A 3500-pF air-gap capacitor question

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The discussion centers on calculating the charge flow from a 32-V battery to a 3500-pF air-gap capacitor when a piece of mica, with a dielectric constant (K) of 7, is introduced between the plates. The initial charge (Q1) is calculated using the formula Q1 = C1V1, while the new charge (Q2) is determined with Q2 = C2V2, where C2 is 7 times C1 due to the mica's dielectric effect. The discrepancy between the calculated charge of 7.84 x 10^-7 Coulombs and the solution manual's value of 6.7 x 10^-7 Coulombs arises from the correct adjustment of the dielectric constant, which accounts for the air gap. The key takeaway is that the question specifically asks for the additional charge that flows from the battery, clarifying the need to calculate the difference Q2 - Q1.

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A 3500-pF air-gap capacitor is connected to a 32-V
battery. If a piece of mica is placed between the plates,
how much charge will flow from the battery?

Homework Equations


C = K€A/d[/B]

The Attempt at a Solution


Q1=C1V1
Q2=C2V2

V1=V2=V
C2=7C1(as only diference between C1 and C2 is the mica and it's K is 7 so it's 7 times C1)
Result: Q2=7.84 x 10^-7Coloumb

**But solution manual say it's 6.7 x 10^-7Columb BECAUSE it calculates the new K by subtracting 1 from 7 (1 is the K of air) ?! Why subtract? totally lost :(
[/B]
 
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The question is how much charge is added to the already existing charge to the capacitor, which is Q2-Q1.
 
Ahh so by saying "how much charge will flow from the battery?" it means "how much ADDITIONAL/EXTRA charge will flow from the battery?" .. OK I've got got it now but the that part if the question is a bit misty, don't you think?

Thank you!
 

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