# Charge on capacitor problem

1. Dec 2, 2015

### gracy

1. The problem statement, all variables and given/known data
find the charge on each plate given area is a and separation between two consecutive plate is d

2. Relevant equations
$Q$=$C$$V$

3. The attempt at a solution
I don't know how to proceed from here.

Last edited: Dec 2, 2015
2. Dec 2, 2015

### Staff: Mentor

You need another relevant equation. What is the capacitance of a parallel plate capacitor with plate area $a$ and plate separation $d$?

3. Dec 2, 2015

### gracy

$C$=$\frac{aε0}{d}$

4. Dec 2, 2015

### Staff: Mentor

Yes. Proceed.

5. Dec 2, 2015

### gracy

Q=C$\frac{aε0}{d}$
Will this be charge on each plate?

6. Dec 2, 2015

### Staff: Mentor

No. That has two capacitance terms. You need a voltage and a capacitance: Q = CV.

7. Dec 2, 2015

### gracy

Haste makes waste!
Q=V$\frac{aε0}{d}$
Will this be charge on each plate?

Last edited: Dec 2, 2015
8. Dec 2, 2015

### Staff: Mentor

Still too hasty. You've written the same thing again

Each capacitor plate in your circuit diagram will have the same magnitude of charge on it. You'll have to sort out the signs of the charges. Then determine how to locate those charges on the plates of the original figure -- remember, one physical plate is shared between two capacitors.

9. Dec 2, 2015

### gracy

But hint says plate B would have different charge.

10. Dec 2, 2015

### Staff: Mentor

Yes. I presume "plate B" is the middle plate of the physical setup? You didn't label your diagram.

Note that I said: "Each capacitor plate in your circuit diagram will have the same magnitude of charge on it". There is a distinction between the circuit diagram and the physical arrangement in that the circuit diagram shows two plates in two separate capacitors whereas the physical setup uses a shared plate. Re-read the entire contents of my post #8.

11. Dec 2, 2015

### gracy

12. Dec 2, 2015

### gracy

Should not plate B cntain zero charge as it is earthed?

13. Dec 2, 2015

### Staff: Mentor

No, "earth" is a vast pool of available charges, both positive and negative and overall neutral. It can supply whatever charge is required to respond to electrostatic forces. Or, if you like, it can source or sink any amount of electrons, if you want to go with an atomic model for the plates and conductors.

Your charge placement is not correct. Note that plates A and C are symmetrical and connected to the same potential (V). They should end up with similar charges distributed in the same way. Charges are made available to those plates via their connection to V. So far I think we've been assuming that V is a positive potential with respect to ground (earth), which is fine, but you should state that assumption in your solution.

14. Dec 2, 2015

### gracy

I did not understand that.

15. Dec 2, 2015

### Staff: Mentor

V is taken to be a voltage supply. Voltage is a potential difference. When a terminal is just labeled "V" then it is assumed that that potential difference is between the terminal and the ground reference.

16. Dec 2, 2015

### gracy

I don't quite get the relationship between being symmetrical and connected to the same potential and having same charge distribution.
I distributed charge by only considering charge induction.

17. Dec 2, 2015

### cnh1995

Plates 2 and 3 along with the wire joining them, form plate B in your original question.
That is for one capacitor(1-2 or 3-4). The capacitors are in parallel. So you can find the total charge now.

18. Dec 2, 2015

### Staff: Mentor

If you rotate the drawing on the page, do you expect the charges on the plates to change? Is there any distinction (electrically) between the two outer plates?
You'll have to explain how you ended up with different distributions on the two outer plates.

19. Dec 2, 2015

### cnh1995

A and C are connected to same potential V and you have shown net +ve charge on A and other plates neutral. Why A? Why not C?

20. Dec 2, 2015

### gracy

Is it correct now?