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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1 Answers

Correct Answer - A
Displacement current,
`I_(d) = epsi_(0)A(dE)/(dt)`
We have, `epsi_(0) = 8.85 xx 10^(-12) C^(2)N^(-1)m^(-2)`
Area `A = 1 cm^(2) = 10^(-4)m^(2)` and `(dE)/(dt) = 3 xx 10^(6)Vm^(-1)s^(-1)`
`rArr I_(d) = 8.85 xx 10^(-12) xx 10^(-4) xx 3 xx 10^(6)`
`:.` Displacement current, `I_(d) = 2.7 xx 10^(-9) A`

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