A parallel plate capacitor has circular plates of radius 6 cm and a plate separation of 2 mm . The electric field between the plates varies sinusoidally with time as E(t) = E₀ ( t) , where E₀ = 5 10^5 V/m and = 2 10^4 rad/s . The peak value of the displacement current between the plates is _______ mA . (Given: Permittivity of free space ₀ = 1 36 10^9 F/m )

Step-by-step solution

The displacement current is given by the rate of change of electric flux: i_d = ₀ d _E dt = ₀ A dE dt The area of the circular plates is: A = r^2 = (6 10⁻²)^2 = 36 10⁻⁴ m ^2 Given the electric field E(t) = E₀ ( t) , its rate of change is: dE dt = E₀ ( t) The peak displacement current I_ d0 occurs when ( t) = 1 : I_ d0 = ₀ A E₀ Substituting the given values: I_ d0 = ( 1 36 10^9 ) (36 10⁻⁴) (5 10^5) (2 10^4) I_ d0 = 10⁻¹³ 10¹⁰ = 10⁻³ A Converting to milliamperes: I_ d0 = 1 mA Answer: 1