How to Determine Plate Configuration in Capacitance Calculations?

[Arman777]

Homework Statement

In attachments

Homework Equations

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 don't know how to decide that

Thanks

 

[TSny]

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..

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

I guess I should decide to Are these plates connected in parallel or in series ? But I don't know how to decide that

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.

 

[mfb]

Which part are you working on?

You have a relation between U and C, but you have to determine both.

I am doing something wrong..I guess I should decide to Are these plates connected in parallel or in series ? But I don't know how to decide that

In the same way you determine it for every circuit, but you don't necessarily have to determine that.

 

[Arman777]

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

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 don't know what's the "U".Its like saying $U=U$ (I found like this) Thats why something seemes wrong.

You have a relation between U and C, but you have to determine both.

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

 

[Arman777]

Or its U=3U cause there's 3 segments ? The sum of energy of plates ?

 

[TSny]

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 don't know what's the "U".Its like saying $U=U$ (I found like this) Thats why something seemes wrong.

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

 

[mfb]

Or its U=3U cause there's 3 segments ? The sum of energy of plates ?

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

 

[Arman777]

Something cannot be 3 times itself (unless it is 0 - it is not).

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

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

 

[mfb]

$U_{tot}=3U=3(Ad)υ=3(Ad)\frac 1 2 ε_0E^2$

That part is fine.

$=\frac{3CV^2}{2}$

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.

 

[Arman777]

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.

$C_{tot}=3ε_0 \frac A d$ ?

 

[TSny]

$C_{tot}=3ε_0 \frac A d$ ?

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

 

[Arman777]

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

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

 

[TSny]

$3(Ad)\frac 1 2 ε_0E^2=\frac {CV^2} {2}$

OK.

 

[Arman777]

Thanks

 

[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?

 

[Arman777]

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?

parallel

 

[mfb]

Right.

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