Please help with the second part of this question. A) A parallel-plate air capac
ID: 1525287 • Letter: P
Question
Please help with the second part of this question. A) A parallel-plate air capacitor of area A = 26.1 cm2 and plate separation of d = 3.20 mm is charged by a battery to a voltage of 61.0 V. What is the charge on the capacitor?
a) 4.40×10-1 nC
PHY131S17, Homework #3: Parallel plate capacitor with dielectric material A parallel-plate air capacitor of area A 26.1 cm2 a d plate separation of d 3.20 mm is charged by a battery to a voltage of 61.0 V. at is the charge on the capacitor? d 440x10-1 nC You are correct. Computer's answer now shown above Previous Ties Your receipt no. is 163-8158 If a dielectric material with 35.00 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? Calculate the change in capacitance, and then the change in charge. Submit Answer Incorrect. Tries 1/20 Previous TriesExplanation / Answer
a) C = eo*A/d = Q/V
eo is the permittivity of free space
A = 26.1 cm^2
d = 3.2 mm
Q is the required charge
V = voltage = 61 V
then eo*A/d = Q/V
(8.85*10^-12*26.1*10^-4)/(3.2*10^-3) = Q/61
Q = 4.4*10^-10 C = 4.4*10^-1 nC
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if a dielectirc is introduced then
new capacitance is C_new = k*eo*A/d = k*C = 5*(4.4*10^-10/61) = 36.06*10^-12 F = 36.06 pF
C_new = Q_new/V
Q_new = C_new*V = 36.06*10^-12*61 = 2.19*10^-9 C = 2.19 nC
additional charge is dQ = ((2.19)-(0.44))*10^-9 = 1.75 nC
change in capacitance is dC = (k-1)*(eo*A/d) = (5-1)*(8.85*10^-12*26.1*10^-4)/(3.2*10^-3) =28.8*10^-12 F = 28.8 pF
change in charge is dQ = 1.75 nC