Asked by andrew
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. If a 19.2-V potential difference is applied to these plates, calculate the following:
(a) the electric field between the plates, magnitude (in kV/m) and direction
(b) the capacitance (pF)
(c) the charge on each plate (pC)
(a) the electric field between the plates, magnitude (in kV/m) and direction
(b) the capacitance (pF)
(c) the charge on each plate (pC)
Answers
Answered by
drwls
(a) E = V/d = 19.2 V/.0017m
= 112,900 V/m = 112.9 kV/m
(b) C = (epsilon0)*A/d
= 8.85*10^-12 Farad/m*7.6*10^-4
m^2/0.0017 m
= 3.96*10^-11 F
= 39.6 pF
(c) Q = C*V = 39.6*10^-12F*19.2 V
= 6.41*10^-10 C = _____ pC
= 112,900 V/m = 112.9 kV/m
(b) C = (epsilon0)*A/d
= 8.85*10^-12 Farad/m*7.6*10^-4
m^2/0.0017 m
= 3.96*10^-11 F
= 39.6 pF
(c) Q = C*V = 39.6*10^-12F*19.2 V
= 6.41*10^-10 C = _____ pC
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