Capacitance of Parallel Plate and Spherical Capacitor

Capacitance of Parallel Plate and Spherical Capacitor

A capacitor with \( C = 10 \, \text{pF} \) in air has a dielectric (\( K = 4 \), thickness \( d/2 \)) inserted. What is

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. Potential difference: V = E₀ ( (d/2) ) + (E₀/K) ( (d/2) ) = E₀ d ( (1/2) + (1/2 × 4) ) = E₀ d ( (1/2) + (1/8) ) = (5/8) E₀ d . C = (Q/V) = (Q/(5/8) V₀) = (8/5) × (Q/V₀) = (8/5) × 10 = 16 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d,

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 80 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/5 \)) in

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. Potential difference: V = E₀ ( (4d/5) ) + (E₀/K) ( (d/5) ) = E₀ d ( (4/5) + (1/5 × 5) ) . V = E₀ d ( (4/5) + (1/25) ) = E₀ d × (21/25) . C = (Q/V) = (Q/(21/25) V₀) = (25/21) × 80 ≈ 95.24 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor has plates of area \( 0.05 \, \text{m}^2 \) separated by 1 mm in air. What is its capacitance

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.05/10⁻³) = 4.425 × 10⁻¹⁰ F = 442.5 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 442.5 pF follows, reflecting potential-capacitance relations.

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 150 \, \text{pF} \) has a dielectric (\( K = 6 \), thickness \( d/5 \)) i

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. Potential difference: V = E₀ ( (4d/5) ) + (E₀/K) ( (d/5) ) = E₀ d ( (4/5) + (1/5 × 6) ) . V = E₀ d ( (4/5) + (1/30) ) = E₀ d × (25/30) = E₀ d × (5/6) . C = (Q/V) = (Q/(5/6) V₀) = (6/5) × 150 = 180 pF . Using V = kQ/r, U = k q₁q₂/r, E =

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 50 \, \text{pF} \) has a dielectric (\( K = 2 \), thickness \( d/3 \)) in

**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 difference: V = E₀ ( (2d/3) ) + (E₀/K) ( (d/3) ) = E₀ d ( (2/3) + (1/3 × 2) ) = E₀ d ( (2/3) + (1/6) ) = E₀ d × (5/6) . C = (Q/V) = (Q/(5/6) V₀) = (6/5) × 50 = 60 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

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Capacitance of Parallel Plate and Spherical Capacitor

Why does the potential difference between the plates of a parallel plate capacitor remain constant when a dielectric sla

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. When a capacitor is connected to a battery, the potential difference V across its plates is fixed by the battery. Inserting a dielectric slab (with K > 1 ) increases the capacitance ( C' = K C ), but the battery maintains V . To keep V constant ( Q = C V ), the charge Q on the plates increases ( Q' = C' V = K

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Capacitance of Parallel Plate and Spherical Capacitor

In a charged parallel plate capacitor, why does the electric field remain uniform between the plates even when a dielect

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. The electric field between the plates of a parallel plate capacitor is ideally uniform ( E = (sigma/ε₀) ) in air. When a dielectric slab is partially inserted, the field in the air region remains E₀ = (sigma/ε₀) , and in the dielectric region, it reduces to E = (E₀/K) . However, within each region (air or dielectric), the field remains uniform because the plates are large and

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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.

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Capacitance of Parallel Plate and Spherical Capacitor

Why does the electric field between two large parallel plates with opposite charges remain uniform even if one plate has

**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. For large parallel plates with charges +sigma and -sigma , the field between them is E = (sigma/ε₀) , uniform due to symmetry. A dielectric coating on one plate polarizes, inducing bound charges, but does not alter the free charge on the plates. The net field between the plates remains determined by the free charge density sigma , and the uniformity is preserved by the geometry,

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Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 10 cm has a charge of \( 2 \times 10^{-7} \, \text{C} \). What is the electric field at

**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. For r = 0.15 m > R = 0.1 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (2 × 10⁻⁷/(0.15)²) = 9 × 10⁹ × (2 × 10⁻⁷/0.0225) = 8 × 10⁴ 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

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with \( C = 50 \, \text{pF} \) in air has a dielectric (\( K = 4 \)) inserted fully between p

**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. C' = K C = 4 × 50 = 200 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 200 pF follows, reflecting potential-capacitance relations.

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Capacitance of Parallel Plate and Spherical Capacitor

Why does the electric field inside the dielectric of a parallel plate capacitor decrease when the dielectric is inserted

**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. When a dielectric ( K > 1 ) is inserted into a disconnected capacitor, the charge Q remains constant. The dielectric polarizes, creating an induced field opposing the applied field. The effective field inside the dielectric becomes E = (E₀/K) , where E₀ = (sigma/ε₀) is the field without the dielectric ( sigma = Q/A ). Since K > 1 , the field decreases

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