The Electrical energy of a conducting cylinder

[doktorwho]

1. The problem stsatemesnt, all variables and given/known data
A cylinder conductor of length $l$, inner radius $a$ and outer radius $b$ is half-filled with a liquid dielectric as shown in the picture. The cylinder is attached to a battery of constant voltage $U$ and then separated from it. Imagine that the dielectric is leaking and calculate the change of electrical energy when there is no dielectric present. ( meaning the difference between the beginning state and the one were there is no dielectric)
There is a picture below.

Homework Equations

3. The Attempt at a Solution
I'm going to guide you through my work and thinking so you can correct me and help me finish the problem.[/B]
$ΔW_e=W_{e2}-W_{e1}$ ( The state 2 is when its empty)
$W_{e1}=\frac{1}{2}\frac{Q_1^2}{C_1}$ ( $C_1,Q_1$ are the total charge and capacitance of the system 1)
Because there is a dielectric in the first system i have two electric fields, the one above the $h$ and the one below going radial.
$E_1=\frac{Q'_1}{2 \pi ε_0 r}$
$U_1=\frac{Q'_1}{2 \pi ε_0}\ln (b/a)$
$E_2=\frac{Q'_2}{2 \pi ε_0 ε_r r}$
$U_2=\frac{Q'_2}{2 \pi ε_0 ε_r} \ln (b/a)$
Then i express the linear charge densities $Q'_1,Q'_2$
$Q'_1=\frac{2 \pi ε_0 U_1}{\ln (b/a)}$
$Q'_2=\frac{2 \pi ε_0 ε_r U_2}{\ln (b/a)}$
$Q_{total}=Q'_1*(l/2) + Q'_2*(l/2)$
Something does not seem right here, can you help me with what i got wrong?

 

[BvU]

Hi,

I'm missing the relevant equations in section 2 like $Q_{\rm total}\$ conserved, $U_1 = U_2$ and such...
And what about $\ C_1, C_2\$ ?

 

[doktorwho]

Hi,

I'm missing the relevant equations in section 2 like $Q_{\rm total}\$ conserved, $U_1 = U_2$ and such...
And what about $\ C_1, C_2\$ ?

It's not stated but I am guessing that $Q$ must be conserved and $C$ and $U$ are changing. :) The whole work is my guess so i don't know the answer if anything is conserved xD I was thinking someone can provide an insight.

 

[rude man]

I would forget about E fields. Just compute C1 and C2, then take the difference in 1/2 CV^2. Yes, Q is conserved; removing the dielectric does not remove any charge. So you know that V1 and V2 will be different also.

 

[doktorwho]

I would forget about E fields. Just compute C1 and C2, then take the difference in 1/2 CV^2. Yes, Q is conserved; removing the dielectric does not remove any charge. So you know that V1 and V2 will be different also.

But to compute C i need to have V and to have V i need E

 

[rude man]

But to compute C i need to have V and to have V i need E

No you don't. C is a property of the geometry etc etc and is independent of V, E or Q. (We are not yet dealing with nonlinear capacitors at your stage of the game).

 

[doktorwho]

No you don't. C is a property of the geometry etc etc and is independent of V, E or Q. (We are not yet dealing with nonlinear capacitors at your stage of the game).

$C=\frac{Q}{V}$ by definition. What do you mean its not dependant on $Q$ or $V$?

 

[rude man]

$C=\frac{Q}{V}$ by definition. What do you mean its not dependant on $Q$ or $V$?

$C=\frac{Q}{V}$ by definition. What do you mean its not dependant on $Q$ or $V$?

OK take two large parallel plates. Area = A, separation = d. What's the capacitance?

 

[doktorwho]

OK take two large parallel plates. Area = A, separation = d. What's the capacitance?

$C=\epsilon_0\frac{A}{d}$
I see your point, it is the conductors geometric property.
For the cylindrical system i then have two capacitances $C_1=\frac{2\pi\epsilon_0L/2}{ln(b/a)}$, $C_2=\frac{2\pi\epsilon_0\epsilon_rL/2}{ln(b/a)}$ Now i need to know the $V_1,V_2$ for the system in case it's half filled and for that i need $E$. When i find that what is my $C$ for the case where its half filled? I have two C's?

 

[rude man]

$C=\epsilon_0\frac{A}{d}$
I see your point, it is the conductors geometric property.
For the cylindrical system i then have two capacitances $C_1=\frac{2\pi\epsilon_0L/2}{ln(b/a)}$, $C_2=\frac{2\pi\epsilon_0\epsilon_rL/2}{ln(b/a)}$ Now i need to know the $V_1,V_2$ for the system in case it's half filled and for that i need $E$. When i find that what is my $C$ for the case where its half filled? I have two C's?

You don't need E.
You have two capacitors in "parallel", do you see that? Call them CA and CB. So C = CA + CB. Say CA has the dielectric, initially. You can compute CA1; also CB1 = CA2 = CB2.
You know V1, = same for CA and CB. You then can find Q = V1(CA1 + CB1).
For V2 you invoke the fact that Q does not change. This enables you to compute V2.
Finally, as I said, energy = 1/2 Δ{CV²}, or = 1/2 Q ΔV.
I can say no more.

 

[BvU]

All a matter of writing down the relevant equations, including the conservation relationships ...

I'm missing the relevant equations in section 2 like $Q_{\rm total}\$conserved, $U_1 = U_2$ and such...

(not to rub it in, but to make clear the ingredients of a solid problem solving approach ... )

It's not stated but I'm guessing that $Q$ must be conserved and $C$ and $U$ are changing.

No guessing. Physics. Where would Q go ?

 

[EricGetta]

Dielectric polarization can make Q change at conductor surfaces. The net change in Q is 0 globally.

 

[rude man]

Dielectric polarization can make Q change at conductor surfaces. The net change in Q is 0 globally.

Don't think so. The dielectric does not assume net charge. All charge stays on the metal surfaces.