#point charges

Energy Stored in Capacitor and Energy Density

Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (-6, 0, 0) \) and \( (6, 0, 0) \, \text{cm} \)

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Mutual energy: U₁₂ = 9 × 10⁹ × (8 × 10⁻⁶ × (-4 × 10⁻⁶)/0.12) = -2.4 J . External potential: V(r) = (10⁵/r) , at r = 0.06 m , V = (10⁵/0.06) = 1.67 × 10⁶ V . External energy: 8 × 10⁻⁶ × 1.67 × 10⁶ + (-4 × 10⁻⁶) ×

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Energy Stored in Capacitor and Energy Density

Two charges \( 32 \, \mu\text{C} \) and \( -16 \, \mu\text{C} \) are placed 32 cm apart. What is the potential energy of

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (32 × 10⁻⁶ × (-16 × 10⁻⁶)/0.32) . U = 9 × 10⁹ × (-512 × 10⁻¹²/0.32) = -14.4 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 common potential V = Q_total/C_total, result -14.4 J follows, reflecting potential-capacitance relations.

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Energy Stored in Capacitor and Energy Density

Two charges \( 16 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (4, 0, 0) \) and \( (-4, 0, 0) \, \text{cm} \)

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. Distance to midpoint = 0.04 m. V = 9 × 10⁹ ( (16 × 10⁻⁶/0.04) + (-4 × 10⁻⁶/0.04) ) = 9 × 10⁹ × (12 × 10⁻⁶/0.04) . V = 9 × 10⁹ × (12 × 10⁻⁶/0.04) = 2.7 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r,

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Conductors, Electrostatic Shielding and Dielectrics

Three charges \( +11 \, \mu\text{C} \), \( -8 \, \mu\text{C} \), and \( +6 \, \mu\text{C} \) are at \( (0, 0, 0) \), \(

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Distances: r₁ = √(11² + 11²) = 11√(2) m , r₂ = 11 m , r₃ = 11 m . V = 9 × 10⁹ ( (11 × 10⁻⁶/11√(2)) + (-8 × 10⁻⁶/11) + (6 × 10⁻⁶/11) ) . V = 9 × 10⁹ ( (11 × 10⁻⁶/15.556) - (8 × 10⁻⁶/11) + (6 × 10⁻⁶/11) ) . V = 9 × 10⁹ (

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Conductors, Electrostatic Shielding and Dielectrics

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (10, 0, 0) \) and \( (-10, 0, 0) \, \text{cm} \

**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. Distance to midpoint = 0.1 m. V = 9 × 10⁹ ( (6 × 10⁻⁶/0.1) + (-3 × 10⁻⁶/0.1) ) = 9 × 10⁹ × (3 × 10⁻⁶/0.1) = 2.7 × 10⁵ 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

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Conductors, Electrostatic Shielding and Dielectrics

Three charges \( +10 \, \mu\text{C} \), \( -7 \, \mu\text{C} \), and \( +5 \, \mu\text{C} \) are at \( (0, 0, 0) \), \(

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Distances: r₁ = √(10² + 10²) = 10√(2) m , r₂ = 10 m , r₃ = 10 m . V = 9 × 10⁹ ( (10 × 10⁻⁶/10√(2)) + (-7 × 10⁻⁶/10) + (5 × 10⁻⁶/10) ) . V = 9 × 10⁹ ( (10 × 10⁻⁶/14.142) - (7 × 10⁻⁶/10) + (5 × 10⁻⁶/10) ) . V = 9 × 10⁹ (

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Conductors, Electrostatic Shielding and Dielectrics

Two charges \( 5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 10 cm apart. What is the potential energy of t

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (5 × 10⁻⁶ × (-5 × 10⁻⁶)/0.1) = 9 × 10⁹ × (-25 × 10⁻¹²/0.1) = -2.25 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 common potential V = Q_total/C_total, result -2.25 J follows, reflecting potential-capacitance relations.

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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