Change in Energy Stored in a Capacitor Due to Removal of Dielectric

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SUMMARY

The discussion focuses on calculating the change in energy stored in a parallel plate capacitor when the dielectric medium is removed. The electric field between the plates is calculated as 22,500 Vm-1, and the charge per unit area is determined to be 3.98 x 10-6 Cm-2. The capacitance of the capacitor with a dielectric is found to be 1.15 x 10-8 F. The key challenge discussed is understanding how the potential difference changes when the dielectric is removed, affecting the energy calculations.

PREREQUISITES
  • Understanding of parallel plate capacitor equations, specifically C=q/V and U=0.5CV2.
  • Knowledge of electric field calculations, particularly E=V/d.
  • Familiarity with dielectric materials and their effect on capacitance.
  • Basic principles of energy storage in capacitors.
NEXT STEPS
  • Learn about the effects of dielectric removal on capacitor voltage and energy storage.
  • Study the relationship between capacitance and dielectric constant in detail.
  • Explore the concept of electrically isolated capacitors and their implications on charge conservation.
  • Investigate advanced capacitor configurations, including partial dielectric removal scenarios.
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in capacitor design and analysis will benefit from this discussion.

teme92
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Homework Statement



(a)
The separation between the plates of a parallel plate capacitor is 2 mm and the potential diff erence between them is 45 V. Find the electric field between the plates. If the permittivity of the dielectric medium between the plates is 20 [itex]\epsilon[/itex]0, fi nd the charge per unit area on each plate.

(b)

If the area of each plate in part (a) is 0.13 m2, fi nd the capacitance of the capacitor.

(c)
If the dielectric medium between the plates in parts (a) and (b) is removed, so that air now fi lls
the region between the plates, fi nd the resulting change in the energy stored in the capacitor if
the plates are electrically isolated.

(d)
If the medium between the plates in parts (a) and (b) is partially removed, so that half of the
area has air between the plates and the other half has the original dielectric medium between the
plates, find the capacitance of the capacitor.

Homework Equations



C=q/V=[itex]\epsilon[/itex]A/d
E=[itex]\sigma[/itex]/[itex]\epsilon[/itex]
[itex]\sigma[/itex]=q/A=E[itex]\epsilon[/itex]
V=Ed
U=0.5CV2

The Attempt at a Solution



(a)
For Electric Field:
E=V/d=45/0.002=22500 Vm-1

For charge per unit area:
[itex]\sigma[/itex]=E[itex]\epsilon[/itex]=22500(20(8.85x10-12))=3.98x10-6 Cm-2

(b)
C=[itex]\epsilon[/itex]A/d=20(8.85x10-12)(0.13)/0.002=1.15x10-8 F

(c)
So here's where I'm stuck a bit.

So the change in energy will be [itex]\Delta[/itex]U=U1 - U2

I have worked out:

U1=1.16x10-5 J

I don't understand what is meant when the question says its electrically isolated. I assumed it was something to do with I=0 but I can't see where this effect my equation so I must be missing something. Any help would be greatly appreciated.
 
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"electrically isolated" just means the charge on the plates doesn't go anywhere.
 
teme92 said:
F

(c)
So here's where I'm stuck a bit.

So the change in energy will be [itex]\Delta[/itex]U=U1 - U2

I have worked out:

U1=1.16x10-5 J

I don't understand what is meant when the question says its electrically isolated. I assumed it was something to do with I=0 but I can't see where this effect my equation so I must be missing something. Any help would be greatly appreciated.


You didn't show how you got U1. Was it 1/2 C V^2? If so you'll need to think about how V changes when the dielectric is removed.
 

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