#electric potential

Energy Stored in Capacitor and Energy Density

A point charge \( Q = 5 \times 10^{-9} \, \text{C} \) is placed at the origin. What is the potential at a point 10 m awa

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (5 × 10⁻⁹/10) = 9 × 10⁹ × 0.5 × 10⁻⁹ = 4.5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

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Energy Stored in Capacitor and Energy Density

A charge of \( 7 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 100 \, \text{V} \). What

**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. Work done = Potential energy = q V . W = 7 × 10⁻⁶ × 100 = 7 × 10⁻⁴ J = 0.7 mJ . 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 0.7 mJ 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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Energy Stored in Capacitor and Energy Density

When a conductor is placed in an external electric field, why does the potential throughout its volume become constant i

**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. In electrostatic equilibrium, the electric field inside a conductor is zero because free charges rearrange to cancel any internal field. Since the electric field is the negative gradient of potential ( E = -(dV/dr) ), if E = 0 , the potential gradient must be zero, implying the potential V is constant

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Energy Stored in Capacitor and Energy Density

A point charge \( Q = 21 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 7 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (21 × 10⁻⁹/7) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 27 V follows, reflecting potential-capacitance relations.

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Energy Stored in Capacitor and Energy Density

A point charge \( Q = 6 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 2 m aw

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (6 × 10⁻⁹/2) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq

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Combination of Capacitors - Series and Parallel

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 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

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Combination of Capacitors - Series and Parallel

An electric dipole with moment \( p = 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, 3,

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (10⁻⁹/3²) = 9 × 10⁹ × (10⁻⁹/9) = 1 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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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

A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 3 m aw

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (9 × 10⁻⁹/3) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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

A point charge \( Q = 18 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (18 × 10⁻⁹/6) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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

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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