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Allisa [31]
3 years ago
13

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

lly uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

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Putting the values, in above equation, we get:

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