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
