Cylindrical capacitor with a dielectric

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Homework Help Overview

The discussion revolves around calculating the capacitance of a cylindrical capacitor consisting of two concentric cylinders, with one cylinder having a dielectric inserted. The problem includes determining the change in energy when the dielectric is removed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the capacitance without the dielectric and then with the dielectric, expressing concerns about potential errors in their calculations. They also explore the relationship between charge, capacitance, and energy stored in the capacitor.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the implications of removing the dielectric and the effects on charge and energy. There is a discussion about the correctness of the original poster's calculations, with some participants asserting that there are no errors in the initial approach.

Contextual Notes

Participants note that the capacitor is not connected to a source of emf, which affects how charge is treated when the dielectric is removed. There is also mention of a potential discrepancy with external resources that may arise from different assumptions about constants used in calculations.

fluidistic
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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 \varepsilon _0 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.
V=\frac{Q}{C}=\int _a ^b \vec E d \vec l.
I'm looking for E : \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}.Hence \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} from which C=\frac{2 \pi \varepsilon _0 L}{\ln \left ( \frac{b}{a} \right )}.
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 \frac{2\pi \varepsilon _0}{\ln \left( \frac{b}{a} \right) } \cdot (L-d+ \kappa d) 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 \frac{Q^2}{2C} and I wanted to find Q_i-Q_f but I got it all wrong, a ? was marked by the professor. So how would I do it?

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

Did you see my error in part 1) ? I cannot find it.
 
There is no error in the first part.
 
rl.bhat said:
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.
 

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