JEE Main · Physics → Steady

Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Displacement Current.

Year 2026 2025 2024 Total
Questions 10 12 9 31

A parallel plate capacitor has a capacitance C = 200~pF. It is connected to 230~V ac supply with an angular frequency 300~rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are:

Solution & Explanation

Related Formula

Capacitive reactance:

XC = (1)/(ω C)

Conduction Current (Irms):

Irms = VrmsXC = Vrms · ω C

Continuity relation: Ic = Id

Core Logic

Given parameters: C = 200~pF = 200 × 10⁻¹²~F, Vrms = 230~V, ω = 300~rad/s.

Calculate current:

I = Vrms · ω C = 230 × 300 × 200 × 10⁻¹²
Step 1: Simplify Numerical Calculation

I = 230 × 60000 × 10⁻¹² = 13.8 × 10⁻⁶~A = 13.8~μ A

By Maxwell's electromagnetic formulation, the displacement current Id within the dielectric space matches the exterior conduction current Ic seamlessly in magnitude.

Pattern Recognition

Factual invariant: Conduction current always equals displacement current inside a standard layout loop context (Ic = Id).

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Alternating Current

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz and an amplitude of 50 ~Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm² ]
  • A. 1.106 × 10⁻⁸ Jm⁻³
  • B. 4.425 × 10⁻⁸ ~Jm⁻³
  • C. 2.212 × 10⁻⁸ ~Jm⁻³
  • D. 2.212 × 10⁻¹⁰ ~Jm⁻³

Solution

Related Formula
Utotal average = (1)/(2)ε₀ E₀²
Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where E₀ is the amplitude of the electric field.

Step 2: Substitution

Given: E₀ = 50 V/m ε₀ = 8.85 × 10⁻¹² C²/(N· m²)

Uavg = (1)/(2) × (8.85 × 10⁻¹²) × (50)² Uavg = (1)/(2) × 8.85 × 10⁻¹² × 2500 Uavg = 1.10625 × 10⁻⁸ J/m³
Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Questions — jee_main_2024_01_february_morning

Practice all Electromagnetic Waves previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)