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Dielectrics and Capacitors, Very confused.

  • Thread starter ryan8642
  • Start date
  • #1
24
0

Homework Statement


A 25pF parallel plate capacitor with an air gap b/w the plates is connected to a 100v battery. A Teflon slab is then inserted b/w the plates and completely fills the gap. What is the change in charge on the positive plate when the Teflon is inserted?


Homework Equations


Q=CV
C=eA/d
C=keA/d

e=permittivity constant



The Attempt at a Solution



I am very confused and dont know what to do, none of the formulas make sense for this question... :(
But

C=25x10^-12F
Kteflon=3.0
permittivity constant=8.85x10^-12
v=100v

what do i do??? soo confused

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,137
0
find the two capacitances after and before the plate is inserted

find the charge using Q=CV as you wrote

then just subtract the charge!!!!
 
  • #3
24
0
i dont get it, how do i find the capacitance when the teflon is inserted? do i just multiply the C (up there) by 2.0?

ktef should be 2.0 not 3.0
 
  • #4
berkeman
Mentor
57,306
7,291
i dont get it, how do i find the capacitance when the teflon is inserted? do i just multiply the C (up there) by 2.0?

ktef should be 2.0 not 3.0
Yes. By 2.0
 

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