n air-filled capacitor consists of two parallel plates, each with an area of 7.6
ID: 1872715 • Letter: N
Question
n air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.20 mm. A 15.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates. 68181 kv/m (b) Calculate the surface charge density. There are a number of ways to get to this answer. How do you calculate the electric field due to a uniform surface charge density? nC/m2 (c) Calculate the capacitance. pF (d) Calculate the charge on each plate. pC Need Help? Read it MasterItExplanation / Answer
a) E = V/d = 6.818 kV/m
b)
surface charge density = e0*E = (8.85*10^(-12))*6.818*10^3 = 6.0339*10^(-8) = 60.34 nCm^2
c)
capacitance C = e0*A/d = (8.85*10^(-12))*7.6*10^(-4)/(2.2*10^(-3)) = 3.057*10^(-12) F = 3.057 pF
d)
Q = C*V = 3.057*15 = 45.85 pC
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