# Charge transfer Problem

1. Jun 5, 2012

Hi friends please help me in solving this. I have great issue in solving this since a very longtime.
Thank you in advance for helping me.
The problem is as follows.
1. The problem statement, all variables and given/known data

https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-prn1/521479_2246934070146_623256457_n.jpg [Broken]

Three large plates are arranged as shown as shown in the figure. How much charge will flow through the key "K" if it is closed ?

2. Relevant equations&
3. The attempt at a solution
For solving this I thought that On the first plate charge Q will be distributed as Q/2 & Q/2 on the either sides. And on the second plate charge 2Q will be distributed as Q & Q on the either sides. Now induction will take place due to the higher charge Q on the lower magnitude charge. Hence -Q will generate on the the right side of 1st plate and + Q on its right side and on the third plate -Q on the left side and + Q on the right side.
Now 1 and three are connected and charge will flow from 1 to another.

inner sides of 1st and 2nd plate should have same but opposite signs. Let 2nd plate has
(Q + x) and 1st plate has (Q/2 - y) charges. These charges should be equal. It is giving the equation : x + y = - Q/2.

Now the outer side of 2nd plate will have (Q/2 + y) charge due to conservation of charge. And the right side of plate 2 will have (Q - x) charge.
Induction will also take place on the 3 rd plate. It'll have -(Q-x) and (Q-x) on the left and right sides.
When the key becomes close the inner charges will remain at the same condition due to iteration of the opposite charges. But the outer charges of the 1st and 3rd plate will move for to make the potential equal. Now I am stuck in this situation how the charge will flow now?
Even if I get another equation It'll be easy to solve the equations but no positive outcome.

The answer of this question is 5Q/6.

Last edited by a moderator: May 6, 2017
2. Jun 6, 2012

Please friends help me in solving this.
Many guys just see this problem but didn't replied. I am suffocating here from this problem.
I will be very thankful.

Last edited: Jun 6, 2012
3. Jun 6, 2012

### pcm

for this problem use the fact that two facing surfaces have equal and opposite charges(a consequence of gauss's theorem).when the switch is closed first assume certain charge distribution on each face and use the above fact.you also have net charge on plate2=2Q,
and the sum of charges on plate1 and plate3=Q.
also potential of plate1 and plate3 are equal.so the change in potential as you go from plate1 to plate2 is equal to the change in potential as you go from plate3 to plate2.

this should give you enough equations to determine charge on each surface.

4. Jun 7, 2012

I have got the trick. But Still facing problem in finding the P.D. from plate 1 to plate2 and plate3 to plate2.

5. Jun 7, 2012

Hey pcm,
Thanks friend I've got the exact answer with the help of your statement.
But still here is a small confusion if you could solve it.
While taking the P.D. of plate 1 and 2, I am equating it to the P.D. of plate 3 and 2. But for to get answer I've to take the negative sign of P.D. of plate 3and 2. Why this is happening?
Why I am sup[posed to take the negative sign of this P.D.
Is it have to do something with the relation, V = - ∫E. dr ?

6. Jun 7, 2012

### pcm

have you checked the direction of electric field between plates?
going against the field ,potential increases.

7. Jun 9, 2012