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#electrostatic force

13 public questions tagged with this topic.

Atomic Models - Rutherford, Thomson and Bohr

In Rutherford’s model, what is the primary force keeping electrons in orbit around the nucleus?

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provides the centripetal force for orbiting. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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Superposition Principle and Equilibrium of Charges

Two point charges \( 5 \times 10^{-7} \, \text{C} \) and \( -7 \times 10^{-7} \, \text{C} \) are 150 cm apart in vacuum.

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 5 × 10⁻⁷ C , q₂ = -7 × 10⁻⁷ C , r = 1.5 m . |q₁ q₂| = 5 × 7 × 10⁻¹⁴ = 35 × 10⁻¹⁴ C² . r² = (1.5)² = 2.25 m² . F = 9 × 10⁹ × (35

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +2 \, \mu\text{C} \) each are at the vertices of an equilateral triangle of side 2 m. What is the force

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. Force between two charges: F = 9 × 10⁹ × ((2 × 10⁻⁶)²/(2)²) = 9 × 10⁻³ N . Two forces at 60°. Net force: Fₙₑt = √(F² + F² + 2 F² cos 60°) = √(3) × 9 × 10⁻³ = 1.56 × 10⁻² N . Substituting values gives 1.56 × 10⁻² N, 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 > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

What property of electric charges explains why two objects with identical charges repel each other?

**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. The polarity of charge (positive or negative) determines interaction: like charges repel due to the repulsive force described by Coulomb’s law, where the force direction depends on the sign of the charges, causing repulsion for identical signs. Substituting values gives Polarity, which matches expected magnitude for this electrosta

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Coulomb's Law and Force Between Point Charges

Two point charges \( 7 \times 10^{-7} \, \text{C} \) and \( -2 \times 10^{-7} \, \text{C} \) are 70 cm apart in vacuum.

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 7 × 10⁻⁷ C , q₂ = -2 × 10⁻⁷ C , r = 0.7 m . |q₁ q₂| = 7 × 2 × 10⁻¹⁴ = 14 × 10⁻¹⁴ C² . r² = (0.7)² = 0.49 m² . F = 9 × 10⁹ × (14 × 10⁻¹⁴/0.49) = 9 × 10⁹ × 2.857 × 10⁻¹³ = 0.00257

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Coulomb's Law and Force Between Point Charges

Two point charges \( 6 \times 10^{-7} \, \text{C} \) and \( 10 \times 10^{-7} \, \text{C} \) are 120 cm apart in vacuum.

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 6 × 10⁻⁷ C , q₂ = 10 × 10⁻⁷ C , r = 1.2 m . |q₁ q₂| = 6 × 10 × 10⁻¹⁴ = 60 × 10⁻¹⁴ C² . r² = (1.2)² = 1.44 m² . F = 9 × 10⁹ × (60 × 10⁻¹⁴/1.44) = 9 × 10⁹ × 4.167 × 10⁻¹³ = 0.00375

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +5 \, \mu\text{C}, -3 \, \mu\text{C}, +4 \, \mu\text{C} \) are at the vertices of an equilateral triang

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. F₁ = 9 × 10⁹ × (5 × 3 × 10⁻¹²/(1.5)²) = 0.06 N (attractive). F₂ = 9 × 10⁹ × (5 × 4 × 10⁻¹²/(1.5)²) = 0.08 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.06² + 0.08² + 0.0048) = 0.108 N . Substituting values gives 0.108 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quan

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +5 \, \mu\text{C}, -5 \, \mu\text{C}, +2 \, \mu\text{C} \) are at the vertices of an equilateral triang

**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. F₁ = 9 × 10⁹ × (5 × 5 × 10⁻¹²/(1.5)²) = 0.1 N (attractive). F₂ = 9 × 10⁹ × (5 × 2 × 10⁻¹²/(1.5)²) = 0.04 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.1² + 0.04² + 0.004) = 0.129 N . Substituting values gives 0.129 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb'

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Coulomb's Law and Force Between Point Charges

Two point charges \( 8 \times 10^{-7} \, \text{C} \) and \( -4 \times 10^{-7} \, \text{C} \) are 60 cm apart in vacuum.

**Inverse-square law** for charges states F ∝ 1/r² while increasing with charge product. Using k = 9×10⁹ N·m²/C², force at distance r follows F = k q₁q₂/r², forming basis for pairwise force calculation. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 8 × 10⁻⁷ C , q₂ = -4 × 10⁻⁷ C , r = 0.6 m . |q₁ q₂| = 8 × 4 × 10⁻¹⁴ = 32 × 10⁻¹⁴ C² . r² = (0.6)² = 0.36 m² . F = 9 × 10⁹ × (32 × 10⁻¹⁴/0.36) = 9 × 10⁹ × 8.89 ×

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +4 \, \mu\text{C}, -2 \, \mu\text{C}, +5 \, \mu\text{C} \) are at the vertices of an equilateral triang

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. F₁ = 9 × 10⁹ × (4 × 2 × 10⁻¹²/(1.2)²) = 0.05 N (attractive). F₂ = 9 × 10⁹ × (4 × 5 × 10⁻¹²/(1.2)²) = 0.125 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.05² + 0.125² + 0.00625) = 0.144 N . Substituting values gives 0.144 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge q

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +7 \, \mu\text{C}, -5 \, \mu\text{C}, +3 \, \mu\text{C} \) are at the vertices of an equilateral triang

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. F₁ = 9 × 10⁹ × (7 × 5 × 10⁻¹²/(2)²) = 0.07875 N (attractive). F₂ = 9 × 10⁹ × (7 × 3 × 10⁻¹²/(2)²) = 0.04725 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.07875² + 0.04725² + 0.00372) = 0.098 N . Substituting values gives 0.098 N, which matches expected magnitude

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

Three charges \( +6 \, \mu\text{C}, -3 \, \mu\text{C}, +3 \, \mu\text{C} \) are at the vertices of an equilateral triang

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. F₁ = 9 × 10⁹ × (6 × 3 × 10⁻¹²/1²) = 0.162 N (attractive). F₂ = 9 × 10⁹ × (6 × 3 × 10⁻¹²/1²) = 0.162 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.162² + 0.162² + 0.026244) = 0.28 N . Substituting values gives 0.28 N, which matches expected magnitude

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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