# Homework Help: Find the Capacitance -- Plates

1. Mar 26, 2017

### Arman777

1. The problem statement, all variables and given/known data
In attachments

2. Relevant equations

3. The attempt at a solution
This is my charge distrubition.From there I draw electric field lines.And then from potential difference I wrote
$E=\frac {V} {d}$ and then $υ=\frac 1 2ε_0E^2$ then $U=\frac {CV^2}{2}$

I am doing something wrong..I guess I should decide to Are these plates connected in parallel or in series ? But I dont know how to decide that

Thanks

2. Mar 26, 2017

### TSny

So far, I don't see anything wrong. What did you do from this point?

Using the approach of this problem, you do not need to decide whether the plates are in parallel or in series. Your answer will tell you if the plates are in series or parallel.

3. Mar 26, 2017

### Staff: Mentor

Which part are you working on?

You have a relation between U and C, but you have to determine both.
In the same way you determine it for every circuit, but you don't necessarily have to determine that.

4. Mar 26, 2017

### Arman777

Welli I found (a) and it is $E=\frac {V} {d}$ then Energy density will be $υ=\frac 1 2ε_0E^2$ .
I am stucked here I wrote for (b) $U=\frac {CV^2}{2}$ but I dont know whats the "U".Its like saying $U=U$ (I found like this) Thats why something seemes wrong.

Thats where I am stucked I think here $E$ will be different then $\frac {V} {d}$ otherwise as I said it becomes $U=U$ and $C=C$ which sounds nonsense in some sense

5. Mar 26, 2017

### Arman777

Or its U=3U cause theres 3 segments ? The sum of energy of plates ?

6. Mar 26, 2017

### TSny

How would you express the total U of the system in terms of electric field energy?

7. Mar 26, 2017

### Staff: Mentor

Something cannot be 3 times itself (unless it is 0 - it is not).
See TSny's post for a good hint.

8. Mar 26, 2017

### Arman777

Not like that I meant $U_{tot}=3U=3(Ad)υ=3(Ad)\frac 1 2 ε_0E^2=\frac {3CV^2} {2}$

9. Mar 26, 2017

### Staff: Mentor

That part is fine.
That would need a "capacitance per layer" or something like that. Don't do that.
You have the total energy, you know the voltage, that is sufficient to determine the total capacitance.

10. Mar 26, 2017

### Arman777

$C_{tot}=3ε_0 \frac A d$ ?

11. Mar 26, 2017

### TSny

Yes. But I can't tell how you got that.

12. Mar 26, 2017

### Arman777

$3(Ad)\frac 1 2 ε_0E^2=\frac {CV^2} {2}$ and $E=\frac V d$

13. Mar 26, 2017

### TSny

OK.

14. Mar 26, 2017

### Arman777

Thanks

15. Mar 26, 2017

### TSny

Can you tell from your answer if the three gaps in the system act like capacitors in series or do they act like capacitors in parallel?

16. Mar 26, 2017

### Arman777

parallel

17. Mar 26, 2017

### Staff: Mentor

Right.

Another way to see this: All three gaps have the same electrodes.