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The area of each plate of parallel plate...

The area of each plate of parallel plated condenser is `144 cm^(2)`. The electrical field in the gap between the plates changes at the rate of `10^(12)Vm^(-1)s^(-1)`. The displacement current is

A

`(4)/(pi)A `

B

`(0.4)/(pi)`

C

`(40)/(pi)`

D

`(1)/(10pi)A`

Text Solution

Verified by Experts

The correct Answer is:
B

`I_(d)=epsilon_(0)A (dE)/(dt)`
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Knowledge Check

  • A parallel plate condenser is immersed in an oil of dielectric constant 2 . The field between the plates is

    A
    increased proportional to `2`
    B
    decreased proportional to `(1)/(2)`
    C
    increased proportional to `sqrt(2)`
    D
    decreased proportional to `(1)/(sqrt(2))`
  • What is the displacement current between the square plate of side 1 cm is of a capacitor, if electric field between the plates is changing at the rate of 3 xx 10^(6) Vm^(-1)s^(-1) ?

    A
    `2.7 xx 10^(-9)A`
    B
    `3.2 xx 10^(-5)A`
    C
    `4.2 xx 10^(-6) A`
    D
    `4.0 xx 10^(-5) A`
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