#electrostatics problem

Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor is charged to \( 50 \, \text{V} \) and then connected to an uncharged \( 2 \, \mu\tex

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial charge: Q = 4 × 10⁻⁶ × 50 = 2 × 10⁻⁴ C . Total capacitance: 4 + 2 = 6 μF . Final voltage: V = (Q/C) = (2 × 10⁻⁴/6 × 10⁻⁶) = 33.33 V . Final energy: U = (1/2) C V² = (1/2) × 6 × 10⁻⁶ × (33.33)² = 3.33 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Equipotential Surfaces and Relation Between Field and Potential

Two charges \( 14 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are at \( (3, 0, 0) \) and \( (-3, 0, 0) \, \text{cm} \)

**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. Distance to midpoint = 0.03 m. V = 9 × 10⁹ ( (14 × 10⁻⁶/0.03) + (-6 × 10⁻⁶/0.03) ) = 9 × 10⁹ × (8 × 10⁻⁶/0.03) . V = 9 × 10⁹ × (8 × 10⁻⁶/0.03) = 2.4 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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

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Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 4 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (4 × 10⁻⁸/0.04) = 9 × 10⁹ × 10⁻⁶ = 9000 V . 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 common potential V = Q_total/C_total, result 9000 V follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor has plates of area \( 0.08 \, \text{m}^2 \) and separation 0.4 mm in air. What is its capacit

**Capacitance depends on geometry** not charge, C = Q/V constant for given arrangement. Parallel plate C ∝ A/d, so larger area and smaller separation increase capacitance, principle used to increase storage by using large area foils with thin dielectric. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.08/0.4 × 10⁻³) = 1.77 × 10⁻⁹ F = 1770 pF . 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 common potential V = Q_total/C_total, result 1770 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Superposition Principle and Equilibrium of Charges

Three charges \( +3 \, \mu\text{C}, +3 \, \mu\text{C}, -6 \, \mu\text{C} \) 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 = F₂ = 9 × 10⁹ × (3 × 6 × 10⁻¹²/(2)²) = 0.0405 N (attractive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.0405² + 0.0405² + 0.0016425) = 0.0702 N . Substituting values gives 0.07 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charg

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

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