#charge interaction

Equipotential Surfaces and Relation Between Field and Potential

Two charges \( 18 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are placed 18 cm apart. What is the potential energy of

**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (18 × 10⁻⁶ × (-6 × 10⁻⁶)/0.18) . U = 9 × 10⁹ × (-108 × 10⁻¹²/0.18) = -5.4 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 V² and

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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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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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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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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