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A parallel- plate capacitor with plate a...

A parallel- plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn summetrically between the plates. Find the displacement current through this area.

Text Solution

Verified by Experts

The correct Answer is:
B
`i_d = epsilon_0 (d phi_E)/(d t)`
`=epsilon_0 (d)/(d t)(E A/2)`
`=(epsilon_0 A)/2 (d)/(d t) (E)`
`=(epsilon_0 A)/2 (d)/(d t)((sigma)/(epsilon_0))`
`= (epsilon_0 A)/2 (d)/ (d t)((d)/(A epsilon_0))`
`= 1/2 (d q)/(d t)`
`i/2 (because (d q)/(d t)= i)`.
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Knowledge Check

  • 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)`
    C
    `10^(8) m^(2)`
    D
    `1.13 m^(2)`

  • In a parallel plate capacitor with plate area A and charge Q, the force on one plate because of the charge on the other is equal to

    A
    `Q^(2)/(epsi_(0)A^(2))`
    B
    `Q^(2)/(2epsi_(0)A^(2))`
    C
    `Q^(2)/(epsi_(0)A)`
    D
    `Q^(2)/(2 epsi_(0) A)`