Physics Homework problem: Dielectrics

AI Thread Summary
The discussion revolves around a physics homework problem involving a 3500 air-gap capacitor connected to a 32V battery and the effect of inserting mica as a dielectric. Participants clarify the initial charge calculation for the air-gap capacitor, with the charge determined to be 1.12E-4 C. The dielectric constant for mica (K=7) is discussed, highlighting its impact on capacitance. There is some confusion regarding the capacitor's specifications, specifically whether it is measured in microfarads or nanofarads. Ultimately, the original poster resolves the problem independently and expresses gratitude for the assistance.
Jscrub
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I cannot figure out this problem...can someone help me?

A 3500 air-gap capacitor is connected to a 32 battery.

If a piece of mica fills the space between the plates, how much charge will flow from the battery?


Express your answer using two significant figures.
 
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Welcome to PF!

Hi Jscrub! Welcome to PF! :wink:

First, how much charge will flow from the battery when the gap is just air?

Second, what equations do you know about how a dielectric affects a field or a capacitor? :smile:
 
c= KCo

c=Q/V

Q Air = 1.12E-4
 
v= 32 V

K(mica) = 7

K (air) = 1.0006

so, Co= 1.12E-4
 
Jscrub said:
A 3500 air-gap capacitor is connected to a 32 battery.

What is a 3500 air-gap capacitor?

Is that a 3500μf capacitor or 3500nf?

What is the effect of the dielectric on the capacitance?

I don't follow your conclusion
so, Co= 1.12E-4

Haven't you added a dielectric to your capacitor?
 
The above was not a conclusion..it was just a reply to the previous person...

Co = 3500E-9...

But it is fine..I don't need anymore help...I figured it out...

Thanks
 
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