90 µF capacitor is charged by a 15 V battery. While remaining on the battery, a layer of Teflon tric constant of 2) is removed from between the plates, leaving air. at is the new capacitance of the plates? [145 µF] nsider just charge on the lates and the voltage across the plates Which of them is the same before and
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Just need help with parts d and e, please.
![3. A 290 µF capacitor is charged by a 15 V battery. While remaining on the battery, a layer of Teflon
(dielectric constant of 2) is removed from between the plates, leaving air.
a. What is the new capacitance of the plates? [145 µF]
b. Consider just charge on the plates and the voltage across the plates. Which of them is the same before and
after the Teflon is removed? Why?
c. Find the new value for the one that changed.
d. Did the electric field increase, decrease, or stay the same when the Teflon was removed? Explain how you
know.
e. Calculate the new potential energy stored by the capacitor. [0.016 J]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F266e7f29-ba1d-4acb-9482-ffb76dbeef2e%2Fff5ce0b7-179b-4bcd-b041-1bccf71d9e6d%2Fd3hfml_processed.png&w=3840&q=75)
Capacitance with air
C0 =ε₀A/d
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