How much charge will flow from the battery?

AI Thread Summary
A 3500 pF air-gap capacitor connected to a 32 V battery has its capacitance affected by inserting mica, which has a dielectric constant (K) of 7. The correct calculation for the charge flow from the battery after inserting the mica involves determining the change in charge due to the dielectric. The formula used is delta Q = (K - 1) * C0 * V, where C0 is the original capacitance and V is the voltage. The final answer for the charge that flows from the battery, accounting for the dielectric, is calculated to be 6.4 * 10^-7 C. This approach clarifies the misunderstanding regarding the charge already present in the capacitor.
gillyr2
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Homework Statement



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

Homework Equations



C= K*C_0
V= V_0/ K
Q= C*V
K of mica is 7

The Attempt at a Solution



solved for the _0's and plugged into Q=CV
i used (C/K) * (V*K) = (3500 *10^-12F)/7 * 32V*7 = 1.12*10^-7C

its wrong but i don't know what else to do. help.
 
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well you've canceled out the mica with your equation.
You know that capacitance depends linearly on the k value of the dielectric (assuming area and distance between parallel plates remains constant).
So effectively the capacitance of with the mica inserted is 7 * 3500E-12.
Q = CV.
Do they want the answer in C or e?
 


ok so it would be

Q = (7*3500*10^-12) * 32V

they want it in C. that gives 7.84*10^-7C which was my first answer but it is wrong.
 


can someone help?
 


The question is "how much charge flows from the battery?" (when you insert the mica).
Your (above) calculation gives the charge after inserting the mica. This includes the charge that was there already.
You just have to replace 7 by 6.

delta Q = kCo*V-Co*V = (k-1)Co*V

Is this the right answer?
 

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