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Potential Energy of System of Charges

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Potential Energy of System of Charges

Two charges \( 20 \, \mu\text{C} \) and \( -10 \, \mu\text{C} \) are at \( (1, 0, 0) \) and \( (-1, 0, 0) \, \text{cm} \

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. Distance to midpoint = 0.01 m. V = 9 × 10⁹ ( (20 × 10⁻⁶/0.01) + (-10 × 10⁻⁶/0.01) ) = 9 × 10⁹ × (10 × 10⁻⁶/0.01) . V = 9 × 10⁹ × (10 × 10⁻⁶/0.01) = 9 × 10⁶ 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

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Potential Energy of System of Charges

Two charges \( 18 \, \mu\text{C} \) and \( -9 \, \mu\text{C} \) are at \( (2, 0, 0) \) and \( (-2, 0, 0) \, \text{cm} \)

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Distance to midpoint = 0.02 m. V = 9 × 10⁹ ( (18 × 10⁻⁶/0.02) + (-9 × 10⁻⁶/0.02) ) = 9 × 10⁹ × (9 × 10⁻⁶/0.02) . V = 9 × 10⁹ × (9 × 10⁻⁶/0.02) = 4.05 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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Potential Energy of System of Charges

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻⁹ × 1 × 10⁵ × (cos 0° - cos 180°) . W = 8 × 10⁻⁹ × 1 × 10⁵ × (1 - (-1)) = 8 × 10⁻⁹ × 1 × 10⁵ × 2 = 1.6 × 10⁻³ J . Using V = kQ/r,

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Potential Energy of System of Charges

Two charges \( 12 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (5, 0, 0) \) and \( (-5, 0, 0) \, \text{cm} \)

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. Distance to midpoint = 0.05 m. V = 9 × 10⁹ ( (12 × 10⁻⁶/0.05) + (-3 × 10⁻⁶/0.05) ) = 9 × 10⁹ × (9 × 10⁻⁶/0.05) . V = 9 × 10⁹ × (9 × 10⁻⁶/0.05) = 1.62 × 10⁶ 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

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Potential Energy of System of Charges

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 2 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 2 × 10⁵ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 2 × 10⁵ × 0.707 = -8.48 × 10⁻⁴ 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

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Potential Energy of System of Charges

A conductor has a surface charge density of \( 1 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outs

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. E = (sigma/ε₀) = (1 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 1.13 × 10⁵ N/C . 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 1.13 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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Potential Energy of System of Charges

In a system where a charged conductor is placed inside a larger uncharged hollow conductor, why does the inner conductor

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. When a charged conductor (charge Q ) is placed inside a larger uncharged hollow conductor, the electric field from the inner charge induces -Q on the inner surface of the hollow conductor to ensure the field inside the conductor's material is zero. Since the hollow conductor is initially uncharged, its total charge must remain zero, so +Q appears on its outer surface to balance the induced -Q

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Potential Energy of System of Charges

Two charges \( 15 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 12 cm apart. What is the potential energy of

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (15 × 10⁻⁶ × (-5 × 10⁻⁶)/0.12) . U = 9 × 10⁹ × (-75 × 10⁻¹²/0.12) = -5.625 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 -5.625 J follows, reflecting potential-capacitance relations.

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Potential Energy of System of Charges

An electric dipole with moment \( p = 5 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (5 × 10⁻⁹/3²) = 9 × 10⁹ × (5 × 10⁻⁹/9) = 5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U =

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