Conversation started Apr 27, 2020 at 15:44.
Apr 27, 2020 15:44
Question:
Three identical, parallel conducting plates A, B and C are placed as shown. Switches $S_1$ and $S_2$ are open, and can connect A and C to earth when closed. +Q charge is given to B, then:
(a) if $S_1$ is closed with $S_2$ open, a charge of amount Q will pass through $S_1$
(b) if $S_2$ is closed with $S_1$ open, a charge of amount Q will pass through $S_2$
(c) if $S_1$ and $S_2$ are closed together, a charge of amount Q/3 will pass through $S_1$ and a charge of amount $2Q/3$ will pass through $S_2$.
(d) All of the above.
My first step:
@JohnRennie: Could you please tell how to proceed from this? I understand earthing a conductor makes its potential equal to that of earth close to 0 volts. But I think we need to determine the potentials of the individual plates for which I don't know any method.
Your initial diagram is wrong.
Ok sir. I came to such charge distribution by equating the electric field inside the plate B to be zero. May I know why is it incorrect?
The charge on B has to add to $+Q$ and the way you have drawn it the charges add up to zero i.e. you've drawn $+Q/2$ on one face and $-Q/2$ on the other.
Ah! Fine. That was an error while drawing that on my computer. It must be Q/2 on either sides.
Yes.
that obeys the two rules:
- equal and opposite charges on opposing faces
- equal charges on the two outer faces
Apr 27, 2020 15:53
Yes sir.
The third rule is that when any plate is earthed the charges on the outer faces are zero.
Ok sir. I remember you mentioned about the third rule yesterday. Is it possible to tell why is that so? Or is that a pure experimental result?
@GuruVishnu if I'm honest I don't know why it is the case. I even asked about it on the main site.
5
Q: Why are the two outer charge densities on a system of parallel charged plates identical?

John RennieOne of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this: the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are ass...

So shall we work out the charges when S1 is closed?
Thank you for sharing that question. I think I can read it when you're not available. It looks like a big post.
@JohnRennie Yes sir.
OK if S2 remans open the charge on plate C cannot change i.e. it must be zero.
Apr 27, 2020 15:58
Yes sir.
And we know the charge on the outer surface has to be zero because we have one plate (A) earthed. That means the inner face of C must also have zero charge.
And that means the charge on the right face of plate B must also be zero. OK so far?
Ok sir.
Ok sir. So far I understood.
The total charge on B is $+Q$ so the charge on the left face of B must be $+Q$ and therefore the charge on the right face of A must be $-Q$.
Apr 27, 2020 16:03
Oops. You were faster :-)
:-) But yes you're correct. So now you know the charge on A and therefore you know how much charge flowed through S1 when it was closed.
Yes sir. It's Q. Or in other words, option (b) is justified.
$-Q$ ...
Ok sir. $-Q$ from the ground to the plate is same as $+Q$ from the plate to the ground. I usually like to use positron picture rather than the electron picture as I like to be positive :-)
Yes, true, the question doesn't specify the direction the charge flows.
The trouble is that if we leave S1 open and close S2 we get exactly the same result so both (a) and (b) will be true.
Apr 27, 2020 16:08
Yes :-)
:-)
Ok sir. Now the problem is so easy. It seems the key fact is:
Actually (c) is going to be true as well isn't it.
15 mins ago, by John Rennie
The third rule is that when any plate is earthed the charges on the outer faces are zero.
Apr 27, 2020 16:10
John Rennie sir's third law analogous to Newton's third law :-)
Sadly it isn't my law :-)
I think it was a JEE student taught it to me.
Fine sir :-)
Can you see why (c) is also true?
@JohnRennie I think I need to verify that.
There's an extra factor to consider when both switches are closed. Can you see what it is?
Apr 27, 2020 16:12
I think as per our previous method, the outer surface of plates A and C must be zero. Does it count as an extra factor sir?
Or do you refer to the differential capacitance?
Plates A and C must be at the same potential because they are both connected to earth. Yes?
Yes sir.
And the two faces of plate B must be at the same potential because B is a conductor.
@JohnRennie Fine sir. I get it. Are we going to assume the capacitors AB and BC are in parallel?
And we could make use of this $Q/C=V=$constant
The potential change between two surfaces is $V = Ex$ where $x$ is the distance between the surfaces and $E$ is the field between them.
And the potential changes shown by the two red arrows have to be the same.
Apr 27, 2020 16:18
Yes sir. I understood that after seeing your diagram (it was so clear even without explanation) :-)
:-)
And now I'm going to call it a day.
I'll be around tomorrow as usual.
Ok sir. Good bye :-)
Even though I completely got the answer, now, I wish to discuss the other method with you tomorrow.
OK, see you tomorrow.
 
13 hours later…
Apr 28, 2020 05:08
@JohnRennie: Hi sir. Good morning :-)
@GuruVishnu hi :-)
Did you have your breakfast?
Yes, and my first coffee of the day :-)
Fine sir :-)
May I explain my alternate method for yesterday's capacitor question?
Yes, go ahead.
Apr 28, 2020 05:15
That's it. I got the same result as of your method of using electric field density.
The two capacitors AB and BC are in parallel and the potential difference across the terminals is equal. Using this I found the charge on individual plates.
Yes, that works.
 
Conversation ended Apr 28, 2020 at 5:17.