A 2400 pF air-gap capacitor is connected to a 8.0 V battery. ?
A 2400 pF air-gap capacitor is connected to a 8.0 V battery. If a piece of mica is placed between the plates, how much charge will flow from the battery?
- billrussell42Lv 71 decade agoFavorite Answer
Initial charge is Q = CV = 19.2 nC
Mica has a dielectric constant of about 7 (varies a lot, depending on type and source)
so the capacitance changes to 2.4 nF x 7 = 16.8 nF
new charge is 16.8 nF x 8 = 134 nC
difference = 134 – 19 = 115 nC
- messengerLv 44 years ago
power soterd is given via E = a million/2 CV^2 the place C = efficient capacitance. so which you will desire to compute C for each variety of circuit. For the parallel circuit C = C1 + C2 = 6 uF so which you plug into th eenergy formulation and you're achieved FOr a chain circuit: a million/C = a million/c1 + a million/C2 ---> C = C1*C2/(C1+C2) = 8/6 uF = a million.33 uF Plug this fee into the flexibility euqation and you're achieved.