Question

A sinusoidally-varying voltage V(t)=V0sin(2pift) with amplitude V0 = 10 V and frequency f = 100 Hz is impressed across the plates of a circular-shaped parallel plate air-gap capacitor of radius a = 1.0 cm and plate separation d = 0.01 mm. The amplitude of Maxwell's displacement current ID flowing across the gap between the plates of this capacitor is?

Answer 1

Answer:

Id=1.75×10⁻⁶A

Explanation:

The flux formula

Iₐ=ε(dФ/dt)

The electric flux is

ФE=E.A

ФE=EA={V(t)/d}πa²

The electric field inside of a parallel plate capacitor points in straight lines of constant field from one plate to the other. Using this form, we can take a derivative to quickly obtain

Id=(2π²a²ε₀V₀f/d)cos(2πft)

putting values

we get Id=1.75×10⁻⁶A