Cylindrical capacitor with a dielectric

[fluidistic]

Homework Statement

I must calculate the capacitance of the following capacitor : A cylindrical capacitor made of 2 shells (not sure if shell is the right word in English. Made of 2 cylinders maybe), one of radius a and the other of radius b>a. It has a length of L.
We introduce entirely inside the capacitor a dielectric (whose permitivity is [tex]\varepsilon _0[/tex] of length d<L.

Then I must calculate the change of energy if we remove the dielectric.

Homework Equations

None given.

The Attempt at a Solution

I first calculated the capacity of such a capacitor without dielectric.
[tex]V=\frac{Q}{C}=\int _a ^b \vec E d \vec l[/tex].
I'm looking for E : [tex]\oint \vec E d \vec A =4 \pi k Q \Rightarrow E \cdot 2 \pi rL=4\pi kQ \Rightarrow E=\frac{2kQ}{Lr}=\frac{Q}{2 \pi \varepsilon _0 Lr}[/tex].Hence [tex]\int _a^b \vec E d \vec l = \frac{Q}{2 \pi \varepsilon _0 L} \int _a^b \frac{dr}{r}=\frac{Q}{2 \pi \varepsilon _0 L} \ln \left ( \frac{b}{a} \right )=\frac{Q}{C}[/tex] from which [tex]C=\frac{2 \pi \varepsilon _0 L}{\ln \left ( \frac{b}{a} \right )}[/tex].
I see that I made an error, however I've been told it's right. (I know I made an error because of this website : http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capcyl.html)From what I've just done, I deduced that the answer to the original first question is [tex]\frac{2\pi \varepsilon _0}{\ln \left( \frac{b}{a} \right) } \cdot (L-d+ \kappa d)[/tex] which once again seemed to make the corrector agreed. (But it can't be right if I have made an error earlier).
For the second question I've wrote that the energy stored in a capacitor is [tex]\frac{Q^2}{2C}[/tex] and I wanted to find [tex]Q_i-Q_f[/tex] but I got it all wrong, a ? was marked by the professor. So how would I do it?

Thank you.

 

[rl.bhat]

If the capacitor is not connected to the source of emf, the charge doers not change when you remove the dielectric. Only capacitor changes and hence the energy changes.

 

[fluidistic]

Thanks!
Ok so for part 2) I'd just use the formula [tex]\frac{Q^2}{C_1}-\frac{Q^2}{C_0}[/tex] where [tex]C_1[/tex] is the capacitance with the dielectric and [tex]C_0[/tex] without it. Is that right?

Did you see my error in part 1) ? I cannot find it.

 

[rl.bhat]

There is no error in the first part.

 

[fluidistic]

There is no error in the first part.

Ah thank you. The discrepancy from hyperphysics can be explained that they can take k=1, so there's no discrepancy.
Thank you once again.