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A parallel plate capacitor of plate sepa...

A parallel plate capacitor of plate separation `2 mm` is connected in an electric circuit having source voltage `400 V`. If the plate area is `60 cm^(2)`, then the value of displacement current for `10^(-6) sec` will be

A

`1.062 amp`.

B

`1.062xx10^(-2) amp`

C

`1.062xx10^(-3) amp`

D

`1.062xx10^(-4) amp`

Text Solution

Verified by Experts

The correct Answer is:
B
`I_(D)=epsilon_(0)(delphi_(E))/(delt)=epsilon_(0)(EA)/(t)=(epsilon_(0)A)/(t)(v)/(d)`
`I_(D)=(8.85xx10^(-12)xx400xx60xx10^(-4))/(2xx10^(-3)xx10^(-6))`
`=1.062xx10^(-2)amp`
Hence the corrent answer will be (`4`).
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Knowledge Check

  • A parallel plate capacitor of plate seperation 2mm is connected in an electric circuit having source voltage 400V. If the plate area is 60 cm^(2) , then the value of displacement current for 10^(-6) sec will be -

    A
    1.062 amp.
    B
    `1.062 xx 10^(-2)` amp
    C
    `1.062 xx 10^(-3)` amp
    D
    `1.062 xx 10^(-4)` amp

  • A parallel plate capacitor having a plate separation of 2mm is charged by connecting it to a 300v supply. The energy density is

    A
    `0.01 J//m^(3)`
    B
    `0.1 J//m^(3)`
    C
    `1.0 J//m^(3)`
    D
    `10 J//m^(3)`

  • Area of a parallel plate capacitor of capacitance 2F and separation between the plates 0.5cm will be

    A
    `1.13xx10^(9) m^(2)`
    B
    `1.13xx10^(6) m^(2)`
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