Dielectric and Parallel Plate Capacitor

  • Thread starter opprobe
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
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...
 
  • #3
17
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Thank you so much! I can't believe I completely forgot the definition of a capacitor...
 

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