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