Question

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart.The switch is closed at t =0s.
Part A.

After the switch is closed, find the maximum electric flux through the capacitor. (Answer is in V * m)

Part B.

After the switch is closed, find the maximum displacement current through the capacitor. (Answer is in A)

Part C.

Find the electric flux at t =0.50ns. (Answer is in V * m)

Part D.

Find the displacement current at t =0.50ns. (Answer is in A)

Answer 1

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF=$2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current,$I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

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

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$