Related Formula
Energy Density = (1)/(2) ε₀ E²$$\text{Energy Density} = \frac{1}{2} \epsilon_0 \mathrm{E}^2$$
Total Energy = Energy Density × Volume$$\text{Total Energy} = \text{Energy Density} \times \text{Volume}$$
Core Logic
Since both cylinders hold equal amounts of electromagnetic energy:
(Energy)₁ = (Energy)₂$$\left(\text{Energy}\right)_1 = \left(\text{Energy}\right)_2$$
(1)/(2) ε₀ E₁² · c π R₁² × L₁ = (1)/(2) ε₀ E₂² · c π R₂² × L₂$$\frac{1}{2} \epsilon_0 \mathrm{E}_1^2 \cdot c \pi \mathrm{R}_1^2 \times \mathrm{L}_1 = \frac{1}{2} \epsilon_0 \mathrm{E}_2^2 \cdot c \pi \mathrm{R}_2^2 \times \mathrm{L}_2$$
Since the lengths are identical (L₁ = L₂$\mathrm{L}_1 = \mathrm{L}_2$), this simplifies to:
E₁² R₁² = E₂² R₂² E₁ R₁ = E₂ R₂$$\mathrm{E}_1^2 \mathrm{R}_1^2 = \mathrm{E}_2^2 \mathrm{R}_2^2 \implies \mathrm{E}_1 \mathrm{R}_1 = \mathrm{E}_2 \mathrm{R}_2$$
Given the second cylinder has half the diameter (and radius) of the first (R₂ = R₁2$\mathrm{R}_2 = \frac{\mathrm{R}_1}{2}$):
100 × R₁ = E₂ × R₁2$$100 \times \mathrm{R}_1 = \mathrm{E}_2 \times \frac{\mathrm{R}_1}{2}$$
E₂ = 200 N/C$$\mathrm{E}_2 = 200 \mathrm{N/C}$$
Step 1: Final Equation Match
The wave equation adjusts its amplitude factor to 200 (ω t - kx)NC⁻¹$200\sin (\omega t - kx)\mathrm{NC}^{-1}$, which matches option (2).
Pattern Recognition
When energy is constant and volume scales down inversely by a factor of 4 (due to R²$\mathrm{R}^2$), the electric field strength must increase by a factor of √(4) = 2$\sqrt{4} = 2$ to maintain balance.
Chapter Mix
Class 12 Physics: Electromagnetic Waves