Question

A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t = 0 s. . (a) After the switch is closed, find the maximum electric flux and the maximum displacement current through the capacitor.. (b) Find the electric flux and the displacement current at t = 0.50 ns.

Answer 1

Find the electric flux and the disp at t=0.50ns 
Given: 
Resistor R = 160 Ω 
Voltage ε = 22.0 V 
Capacitor C = 3.10 pF = 3.10 * 10^-12 F 
time t = 0.5 ns = 0.5 * 10^-9 s 
ε0 = 8.85 * 10^-12 
Solution: 
ELECTRIC FLUX: 
Φ = Q/ε0 
we have ε0, we need to find Q the charge 
STEP 1: FIND Q 
Q = C ε ( 1 - e^(-t/RC) ) 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } 
Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } 
Q = 43.31 * 10^-12 C 
STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> 
Φ = Q/ε0 
Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } 
Φ = 4.8937 = 4.9 V.m 
DISPLACEMENT CURRENT 
we use the following equation: 
I = { ε / R } { e^(-t/RC) } 
I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
I = { 0.1375 } { 0.365 } 
I = 0.0502 A = 0.05 A