Question

A 3600-pF air-gap capacitor is connected to a 30-V battery.

If a piece of mica (K= 7) is placed between the plates, how much charge will flow from the battery?

**I've tried this so many ways and I keep getting the wrong answer! Thanks for the help!

Answer 1

Given that

voltage v= 30v

capacitance c=3600PF = 3600*10^-12 F

we know that

Q=CV

initial charge Q1= (3600*10^-12)(30) C

Q1= 1.08*10^-7 C

Final charge when piece of mica placed between plates

Q2= 7*(3600*10^-12)(30) C

Q2=7.56*10^-7

so the charge flow from battery will be

Q2-Q1 = (7.56*10^-7)-(1.08*10^-7)

= 0.000000648 C