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Cylindrical capacitor with a dielectric

  1. Nov 11, 2009 #1

    fluidistic

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

    2. Relevant equations
    None given.


    3. 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.
     
  2. jcsd
  3. Nov 11, 2009 #2

    rl.bhat

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    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.
     
  4. Nov 11, 2009 #3

    fluidistic

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    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.
     
  5. Nov 12, 2009 #4

    rl.bhat

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    There is no error in the first part.
     
  6. Nov 12, 2009 #5

    fluidistic

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    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.
     
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