Cylindrical capacitor with a dielectric

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SUMMARY

The discussion focuses on calculating the capacitance of a cylindrical capacitor consisting of two concentric cylinders with radii a and b, where b > a, and a dielectric with permittivity ε₀ is introduced within the capacitor. The capacitance without the dielectric is derived as C = (2πε₀L) / ln(b/a). The change in energy when the dielectric is removed is calculated using the formula ΔU = (Q² / C₁) - (Q² / C₀), where C₁ is the capacitance with the dielectric and C₀ is the capacitance without it. The participants confirm that the initial calculations are correct and clarify discrepancies with external resources.

PREREQUISITES
  • Understanding of cylindrical capacitors and their geometry.
  • Familiarity with capacitance formulas and energy stored in capacitors.
  • Knowledge of dielectric materials and their effect on capacitance.
  • Basic proficiency in calculus for integrating electric fields.
NEXT STEPS
  • Study the derivation of capacitance for cylindrical capacitors in detail.
  • Learn about the effects of different dielectric materials on capacitance.
  • Explore energy storage in capacitors and the implications of removing dielectrics.
  • Investigate the relationship between electric field strength and capacitance in cylindrical geometries.
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Students in physics or electrical engineering, educators teaching capacitor theory, and anyone involved in designing or analyzing capacitors in practical applications.

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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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