A simple series circuit consists of a $150\Omega$resistor, a $25V$battery, a switch, and a $2.5pF$parallel-plate capacitor (initially uncharged) with plates $5.0mm$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 $t=0.50ns.$

We are given that Resistance of resistor = $150\Omega$,voltage localid="1649931269825" $\left(v\right)$= $25V$,

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

Determine the maximum electric flux after the switch is closed

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

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

therefore, $\varphi =\frac{2.5\times {10}^{-12}\times 25}{8.85\times {10}^{-12}}=7.06vm$

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 $t=0.50ns$

$t=0.50ns=0.5\times {10}^{-9}s$

Electric flux$(\varnothing )=\frac{q}{e_o}$, 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 .

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

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

$=62.5\times {10}^{-12}[1-e^{-1.33}]$

$=62.5\times {10}^{-12}[1-0.265]$

$=62.5\times {10}^{-12}[0.736]$

$=46\times {10}^{-12}c$

$\therefore$electric flux $(\varnothing )=\frac{46\times {10}^{-12}}{8.85\times {10}^{-12}}=5.19vm$

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$}\right.$

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

$=0.16e^{-1.33}$

$=0.16\times 0.2644$

$I=0.044A$