#electric field

EM Wave Characteristics - Transverse Nature and E/B Ratio

An electromagnetic wave in vacuum has an electric field amplitude of \( 30 \, \text{V/m} \). What is the magnetic field

**Relationship E and B** in EM wave E₀ = c B₀, B₀ = E₀/c, for vacuum. Fields sustain each other via Maxwell's equations ∇×E = -∂B/∂t, ∇×B = μ₀ ε₀ ∂E/∂t, time-varying E produces B and vice versa, self-sustaining propagation without medium, speed c. Using B₀ = (E₀/c) , we have B₀ = (30/3 × 10⁸) = 1 × 10⁻⁷ T . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1 × 10⁻⁷ T, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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Displacement Current and Ampere-Maxwell Law

A parallel plate capacitor with plate area \( A = 0.01 \, \text{m}^2 \) and separation \( d = 2 \, \text{mm} \) is being

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Displacement current i_d = ε₀ (d Φ_E/dt) . Since Φ_E = (Q/ε₀) , we have (d Φ_E/dt) = (1/ε₀) (dQ/dt) . But in a capacitor, i_d = (dQ/dt) . Thus, i_d = 0.5 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 0.5 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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EM Wave Characteristics - Transverse Nature and E/B Ratio

In electromagnetic theory, what explains the constant ratio of electric to magnetic field amplitudes in a wave?

**Transverse nature** means E and B perpendicular to direction, e.g., wave propagating along z, E along x, B along y, Poynting vector S = E×B/μ₀ along z, energy flow direction. E and B in phase, maxima together, ratio fixed c. The ratio (E₀/B₀) = c , where c is the speed of light, arises from Maxwell’s equations, ensuring energy conservation and wave propagation consistency in vacuum. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Maxwell’s equations, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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EM Wave Characteristics - Transverse Nature and E/B Ratio

An electromagnetic wave has a magnetic field amplitude of \( B_0 = 8 \times 10^{-8} \, \text{T} \). What is the electric

**Relationship E and B** in EM wave E₀ = c B₀, B₀ = E₀/c, for vacuum. Fields sustain each other via Maxwell's equations ∇×E = -∂B/∂t, ∇×B = μ₀ ε₀ ∂E/∂t, time-varying E produces B and vice versa, self-sustaining propagation without medium, speed c. Using E₀ = B₀ c , we have E₀ = (8 × 10⁻⁸) × (3 × 10⁸) = 24 V/m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 24 V/m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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EM Wave Characteristics - Transverse Nature and E/B Ratio

In electromagnetic wave propagation, what ensures that the electric and magnetic fields sustain each other?

**Transverse nature** means E and B perpendicular to direction, e.g., wave propagating along z, E along x, B along y, Poynting vector S = E×B/μ₀ along z, energy flow direction. E and B in phase, maxima together, ratio fixed c. The mutual induction described by Maxwell’s equations (Faraday’s law and Ampere-Maxwell law) ensures that a changing electric field induces a magnetic field, and vice versa, sustaining wave propagation. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Mutual induction via Maxwell’s equations, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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EM Wave Characteristics - Transverse Nature and E/B Ratio

An electromagnetic wave in vacuum has an electric field amplitude of \( 45 \, \text{V/m} \). What is the magnetic field

**Transverse nature** means E and B perpendicular to direction, e.g., wave propagating along z, E along x, B along y, Poynting vector S = E×B/μ₀ along z, energy flow direction. E and B in phase, maxima together, ratio fixed c. Using B₀ = (E₀/c) , we have B₀ = (45/3 × 10⁸) = 1.5 × 10⁻⁷ T . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.5 × 10⁻⁷ T, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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EM Wave Characteristics - Transverse Nature and E/B Ratio

What is the relationship between the amplitudes of electric and magnetic fields in an electromagnetic wave in vacuum? (G

**Relationship E and B** in EM wave E₀ = c B₀, B₀ = E₀/c, for vacuum. Fields sustain each other via Maxwell's equations ∇×E = -∂B/∂t, ∇×B = μ₀ ε₀ ∂E/∂t, time-varying E produces B and vice versa, self-sustaining propagation without medium, speed c. From Maxwell's equations, the amplitudes are related as B₀ = (E₀/c) , where c is the speed of light in vacuum. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields B₀ = (E₀/c), illustrating EM wave transverse nature and Maxwell's displacement current concept.

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Energy Stored in Capacitor and Energy Density

Why does the electric field near the edge of a charged conducting plate differ from the field at the center of the plate

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Near the center of a large charged conducting plate, the field is approximately uniform ( E = (sigma/ε₀) ), as the plate behaves like an infinite sheet. At the edges, the field lines fringe outward due to the finite size of the plate, leading to a non-uniform field. The charge density sigma may also

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Energy Stored in Capacitor and Energy Density

Why does the energy density in an electric field depend on the square of the field strength?

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². The energy density in an electric field is given by u = (1/2) ε₀ E² . This arises from the energy stored in a capacitor ( U = (1/2) C V² ), scaled over the volume. For a parallel plate capacitor, E = (V/d) , C = (ε₀ A/d) , so U = (1/2) (ε₀ A/d) (E d)² = (1/2) ε₀ E² (A d) , where A d is

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Energy Stored in Capacitor and Energy Density

Why does the electric field due to an infinite charged sheet remain constant with distance from the sheet?

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Using Gauss’s law for an infinite charged sheet with surface charge density sigma , a Gaussian surface (e.g., a cylinder perpendicular to the sheet) shows that the electric field E is perpendicular to the sheet and constant. The flux through the cylinder's end caps is E · 2A , and the enclosed charge is

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Combination of Capacitors - Series and Parallel

In an electrostatic field, if a positive charge is moved along an equipotential surface, what can be said about the work

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. An equipotential surface has a constant potential at all points. The work done by the electric field when a charge moves along such a surface is zero because the potential difference between any two points on the surface is zero ( W = q Δ V , and Δ V = 0 ). Additionally,

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Combination of Capacitors - Series and Parallel

A spherical conductor of radius 20 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the electric field at

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. For r = 0.5 m > R = 0.2 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (8 × 10⁻⁸/(0.5)²) = 9 × 10⁹ × (8 × 10⁻⁸/0.25) = 2.88 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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