A parallel-plate air capacitor of area A = 27.5 cm 2 and plate separation of d =
ID: 1414521 • Letter: A
Question
A parallel-plate air capacitor of area A = 27.5 cm2 and plate separation of d = 3.10 mm is charged by a battery to a voltage of 60.0 V. What is the charge on the capacitor?
THE ANSWER 1.29e-8 C IS INCORRECT!
If a dielectric material with = 3.80 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?
THE ANSWER 3.61e-8 C IS INCORRECT!
What is the capacitance of a parallel plate capacitor?Explanation / Answer
A = area = 27.5 cm2
d = distance between plates = 3.10 mm
V = potential = 60 V
C = capacitance
1) C = E0*A / d
= 8.854*10-12*27.5*10-4 / 3.10*10-3
= 7.85*10-12 F
2) Q = charge on plates
Q = CV
= 7.85*10-12 * 60
= 4.71*10-10 C
3) now dilectric plate is inserted with k = 3.80
new capacitance. C' = k*E0A/ d
C' = k*C
= 3.80*7.85*10-12
= 2.983*10-11F
charge after plate is inserted , Q' = C'*V
= 2.983*10-11* 60
= 1.789*10-9C
difference in charge ( this is the additional chrge that will flow to positive plate) = 1.789*10-9 - 4.71*10-10
= 1.318*10-9C