A parallel plate capacitor has two circular plates each of radius 5 cm separated by a distance of 3 m m . The capacitor current of 6 A is delived to it , find
(a) The capacitance
(b ) The rate of change of potential difference across the plates
(c ) the displacement current .

Explanation

Given , radius of capcitor .s plate ,` r=5 cm = 5 xx 10 ^(-2 )m`
Separation between the plates , `d=3 mm = 3xx 10 ^(-3)` m currrent
` I=6A `
(a) capacitance is given by
` C= (epsi_0 A )/( d ) = (epsi_0 (pi r^2 ) )/( d ) `
` = ( 8.85 xx 10 ^(-12) xx 3.14 xx ( 5 xx 10^(-2)) ^2)/( 3 xx 10^(-3))`
`= 23.15 pF `
(b) IF q is the instantaneous charge on capacitor , then
` I= ( dq) /( dt ) = ( d)/(dt) (CV ) = C (dV) /( dt)`
` implies (dV )/(d t) = (I )/(C)`
` = ( 6) /( 23.15 xx 10^(-12) ) = 0.26 xx 10^(12) V//s`
` = 26 xx 10^(10) V//s`
(c ) Displacement current is given by
` I_d= C(dv )/( dt) ` ( refer to example 2 )
`= 23.15 xx 10^(-12) xx 26 xx 10^(-10) = 6.02 A `