Electrostatic Potential and Capacitance

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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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 90 \, \text{pF} \) in air has a dielectric (\( K = 9 \)) inserted fully between p

**Potential difference increase** statement with constant charge is incorrect; V decreases when K>1 inserted. If slab inserted while maintaining constant charge, E reduces to E₀/K, V = E d reduces. If battery connected maintaining constant V, E stays V/d, D = ε E increases, Q increases K times. C' = K C = 9 × 90 = 810 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 810 pF follows, reflecting potential-capacitance relations.

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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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Energy Stored in Capacitor and Energy Density

Why does the electric field near the edge of a charged conducting plate differ from the field at the center of the plate

**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. Near the center of a large charged conducting plate, the field is approximately uniform ( E = (sigma/ε₀) ), as the plate behaves like an infinite sheet. At the edges, the field lines fringe outward due to the finite size of the plate, leading to a non-uniform field. The charge density sigma may also

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Energy Stored in Capacitor and Energy Density

Why does the energy density in an electric field depend on the square of the field strength?

**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. The energy density in an electric field is given by u = (1/2) ε₀ E² . This arises from the energy stored in a capacitor ( U = (1/2) C V² ), scaled over the volume. For a parallel plate capacitor, E = (V/d) , C = (ε₀ A/d) , so U = (1/2) (ε₀ A/d) (E d)² = (1/2) ε₀ E² (A d) , where A d is

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Sharing of Charges, Common Potential and Applications

A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \)

**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 energy: U_i = (1/2) × 2 × 10⁻⁶ × (400)² = 0.16 J . Charge: Q = 2 × 10⁻⁶ × 400 = 8 × 10⁻⁴ C . Total C = 2 + 6 = 8 μF , V = (8 × 10⁻⁴/8 × 10⁻⁶) = 100 V . Final energy: U_f = (1/2) × 8 × 10⁻⁶ × (100)² = 0.04 J . Loss: U_i - U_f =

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Energy Stored in Capacitor and Energy Density

An electric dipole with moment \( p = 8 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**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. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (8 × 10⁻⁹/4²) = -9 × 10⁹ × (8 × 10⁻⁹/16) = -4.5 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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Sharing of Charges, Common Potential and Applications

Three capacitors \( 2 \, \text{pF} \), \( 4 \, \text{pF} \), and \( 8 \, \text{pF} \) are in parallel. What is the total

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. C = 2 + 4 + 8 = 14 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 14 pF follows, reflecting potential-capacitance relations.

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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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