Answered

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 :

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Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF=[tex]2.1\times 10^{-12} F[/tex]

[tex]1pF=10^{-12} F[/tex]

[tex]d=5.0 mm=5\times 10^{-3} m[/tex]

[tex] 1mm=10^{-3} m[/tex]

a.Maximum flux

[tex]\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

[tex]\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm[/tex]

b.Maximum displacement current,[tex]I=\frac{V}{R}=\frac{30}{130}=0.23 A[/tex]

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

[tex]t=0.5ns=0.5\times 10^{-9} s[/tex]

[tex]1ns=10^{-9}s[/tex]

[tex]q=CV(1-e^{-\frac{t}{RC}})[/tex]

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

[tex]q=52.9\times 10^{-12} C[/tex]

[tex]\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm[/tex]

d.Displacement current at t=0.5ns

[tex]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}}}[/tex]

[tex]I=0.037 A[/tex]

batolisis

A) The max electric flux through the capacitor after switch is closed : 7.12 Vm

B) The max displacement current after switch is closed :  0.23 A

C) Electric flux at t = 0.50 ns :  5.98 Vm

D) Displacement current at t = 0.50 ns :  0.037 A

Given data :

Resistance of resistor = 130 Ω

Battery voltage ( P.D ) = 30.0 V

capacitance ( C ) = 2.10 * 10⁻¹²

Distance between plates = 5 * 10⁻³ m

A) Determine the max electric flux After the switch is closed

max electric flux ( ∅ ) = [tex]\frac{CV }{e_{0} }[/tex]  ----- ( 1 )

where : C =  2.10 * 10⁻¹²  ,  V = 30 v ,  [tex]e_{0}[/tex] = 8.85 * 10⁻¹²

Insert values into equation ( 1 )

∴ max electric flux = 7.12 V/m

B) Determine the maximum displacement current

I = V / R

V = 30 v ,  R = 130 Ω

∴ I =  30 / 130 = 0.23 A

C) Determine the electric flux at t = 0.50 ns

t = 0.50 ns = 0.5 * 10⁻⁹ s

electric flux ( ∅ ) = [tex]\frac{q}{e_{0} }[/tex]    ------ ( 2 )

where : q = CV * ( 1 - [tex]e^{-\frac{t}{RC} }[/tex] )  = 52.9 * 10⁻¹² C

∴ electric flux ( ∅ ) at t = 0.5 ns  = ( 52.9 * 10⁻¹² ) / ( 8.85 * 10⁻¹² )

                                                    = 5.98 Vm

D) Determine the displacement current at t = 0.5 ns

Applying the equation below

I = [tex](\frac{V}{R}) e^{-\frac{t}{RC} }[/tex]

 = ( 30 / 130 ) * [tex]e^{-\frac{0.5*10^{-9} }{130*2.1*10^{-12} } }[/tex]

 = 0.037 A

Hence we can conclude that the maximum displacement current and the electric flux at different times are as listed in the answers above

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