Respuesta :
Answer:
A) The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C[/tex]
B) The total charge that accumulates at the upper terminal is 50μC
C) If the current is stopped at t = 0.5 ms then total charge stored on upper terminal is 35.67μC
Explanation:
Given that:
[tex]I(t)=0 \quad \quad \quad \quad \quad t<0\\\\I(t)= 125e^{-2500t} \quad t\geq 0[/tex]
A) The charge that accumulates at the upper terminal for t > 0:
As we know
[tex]q(t)=\int {I(t)} \, dt[/tex]
for t > 0
[tex]q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]\\\\[/tex]
The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C---(1)[/tex]
B) The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C[/tex]
C) If the current is stopped at t = 0.5 ms then
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C[/tex]
