If a dielectric material with = 3.90 is inserted so that it fills the volume bet

Question

If a dielectric material with = 3.90 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

PHY131S17, Homework #3: Parallel plate capacitor with dielectric material A parallel-plate air capacitor of area A 29.5 cm2 and plate separation of d 2.00 mm is charged by a battery to a voltage of 58.0 V What is the charge on the capacitor? 7.57x10 nC You are correct. Computer's answer now shown above. Previous Tries Your receipt no. is 163-7835 If a dielectric material with J390 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery, how much additional charge will flow from the battery onto the positive plate? 2.95 10 -9C Calculate the change in capacitance, and then the change in charge. Submit Answer Incorrect. Tries 11/20 Previous Tries

Explanation / Answer

here,

area, a = 29.5 cm^2 = (29.5*10^-4) m^2

seperation, d = 2 mm = 0.002 m

voltage, v = 58 V

Charge, Q = Capacitance * Volatge

Charge, Q = eo*a/d * V

Charge, Q = ((8.85*10^-12*29.5*10^-4)/0.002) * 58

Charge, Q = 7.57*10^-1 nC

When Dielectric insterted with constant, k = 3.9

Charge, Q = Capacitance * Volatge

Charge, Q = ((k*eo*a)/d) * V

Charge, Q = ((3.9*8.85*10^-12*29.5*10^-4)/0.002) * 58

Charge, Q = 2.953 nC

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