Physics, 07.04.2020 22:27 theirishgold42

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)

Answers: 3

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Answers

Answer from: alfonso55

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 ( ∅ ) = $\frac{CV }{e_{0} }$ ----- ( 1 )

where : C = 2.10 * 10⁻¹² , V = 30 v , $e_{0}$ = 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 ( ∅ ) = $\frac{q}{e_{0} }$ ------ ( 2 )

where : q = CV * ( 1 - $e^{-\frac{t}{RC} }$ ) = 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 = $(\frac{V}{R}) e^{-\frac{t}{RC} }$

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

= 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

Learn more about displacement current : link

Answer from: bnnn

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

Answer from: Quest

connect the batteries in series and hope

Answer from: Quest

i think the answer is a, hope this : )

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A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF paral...

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