Dielectric and Parallel Plate Capacitor

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SUMMARY

The discussion centers on a 4.0-nF parallel plate capacitor filled with Mylar (κ = 3.1) charged to 120 V. The initial capacitance with the dielectric is confirmed to be 4.0 nF. To find the work required to remove the Mylar and the potential difference after its removal, the capacitance without the dielectric is calculated as (4.0 nF)/3.1, and the potential energy with the dielectric is determined to be 28,880 nJ. The user seeks guidance on calculating the potential energy without the dielectric.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and potential energy equations.
  • Familiarity with dielectric materials and their effect on capacitance.
  • Knowledge of the relationship between charge (Q), capacitance (C), and voltage (V).
  • Basic algebra skills for manipulating equations related to capacitors.
NEXT STEPS
  • Calculate the capacitance of a parallel plate capacitor without a dielectric using the formula C = kA(Epsilon_0)/d.
  • Explore the concept of energy stored in capacitors and how it changes with dielectric removal.
  • Learn about the implications of charge conservation in capacitors during dielectric removal.
  • Investigate the physical principles behind dielectrics and their impact on electrical properties.
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Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of capacitors and dielectrics.

opprobe
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Homework Statement



A 4.0-nF parallel plate capacitor with a sheet of Mylar (κ = 3.1) filling the space between the plates is charged to a potential difference of 120 V and is then disconnected. (The initial capacitance including the dielectric is 4.0 nF.)

(i) How much work is required to completely remove the sheet of Mylar from the space between the two plates?

(ii) What is the potential difference between the plates of the capacitor once the Mylar is completely removed?

Homework Equations



Q=CV
C=kA(Epsilon_0)/d
-W=Uf-Ui
U = (Q^2)/2C = C(V^2)/2 = QV/2

The Attempt at a Solution



Capacitance w/ dielectric = 4.0 nF
Capacitance w/o dielectric = (4.0 nF)/κ = (4.0 nF)/3.1

Potential Energy w/ dielectric = [(4.0 nF)(120 V)^2]/2 = 28,880 nJ



And...I don't know how to calculate the Potential Energy w/o the dielectric. Can someone point me to the right direction?

Thanks in advance!
 
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opprobe said:
And...I don't know how to calculate the Potential Energy w/o the dielectric. Can someone point me to the right direction?

If the charge remains the same on the plates (where could it have gone?), and you know the capacitance value...
 
Thank you so much! I can't believe I completely forgot the definition of a capacitor...
 

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