Q42. The intensity of a plane monochromatic wave is 1380 Wm⁻² What is the amplitude of electric fiel

Q: What is the correct answer for CUET UG Physics Q42?

The correct answer is C. The intensity (I) of a plane electromagnetic wave in free space is related to the amplitude of its electric field (E₀) by the formula: I = (1/2)ε₀cE₀², where ε₀ is the permittivity of free space and c is the speed of light. The 'wo' in the question refers to ε₀. Given: Intensity I = 1380 W/m². Speed of light c = 3 x 10⁸ m/s. Permittivity ε₀ = 8.85 x 10⁻¹² C²N⁻¹m⁻². We need to find E₀. Rearrangin

Q: Q42. The intensity of a plane monochromatic wave is 1380 Wm⁻² What is the amplitude of electric field in this wave? (Take speed of light 3 × 10⁸ ms⁻¹; wo = 8.85 x 10⁻¹²C²N⁻¹m⁻²)

The correct answer is C. The intensity (I) of a plane electromagnetic wave in free space is related to the amplitude of its electric field (E₀) by the formula: I = (1/2)ε₀cE₀², where ε₀ is the permittivity of free space and c is the speed of light. The 'wo' in the question refers to ε₀. Given: Intensity I = 1380 W/m². Speed of light c = 3 x 10⁸ m/s. Permittivity ε₀ = 8.85 x 10⁻¹² C²N⁻¹m⁻². We need to find E₀. Rearrangin

Q: What are the options for this CUET UG Physics question?

A. 102 N C⁻¹ · B. 10.2 N C⁻¹ · C. 1020 N C⁻¹ · D. 10200 N C⁻¹

Question

Q42. The intensity of a plane monochromatic wave is 1380 Wm⁻² What is the amplitude of electric field in this wave? (Take speed of light 3 × 10⁸ ms⁻¹; wo = 8.85 x 10⁻¹²C²N⁻¹m⁻²)

FirstInTest · Accepted Answer

The intensity (I) of a plane electromagnetic wave in free space is related to the amplitude of its electric field (E₀) by the formula: I = (1/2)ε₀cE₀², where ε₀ is the permittivity of free space and c is the speed of light. The 'wo' in the question refers to ε₀. 
Given: Intensity I = 1380 W/m². Speed of light c = 3 x 10⁸ m/s. Permittivity ε₀ = 8.85 x 10⁻¹² C²N⁻¹m⁻². 
We need to find E₀. Rearranging the formula: E₀² = (2I) / (ε₀c), so E₀ = √((2I) / (ε₀c)). 
Substitute the values: 
E₀ = √((2 * 1380 W/m²) / (8.85 x 10⁻¹² C²N⁻¹m⁻² * 3 x 10⁸ m/s)) 
E₀ = √((2760) / (26.55 x 10⁻⁴)) 
E₀ = √((2760 / 26.55) x 10⁴) 
E₀ = √(103.9548 x 10⁴) 
E₀ ≈ √(1039548) 
E₀ ≈ 1019.58 N/C. 
Rounding to appropriate significant figures, E₀ ≈ 1020 N/C. This matches option (c).

Answer

102 N C⁻¹

Answer

10.2 N C⁻¹

Answer

1020 N C⁻¹

Answer

10200 N C⁻¹

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