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#uniform electric field

12 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

A dipole with \( p = 4 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 5 \t

**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. U = -p E cos θ = -4 × 10⁻⁹ × 5 × 10⁴ × cos 30° . cos 30° = (√(3)/2) ≈ 0.866 , so U = -4 × 10⁻⁹ × 5 × 10⁴ × 0.866 = -1.732 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. Work done: W = p E (cos θ₀ - cos θ₁) = 9 × 10⁻⁹ × 5 × 10⁴ × (cos 90° - cos 180°) . W = 9 × 10⁻⁹ × 5 × 10⁴ × (0 - (-1)) = 9 × 10⁻⁹ × 5 × 10⁴ × 1 = 4.5 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Potential Energy of System of Charges

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻⁹ × 1 × 10⁵ × (cos 0° - cos 180°) . W = 8 × 10⁻⁹ × 1 × 10⁵ × (1 - (-1)) = 8 × 10⁻⁹ × 1 × 10⁵ × 2 = 1.6 × 10⁻³ J . Using V = kQ/r,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Potential Energy of System of Charges

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 2 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 2 × 10⁵ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 2 × 10⁵ × 0.707 = -8.48 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Potential Energy of System of Charges

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 5 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 5 × 10⁴ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 5 × 10⁴ × 0.707 = -2.121 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Continuous Charge Distribution

Why does the electric field remain constant inside a region where field lines are uniformly spaced and parallel?

**Continuous distribution** uses linear density λ = dq/dl (C/m), surface σ = dq/dA, volume ρ = dq/dV. Field of infinite line with uniform λ is E = 2kλ/r = λ/(2π ε₀ r) radially outward, derived via cylindrical Gaussian surface, showing 1/r dependence. Uniformly spaced, parallel field lines indicate a uniform field, where the field strength and direction do not vary. This occurs in regions like between parallel plates, where the field is constant due to consistent charge distribution. Substituting values gives Uniformity, which matches expected magnitude for this electrostatic configuration, con

Ref: NCERT > Physics Book > Electric Charges and Fields > Continuous Charge Distribution

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Electric Flux

A uniform field \( E = 5 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a circle of radius 2

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Area: A = π (0.2)² = 0.1256 m² . Flux: Φ = E A cos 0° = 5 × 10³ × 0.1256 = 628 N·m²/C . Substituting values gives 628 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux

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Electric Flux

A uniform electric field \( E = 2 \times 10^4 \, \text{N/C} \) is along the x-axis. What is the flux through a square of

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Area vector Δ S = 0.04 m² along x-axis. Flux: Φ = E · Δ S = (2 × 10⁴) × 0.04 = 800 N·m²/C . Substituting values gives 800 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux

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Electric Flux

A uniform electric field \( E = 3 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a square of

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Area vector Δ S = (0.6)² = 0.36 m² along y-axis. Flux: Φ = E · Δ S = 3 × 10³ × 0.36 = 1080 N·m²/C . Substituting values gives 1080 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux

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Electric Flux

A uniform field \( E = 9 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle of 15 cm

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Area: A = 0.15 × 0.2 = 0.03 m² . Flux: Φ = E A cos 0° = 9 × 10³ × 0.03 = 270 N·m²/C . Substituting values gives 270 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux

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