ZY}$ and you know that ‘V=Ed’. If you can pull all that together you, can work out the charges on the inner faces of X and Z – and hence how much charge flowed from X and Z to earth when the switches were closed - which I guess is what the question is getting at.

FWIW, you have used ‘$-q$’ and ‘$Q+q$’ for the charges on the faces of Y. This is (to me) confusing: for example if $Q$ is positive then the value of $q$ would be have to be negative. Consider using $q_1$ and $q_2$ for the charges on the faces of Y, noting that $Q = q_1+ q_2$. For example if $Q$ is positive, then so will be $q_1$ and $q_2$

Edits = Latex problems as Preview not working.