Charge conservation for 3 parallel charged plates

[tellmesomething]

Homework Statement

Three identical metal plates with large surface areas are kept parallel to each other as shown in the figure. The leftmost plate is given a charge Q, the rightmost a charge -2Q and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate

Relevant Equations

None

View attachment 348052
I took a point P on the conductor with charge Q. We know that the field inside a conductor is zero in electrostatic equilibrium
Therefore I took induced charge on the neutral conductor to be x and gave the respective charges to the others. On the side facing conductor with charge Q the neutral one has an induced charge of -x and the induced charge on the inner side Of conductor with charge Q is x so the outer part has Q -x.
Similarly the other side of the neutral sheet has x charge and the inner side of the rightmost sheet has -x charge while the outer side has -2Q + x charge

Field at point P due to all the charges =0

The arrows are the direction of fields

As we can see the direction for field produced by Q-x and -2Q+x would be the same.

But my x ends up cancelling.

I dont know what is going wrong here pls help?

 

[SammyS]

Homework Statement: Three identical metal plates with large surface areas are kept parallel to each other as shown in the figure. The leftmost plate is given a charge Q, the rightmost a charge -2Q and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate
Relevant Equations: None

View attachment 348052
I took a point P on the conductor with charge Q. We know that the field inside a conductor is zero in electrostatic equilibrium
Therefore I took induced charge on the neutral conductor to be x and gave the respective charges to the others. On the side facing conductor with charge Q the neutral one has an induced charge of -x and the induced charge on the inner side Of conductor with charge Q is x so the outer part has Q -x.
Similarly the other side of the neutral sheet has x charge and the inner side of the rightmost sheet has -x charge while the outer side has -2Q + x charge

Field at point P due to all the charges =0

View attachment 348054
The arrows are the direction of fields

As we can see the direction for field produced by Q-x and -2Q+x would be the same.

But my x ends up cancelling.

I don't know what is going wrong here pls help?

Of course your $x$ ends up cancelling, due to charge conservation. In addition to that, your final equation, $\displaystyle \ Q-x -2Q+x = 0 \$, says that $Q=0$ .

I suspect that you need to use Gauss's Law (probably several times) to solve this problem.

By The Way:
Helpers here will not be pleased with the sideways image.

 

[tellmesomething]

Of course your $x$ ends up cancelling, due to charge conservation. In addition to that, your final equation, $\displaystyle \ Q-x -2Q+x = 0 \$, says that $Q=0$ .

I suspect that you need to use Gauss's Law (probably several times) to solve this problem.

By The Way:
Helpers here will not be pleased with the sideways image.

But why so where did I formulate the equations wrongly?
Q shouldnt come to 0...

 

[SammyS]

But why so where did I formulate the equations wrongly?
Q shouldnt come to 0...

Please work on your punctuation.
Which is it?

But why? So, where did I formulate the equations wrongly?​

or ...

But why so? Where did I formulate the equations wrongly?​

The charge, $Q\,,\$ is a given quantity. We would not expect that it must be zero as the solution.

You have not indicated where your equations are from.

If you are attempting to use Gauss's Law, then you have not defined what you are using for the Gaussian surface. Note: A Gaussian surface is the closed surface you are using to enable you to use Gauss's Law in relating the electric flux to the enclosed charge.


"

 

[tellmesomething]

Please work on your punctuation.
Which is it?

But why? So, where did I formulate the equations wrongly?​

or ...

But why so? Where did I formulate the equations wrongly?​

The charge, $Q\,,\$ is a given quantity. We would not expect that it must be zero as the solution.

You have not indicated where your equations are from.

If you are attempting to use Gauss's Law, then you have not defined what you are using for the Gaussian surface. Note: A Gaussian surface is the closed surface you are using to enable you to use Gauss's Law in relating the electric flux to the enclosed charge.


"

Sorry for the punctuation bit.
Yes I used gauss law's results
My gaussian surface is a horizontal cylinder going through the sheet.
Since we know that the sheet is infinitely large, by symmetry we know that the only unique direction of field would be perpendicular to the plane of the sheet.
That would mean on the base of my gaussian surface the field would be same. On the curved surface the field would just graze hence flux would be 0

So if my cylinder has base area A
We get $E(A)+E(A)$= $\frac {\sigma A} {\epsilon_{0}}$
hence E= $\frac{\sigma} {2\epsilon_{0}}$

This is the field outside a non conducting sheet of charge density $\sigma$

If this were conducting as required in my question it would have an infinitesmal thickness and hence it would have two surfaces which would distribute the charge equally in presence of no external field and change its charge density and in this case field would become $\frac{\sigma_{1}} {\epsilon_{0}}$

Then I followed the principle of induction as explained in the post and arrived at the equations

 

[SammyS]

So if my cylinder has base area A
We get $E(A)+E(A)$= $\frac {\sigma A} {\epsilon_{0}}$
hence E= $\frac{\sigma} {2\epsilon_{0}}$

This is the field outside a non conducting sheet of charge density $\sigma$

If this were conducting as required in my question it would have an infinitesimal thickness and hence it would have two surfaces which would distribute the charge equally in presence of no external field and change its charge density and in this case field would become $\frac{\sigma_{1}} {\epsilon_{0}}$

Then I followed the principle of induction as explained in the post and arrived at the equations

What are the locations of the 2 bases of the cylinder that you used? The electric field may be different at the two bases.

As regards your induction arguments:
All that you can reasonably be sure about is that the total charge on the left side of the middle (neutral) plate is equal in magnitude and of opposite sign as the total charge on the right side of the middle plate. It makes good sense to use $x$ as the charge on the right side and $-x$ as the charge on the left as you have done.
You should actually use Gauss's Law to show that the charges on the inner surfaces of the outer plates are $x$ and $-x$ .

As we can see the direction for field produced by Q-x and -2Q+x would be the same.

Upon completing the solution to this exercise, it may surprise you to find that direction of the field to the left of the three plates is in the direction opposite that of the field to the right of the three plates

 

[tellmesomething]

What are the locations of the 2 bases of the cylinder that you used? The electric field may be different at the two bases.

As regards your induction arguments:
All that you can reasonably be sure about is that the total charge on the left side of the middle (neutral) plate is equal in magnitude and of opposite sign as the total charge on the right side of the middle plate. It makes good sense to use $x$ as the charge on the right side and $-x$ as the charge on the left as you have done.
You should actually use Gauss's Law to show that the charges on the inner surfaces of the outer plates are $x$ and $-x$ .



Upon completing the solution to this exercise, it may surprise you to find that direction of the field to the left of the three plates is in the direction opposite that of the field to the right of the three .

Thats the cylinder. sorry I should have attached a diagram earlier. It is symmetrical about the plane of the sheet. Sorry if the diagram is skewed.

By the way how do I use gauss law to show that the charge is equal and opposite on both sides of the plate

 

[tellmesomething]

What are the locations of the 2 bases of the cylinder that you used? The electric field may be different at the two bases.

As regards your induction arguments:
All that you can reasonably be sure about is that the total charge on the left side of the middle (neutral) plate is equal in magnitude and of opposite sign as the total charge on the right side of the middle plate. It makes good sense to use $x$ as the charge on the right side and $-x$ as the charge on the left as you have done.
You should actually use Gauss's Law to show that the charges on the inner surfaces of the outer plates are $x$ and $-x$ .



Upon completing the solution to this exercise, it may surprise you to find that direction of the field to the left of the three plates is in the direction opposite that of the field to the right of the three plates

I think I can take a gaussian pill box between the two plates. We know field inside a conductor is 0 therefore flux through the base of the cylinders would be 0 and flux through the curved surface would be 0 since the field is parallel to it therefore we would get
$q_{enclosed}$=0 which would mean the charges on the inner side of the 3rd plate and the right side of the neutral plate is equal and opposite.

Similarly the other.

.



Upon completing the solution to this exercise, it may surprise you to find that direction of the field to the left of the three plates is in the direction opposite that of the field to the right of the three plates

Im really not sure how to get That. I know I should be getting that but how?

 

[anuttarasammyak]

Applying Gauss's law to each of these colored refgions, I got

Spoiler

$$x=\frac{3}{2}Q$$

Is it right ?

 

[tellmesomething]

Applying Gauss's law to each of these colored refgions, I got

Spoiler

$$x=\frac{3}{2}Q$$

Is it right ?

View attachment 348110

Yes

 

[SammyS]

By the way how do I use gauss law to show that the charge is equal and opposite on both sides of the plate

You don't use Gauss's Law for this. It's simply conservation of charge. - As in the title of this thread.

 

[tellmesomething]

You don't use Gauss's Law for this. It's simply conservation of charge. - As in the title of this thread.

Okay. can you please tell me how do you get this "Upon completing the solution to this exercise, it may surprise you to find that direction of the field to the left of the three plates is in the direction opposite that of the field to the right of the three plates" in post #6. Do I just take cases of positive and negative?

 

[Steve4Physics]

The problem is now solved but it’s worth noting that there is a very simple solution.

In general, for a system of any number of large, parallel conducting plates with total charge $Q_{tot}$, the charge on the outer surface of each outer plate is $\frac {Q_{tot}}2$.

E.g. see here for discussion: https://physics.stackexchange.com/q...ies-on-a-system-of-parallel-charged-plates-id

For the Post #1 question, the total charge is $Q+(-2Q) = -Q$. So both the left and right plates have an outer surface charge of $-\frac Q2$.

(In terms of field direction, external field lines point into left and right plates. At large distances (much bigger than the plate dimensions) the field approximates to that of a point charge $-Q$.)

 

[tellmesomething]

The problem is now solved but it’s worth noting that there is a very simple solution.

In general, for a system of any number of large, parallel conducting plates with total charge $Q_{tot}$, the charge on the outer surface of each outer plate is $\frac {Q_{tot}}2$.

E.g. see here for discussion: https://physics.stackexchange.com/q...ies-on-a-system-of-parallel-charged-plates-id

For the Post #1 question, the total charge is $Q+(-2Q) = -Q$. So both the left and right plates have an outer surface charge of $-\frac Q2$.

(In terms of field direction, external field lines point into left and right plates. At large distances (much bigger than the plate dimensions) the field approximates to that of a point charge $-Q$.)

Thankyou. I was referred to the same article by sir john rennie over stack exchange as well. I'll read it at the earliest. :)

As for my initial doubt I think it was just a stupid mistake as the charge on the outer surface of the leftmost and rightmost plate must be same in sign for the net field at a point P inside the conductor to cancel out.

 

[Steve4Physics]

Thankyou. I was referred to the same article by sir john rennie...

Ah yes. I've used his bridges and indigestion tablets in the past.

 

[anuttarasammyak]

In general, for a system of any number of large, parallel conducting plates with total charge Qtot, the charge on the outer surface of each outer plate is Qtot2.

In other words, electric field generated by a plane charge does not depend on distance. Right wall and left wall have same but opposite electric field in applying Gauss law to planes.

 

[Steve4Physics]

In other words, electric field generated by a plane charge does not depend on distance.

In the ideal case, for an infinite plane of charge, the field is independent of distance from the plane.

For a finite plane of charge (e.g. measuring $L \times L$) the field is approximately independent of distance ($d$) from the plane providing $L \gg d$.

If $d \gg L$ the finite plane will 'look like' a point charge and the field will approximate to that of a point charge.

Right wall and left wall have same but opposite electric field in applying Gauss law to planes.

In the ideal case, yes. But exercise caution depending on the problem. E.g.

from https://en.wikipedia.org/wiki/Electric_field#/media/File:VFPt_capacitor-square-plate.svg

 

[tellmesomething]

Ah yes. I've used his bridges and indigestion tablets in the past.

I dont get it..

 

[Steve4Physics]

I dont get it..

Don't worry about it. It's just my (not very sophisticated) sense of humour. But for information...

Sir John Rennie (1794-1874) was a (fairly well known) British engineer involved in designing/building bridges (and other projects).

'Rennies' are the UK's top selling indigestion tablet.

 

[tellmesomething]

Don't worry about it. It's just my (not very sophisticated) sense of humour. But for information...

Sir John Rennie (1794-1874) was a (fairly well known) British engineer involved in designing/building bridges (and other projects).

'Rennies' are the UK's top selling indigestion tablet.

Oh I see :-)