#charge calculation

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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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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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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Electric Dipole - Moment, Field and Torque

A conducting sphere of radius 21 cm has an electric field of \( 8 \times 10^3 \, \text{N/C} \) at 42 cm from its center.

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. E = (k q/r²) . 8 × 10³ = 9 × 10⁹ × (q/(0.42)²) . q = (8 × 10³ × 0.1764/9 × 10⁹) = 1.57 × 10⁻⁷ C . Substituting values gives 1.57 × 10⁻⁷ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Superposition Principle and Equilibrium of Charges

A conducting sphere of radius 25 cm has an electric field of \( 5 \times 10^3 \, \text{N/C} \) at 50 cm from its center.

**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. E = (k q/r²) . 5 × 10³ = 9 × 10⁹ × (q/(0.5)²) . q = (5 × 10³ × 0.25/9 × 10⁹) = 1.389 × 10⁻⁷ C . Substituting values gives 1.39 × 10⁻⁷ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

A conducting sphere of radius 12 cm has an electric field of \( 6 \times 10^3 \, \text{N/C} \) at 24 cm from its center.

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. E = (k q/r²) . 6 × 10³ = 9 × 10⁹ × (q/(0.24)²) . q = (6 × 10³ × 0.0576/9 × 10⁹) = 3.84 × 10⁻⁸ C . Substituting values gives 3.84 × 10⁻⁸ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Gauss's Theorem and Total Flux

A spherical shell has a net flux of \( 1.13 \times 10^5 \, \text{Nm}^2/\text{C} \) through it. What is the charge enclos

**Gauss's theorem** states total flux through closed surface equals enclosed charge divided by free-space permittivity, Φ_total = q_enc/ε₀, ε₀ = 8.854×10⁻¹² C²/(N·m²). Result independent of shape or size, depends only on net enclosed charge, enabling charge determination from flux. Φ = (q/ε₀) . q = Φ ε₀ = 1.13 × 10⁵ × 8.854 × 10⁻¹² = 1.0 × 10⁻⁶ C = 1 μC . Substituting values gives 1.0 μC, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Gauss's Theorem and Total Flux

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 20 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 40 cm from its center.

**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². For a conductor, E = (k q/r²) outside. 3 × 10³ = 9 × 10⁹ × (q/(0.4)²) . q = (3 × 10³ × 0.16/9 × 10⁹) = 5.33 × 10⁻⁸ C . Substituting values gives 5.33 × 10⁻⁸ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 15 cm has an electric field of \( 4 \times 10^3 \, \text{N/C} \) at 30 cm from its center.

**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². E = (k q/r²) . 4 × 10³ = 9 × 10⁹ × (q/(0.3)²) . q = (4 × 10³ × 0.09/9 × 10⁹) = 4 × 10⁻⁸ C . Substituting values gives 4.0 × 10⁻⁸ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 19 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 38 cm from its center.

**Inverse-square law** for charges states F ∝ 1/r² while increasing with charge product. Using k = 9×10⁹ N·m²/C², force at distance r follows F = k q₁q₂/r², forming basis for pairwise force calculation. E = (k q/r²) . 3 × 10³ = 9 × 10⁹ × (q/(0.38)²) . q = (3 × 10³ × 0.1444/9 × 10⁹) = 4.81 × 10⁻⁸ C . Substituting values gives 4.81 × 10⁻⁸ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Continuous Charge Distribution

A conducting sphere of radius 29 cm has a surface charge density of \( 50 \, \mu\text{C/m}^2 \). What is the total charg

**Continuous distribution** uses linear density λ = dq/dl (C/m), surface σ = dq/dA, volume ρ = dq/dV. Field of infinite line with uniform λ is E = 2kλ/r = λ/(2π ε₀ r) radially outward, derived via cylindrical Gaussian surface, showing 1/r dependence. Surface area: A = 4 π r² = 4 π (0.29)² = 0.335 π m² . Charge: q = sigma A = 50 × 10⁻⁶ × 0.335 π = 5.27 × 10⁻⁵ C . Substituting values gives 5.27 × 10⁻⁵ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Continuous Charge Distribution

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Electric Charge, Quantization and Conservation

A conducting sphere of radius 27 cm has an electric field of \( 6 \times 10^3 \, \text{N/C} \) at 54 cm from its center.

**Fundamental property of charge** includes additivity and quantization, meaning net charge equals algebraic sum of constituents and each is multiple of e. When rod loses charge, electron removal is inferred, and n = q/e gives transferred count. E = (k q/r²) . 6 × 10³ = 9 × 10⁹ × (q/(0.54)²) . q = (6 × 10³ × 0.2916/9 × 10⁹) = 1.944 × 10⁻⁷ C . Substituting values gives 1.944 × 10⁻⁷ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Charge, Quantization and Conservation

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