NEETPhysicsCapacitance
A parallel plate capacitor is being charged such that a constant displacement current of 8.85 mA exists between its plates. If the area of each plate is 100 cm ^2 , what is the rate at which the electric field between the plates is changing? (Given ₀ = 8.85 10⁻¹² C ^2 N ⁻¹ m ⁻² )
Options
- A10¹⁵ V m ⁻¹ s ⁻¹
- B10⁷ V m ⁻¹ s ⁻¹
- C8.85 10¹¹ V m ⁻¹ s ⁻¹
- D10¹¹ V m ⁻¹ s ⁻¹
Correct answer
D. 10¹¹ V m ⁻¹ s ⁻¹
Step-by-step solution
The displacement current I_d is given by the fundamental relation: I_d = ₀ d _E dt Since the electric flux _E = E A , we can write: I_d = ₀ A dE dt Rearranging for the rate of change of electric field dE dt : dE dt = I_d ₀ A Given values: I_d = 8.85 mA = 8.85 10⁻³ A A = 100 cm ^2 = 100 10⁻⁴ m ^2 = 10⁻² m ^2 ₀ = 8.85 10⁻¹² C ^2 N ⁻¹ m ⁻² Substituting the values: dE dt = 8.85 10⁻³ 8.85 10⁻¹² 10⁻² dE dt = 10⁻³ 10⁻¹⁴ = 10¹¹ V m ⁻¹ s ⁻¹ Answer: 10¹¹ V m ⁻¹ s ⁻¹