91Ó°ÊÓ

We are given that Resistance of resistor = $150\mathrm{Î\copyright }$,voltage localid="1649931269825" $\left(v\right)$= $25V$,

capacitance localid="1649931273846" $\left(c\right)$$=$$2.5×{10}^{-12}$, distance between plates = localid="1649931277837" $5.0mm=5×{10}^{-3}$

Determine the maximum electric flux after the switch is closed

max. electric flux ($âˆ\dots$) = $\frac{CV}{e_o}$, where $âˆ\dots$is electric flux , $C$is capacitance ,$V$is voltage

We know $e_o$is the epsilon naught$\left({\mathrm{Î\mu }}_o\right)=8.85×{10}^{-12}$

therefore, $\mathrm{ϕ}=\frac{2.5×{10}^{-12}×25}{8.85×{10}^{-12}}=7.06vm$

Let maximum displacement current through capacitor,$I$is maximum displacement current.

$I=\frac{V}{R}$=$\frac{25}{150}=0.16A$

We are given that the electric flux at $t=0.50ns$

$t=0.50ns=0.5×{10}^{-9}s$

Electric flux$(âˆ\dots )=\frac{q}{e_o}$, where $q$is charge and $e_o$is epsilon naught.

where$q=cv(1-e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$Rc$}\right.})$, $c$is capacitance,$v$is voltage ,$t$is time ,$R$is resistance .

$=2.5×{10}^{-12}×25[1-e\frac{-0.5×{10}^{-9}}{150×2.5×{10}^{-12}}]$

$=62.5×{10}^{-12}[1-e\frac{-0.5×{10}^{-9}}{375×{10}^{-12}}]$

$=62.5×{10}^{-12}[1-e^{-1.33}]$

$=62.5×{10}^{-12}[1-0.265]$

$=62.5×{10}^{-12}[0.736]$

$=46×{10}^{-12}c$

$∴$electric flux $(âˆ\dots )=\frac{46×{10}^{-12}}{8.85×{10}^{-12}}=5.19vm$

Let displacement current at $t=0.50ns$, $I$is displacement current ,$t$ is time , $R$is resistance ,$v$ is voltage ,$c$ is capacitance.

$I=\frac{v}{R}e\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$Rc$}\right.$

$=\left[\frac{25}{150}\right]e^{-\frac{0.5×{10}^{-9}}{150×2.5×{10}^{-12}}}$

$=0.16e^{-1.33}$

$=0.16×0.2644$

$I=0.044A$