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Sea water at frequency v=4 xx 10^(8) Hz ...

Sea water at frequency `v=4 xx 10^(8)` Hz has permittivity `epsilon ~~ 80 epsilon_(0)` permeability `mu= mu_(0)` and resistivity `rho= 0.25`M. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source `V(t) =V_(0) " sin "(2pivt).` What fraction of the conduction current density is the displacement current density ?

A

`2//3`

B

`4//9`

C

`9//4`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:
C
`V(t)=V_(0) sin 2pi upsilont`
Let distance between the plates = d.
Electric field `=(V(t))/(d)=(V_(0))/(d)sin 2piupsilont`
Conduction current density
`J_(c)=(E)/(rho)=(V_(0))/(rhod)sin 2pi upsilont=J_(o_(c)) sin 2pi upsilont`
where `J_(o_(c))` is maximum conduction currect density.
Displacement current density,
`J_(d)=epsilon(d)/(dt)[(V_(0))/(d)sin(2piupsilont)]=(2piupsilonepsilon)/(d)V_(0)cos 2piupsilont`
`J_(d)=j_(o_(d))cos(2piupsilont)`
`(J_(o_(d)))/(J_(o_(c)))=(2piupsilon)/(d)epsilon(V_(0))/(d)=2piupsilonepsilonrho=2pi(4xx10^(8))(80epsilon_(0))(0.25)`
`=(J_(o_(c)))/(J_(o_(d)))=(9xx10^(9))/(4xx10^(9))=(9)/(4)" "((1)/(4piepsilon_(0))=9xx10^(9)"N m"^(2)//C^(2))`
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Knowledge Check

  • A parallel plate capacitor consits of two circular plates of radius 0.5m. If electric field between the plates changed as (dE)/(dt) = 10^(10). (V)/(m-c) , then find displacement current between the plates

    A
    7 mA
    B
    0.07 mA
    C
    0.7 mA
    D
    70 mA

  • A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage V=10 sin (100t) . The peak current in the circuit will be:

    A
    2amp
    B
    1amp
    C
    10amp
    D
    20amp

  • The voltage between the plates of a parallel plate capacitor of capacitance 2 mu F is changing at the rate of 4 V//s . What is the displacement current in the capacitor?

    A
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    B
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    C
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    D
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