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

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

Answer from: Quest

Since the pully is a simple machine so the mechanical advantage is a 1

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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...