Each of the circular plates of a parallel plate capacitor has radius of 4 cm . The capacitor is charged such that the electric field in the gap between the plates rises at a constant rate of `2 xx 10^(12) VM^(-1) s^(-1)` . What is the displacement current ? ` [ epsi_0 = 8.85 xx 10^(-12) C^2 N^(-1) m^(-2)]`

Explanation

Given radius of a plate ,` r=4 cm = 4 xx 10^(-2) m `
Rate of change of electric field `.(dE)/(dt ) = 2 xx 10^(12) V m^(-1) s^(-1)`
Displacement current is given by
`I_d = epsi_0 (dphi_E )/(d t) = epsi_0 A (dE )/(dt )`
where , `phi_E`= electric flux ` =AE ` , [ A=Area of plates ]
`therefore I_d = epsi_0 ( pi r^2) (dE)/(dt ) `
` = 8.85 xx 10^(-12) xx 3.14 xx ( 4xx 10^(-2) xx 2xx 10^(12)`
`= 0.088 A ~~ 0.09A `