#spherical conductor

Combination of Capacitors - Series and Parallel

A spherical conductor of radius 20 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the electric field at

**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. For r = 0.5 m > R = 0.2 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (8 × 10⁻⁸/(0.5)²) = 9 × 10⁹ × (8 × 10⁻⁸/0.25) = 2.88 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Conductors, Electrostatic Shielding and Dielectrics

A spherical conductor of radius 4 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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 at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (8 × 10⁻⁸/0.04) = 9 × 10⁹ × 2 × 10⁻⁶ = 18000 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Conductors, Electrostatic Shielding and Dielectrics

A spherical conductor of radius 3 cm has a charge of \( 3 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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 at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (3 × 10⁻⁸/0.03) = 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.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Equipotential Surfaces and Relation Between Field and Potential

A spherical conductor of radius 8 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the electric field at 1

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. For r = 0.12 m > R = 0.08 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (4 × 10⁻⁸/(0.12)²) = 9 × 10⁹ × (4 × 10⁻⁸/0.0144) = 2.5 × 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

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Equipotential Surfaces and Relation Between Field and Potential

A spherical conductor of radius 15 cm has a charge of \( 9 \times 10^{-8} \, \text{C} \). What is the electric field at

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. For r = 0.2 m > R = 0.15 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (9 × 10⁻⁸/(0.2)²) = 9 × 10⁹ × (9 × 10⁻⁸/0.04) = 2.025 × 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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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.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Capacitance of Parallel Plate and Spherical Capacitor

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

**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. For r = 0.25 m > R = 0.1 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (5 × 10⁻⁸/(0.25)²) = 9 × 10⁹ × (5 × 10⁻⁸/0.0625) = 7.2 × 10³ 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). 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 × (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 V² and common potential V = Q_total/C_total, result 9000 V follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Electric Potential and Potential Difference

A spherical conductor of radius 25 cm has a charge of \( 10 \times 10^{-8} \, \text{C} \). What is the electric field at

**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. For r = 0.6 m > R = 0.25 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (10 × 10⁻⁸/(0.6)²) = 9 × 10⁹ × (10 × 10⁻⁸/0.36) = 2.5 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference

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Electric Potential and Potential Difference

A spherical conductor of radius 30 cm has a charge of \( 12 \times 10^{-8} \, \text{C} \). What is the electric field at

**Electric potential** V = k Q/r, k = 1/(4π ε₀)=9×10⁹ N·m²/C², Q charge (C), r distance (m), scalar, potential at surface of spherical conductor radius R, V = k Q/R. Potential difference ΔV = V_B - V_A = -∫ E·dl, work per unit charge to move charge without acceleration. For r = 0.7 m > R = 0.3 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (12 × 10⁻⁸/(0.7)²) = 9 × 10⁹ × (12 × 10⁻⁸/0.49) ≈ 2.204 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference

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Electric Potential and Potential Difference

A spherical conductor of radius 6 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Spherical conductor** behaves as point charge outside, potential V = k Q/R at surface, field E = k Q/r² for r>R, zero inside for rR, so E = k Q/r² =9×10⁹×6×10⁻⁸/0.16=3375 N/C. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (6 × 10⁻⁸/0.06) = 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference

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