9. (a) How much charge can be placed on a capacitor with air between the plates
ID: 2276301 • Letter: 9
Question
9.
(a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 6.00 cm2?
_______ nC
(b) Find the maximum charge if porcelain is used between the plates instead of air.
_______ nC
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10.
(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.60 cm2 and a plate separation of 0.070 0 mm.
_______ pF
b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.60 cm2 and a plate separation of 0.070 0 mm.
_______ kV
Explanation / Answer
9.
(a)
We know that :
sigma = Q/A
For air e = e0 = 8.85*10^-12 F/m
Q = 3*10^6*0*0.0006*8.85*10^-12
Q = 1.593*10^-8 C
Q = 1.593 nC
(b)
Dielectric strength of porcelain = 12*10^6
Dielectric constant = 6
sigma = k*e0*E
Q = A*sigma = A* k*e0*E
Q = 6*10^-4*6*12*10^6*8.85*10^-12
Q = 3.8232*10^-7
Q = 382 nC
10.
(a)
C = e0*A/d
Also C*d = C*K
C =(e0*A*C)/d
E = 8.85*10^-12
A = 1.6*10^-4
d = 7*10^-5
K (Teflon) = 21
C = (8.85*10^-12*1.6*10^-4*21)/7*10^-5
C = 4.248*10^-10 F
C = 42.48 pF
(b)
Dielectric strength of Teflon = 60 MV/m
V = 60 * 10^6 V/m * 0.07*10^-3 m
V = 4200 V
V = 4.2 kV
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