Question

There is no charge at the upper terminal of the ele- ment in Fig. 1.5 for t 6 0. At t = 0 a current of 125e-2500t mA enters the upper terminal. a) Derive the expression for the charge that accu- mulates at the upper terminal for t 7 0. b) Find the total charge that accumulates at the upper terminal. c) If the current is stopped at t = 0.5 ms, how much charge has accumulated at the upper terminal?

Answer 1

Answer:

A) The charge accumulated on upper plate for t>0 is

$q(t)=50[1-e^{-2500t}]\mu C$

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:

$I(t)=0 \quad \quad \quad \quad \quad t$

A) The charge that accumulates at the upper terminal for t > 0:

As we know

$q(t)=\int {I(t)} \, dt$

for t > 0

$q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]$

The charge accumulated on upper plate for t>0 is

$q(t)=50[1-e^{-2500t}]\mu C---(1)$

B)  The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)

$q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C$

C)  If the current is stopped at t = 0.5 ms then

$q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C$