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#equilateral triangle

13 public questions tagged with this topic.

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

Three charges \( +4 \, \mu\text{C}, -4 \, \mu\text{C}, +2 \, \mu\text{C} \) 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 × 4 × 10⁻¹²/1) = 0.144 N (attractive). F₂ = 9 × 10⁹ × (4 × 2 × 10⁻¹²/1) = 0.072 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.144² + 0.072² + 0.010368) = 0.187 N . Substituting values gives 0.187 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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Coulomb's Law and Force Between Point Charges

Three charges \( +q, +q, -q \) are at the vertices of an equilateral triangle of side 3 m. What is the net force magnitu

**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. Force between +q and +q : F = (k q²/(3)²) = (k q²/9) (repulsive). Force between +q and -q : F = (k q²/9) (attractive). Angle between forces is 60°. Resultant: Fₙₑt = √(F² + F² + 2F² cos 60°) = √(3) F = (√(3) k q²/9) . Substituting values gives (√(3) k q²/9), 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 > 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 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

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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Electric Field and Electric Field Lines

Two charges \( +4 \, \mu\text{C} \) and \( -8 \, \mu\text{C} \) are 40 cm apart. What is the electric field magnitude at

**Electric field concept** visualizes influence of source charge. Uniform field exerts constant force F = qE, and flux Φ = E·A = E A cosθ links field to area orientation, maximum when field normal to surface. E₁ = (k |q₁|/r²) = 9 × 10⁹ × (4 × 10⁻⁶/(0.2)²) = 9 × 10⁵ N/C (away). E₂ = 9 × 10⁹ × (8 × 10⁻⁶/(0.2)²) = 1.8 × 10⁶ N/C (towards). Angle between E₁ and E₂ is 60°. Net E = √(E₁² + E₂² + 2 E₁ E₂ cos 60°) . E = √((9 × 10⁵)² + (1.8 × 10⁶)² + 2 × 9 × 10⁵ ×

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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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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Superposition Principle and Equilibrium of Charges

Three equal charges \( q = 3 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1 m. What is the e

**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. Distance from vertex to centroid: r = (l/√(3)) = (1/√(3)) m . E = (k q/r²) = 9 × 10⁹ × (3 × 10⁻⁶/(1/√(3))²) = 81 × 10⁴ N/C per charge. By symmetry (all +q ), vectors cancel, so Eₙₑt = 0 . Substituting values gives 0 N/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 > Superposition Principle and Equilibrium of Charges

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Gravitational Effects and Applications

Three masses of 4kg each are at the vertices of an equilateral triangle with side 3m. What is the net force on one mass?

Force between two masses: F = Gm1m2r2 = 6.67×10−114×432 = 1.185×10−10N. Two forces act at 60°: FR = F2+F2+2F2cos⁡60∘. FR = (1.185×10−10)2(1+1+1) = 1.185×10−103. FR≈2.05×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.2 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Energy of Satellites and Binding Energy

Two identical masses of 5kg each are placed at the vertices A and B of an equilateral triangle ABC with side length 2m.

Force due to A on C: FAC = GmAmCr2 = 6.67×10−115×1022 = 8.34×10−10N. Force due to B on C: FBC = 8.34×10−10N (equal magnitude, 60° apart). Resultant: FR = FAC2+FBC2+2FACFBCcos⁡60∘. FR = (8.34×10−10)2+(8.34×10−10)2+2(8.34×10−10)2(0.5). FR = 8.34×10−101+1+1 = 1.44×10−9N.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Escape Velocity and Bound Systems

Three masses of 12kg each form an equilateral triangle with side 9m. What is the net force on one mass? (G\=6.67×10−11N

Force between two masses: F = Gm2r2 = 6.67×10−1112×1292 = 1.185×10−10N. Two forces at 60°: FR = F2+F2+2F2cos⁡60∘. FR = (1.185×10−10)2(1+1+1) = 1.185×10−103. FR≈2.05×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Orbital Velocity and Time Period

Three masses of 8kg each form an equilateral triangle with side 6m. What is the net force on one mass? (G\=6.67×10−11N m

Force between two masses: F = Gm2r2 = 6.67×10−118×862 = 1.185×10−10N. Two forces at 60°: FR = F2+F2+2F2cos⁡60∘. FR = (1.185×10−10)2(1+1+1) = 1.185×10−103. FR≈2.05×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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