A 3000 pF air-gap capacitor is connected to a 16 V battery. If a ceramic dielect

Question

A 3000 pF air-gap capacitor is connected to a 16 V battery. If a ceramic dielectric material (K = 5.8) fills the space between the plates, how much charge will flow from the battery? An air-gap capacitor (C1 = 3000 x 10^-12 F) accumulates a charge Qi when it is connected to a 16 V battery. While the battery is still connected, a dielectric material (K = 5.8) is placed between the plates of the capacitor, which increases the capacitance to Cf = kCi. The final charge on the capacitor is Qf. We can use the definition of capacitance to calculate the amount of charge that flowed from the battery to the capacitor upon adding the dielectric. How much charge(Q) will flow from the battery?

Explanation / Answer

here,

capacitance , C = 3000 pF

C = 3 * 10^-9 F

voltage of battery , V = 16 V

initial charge flowed from battery , Qi = C * V

Qi = 3 * 10^-9 * 16

Qi = 4.8 * 10^-8 C

capacitance after adding the dielectric , Cf = 5.8 * C

Cf = 1.74 * 10^-8 F

final charge flowed from battery , Qf = Cf * V

Qf = 1.74 * 10^-8 * 16

Qf = 27.84 * 10^-8 C

the amount of charge that flow from the battery to the capacitor upon adding the dielectric , Q = Qf- Qi

Q = 27.84 * 10^-8 - 4.8 * 10^-8

the amount of charge that flow from the battery to the capacitor upon adding the dielectric is 23.04*10^-8 C

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