Question

In a parallel plate air capacitor, intensity of electric field is changing at the rate of 2 × 10¹¹ V/ms. If the area of each plate is 20 cm², calculate the displacement current.

Answer 1

Given:

Rate of change of electric field: `(dE)/dt` = 2 × 10¹¹ V/ms = 2 × 10¹⁴ V/s

Area of each plate: A = 20 cm²

= 20 × 10^-4 m²

= 2 × 10^-3 m²

Permittivity of free space: 

ε₀ = 8.854 × 10^-12 F/m

Formula:

Displacement current (I_d) is given by:

I_d = `ε_0A(dE)/dt`

I_d = (8.854 × 10^-12 ) × (2 × 10^-3) × (2 ×10¹⁴)

⇒ I_d = (8.854 × 10^-12 ) × (4 × 10¹¹)

⇒ I_d = 3.54 × 10⁰

⇒ I_d = 3.54 A

∴ The displacement current is 3.54 A.