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#charge distribution

26 public questions tagged with this topic.

Sharing of Charges, Common Potential and Applications

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decre

**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. When a charged capacitor ( C , charge Q , voltage V ) is connected in parallel with an uncharged capacitor ( C ), the total capacitance becomes 2C , and the charge redistributes to a final voltage V' = Q/(2C) = V/2 . Initial energy is U_i = (Q²/2C) , while final energy

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Equipotential Surfaces and Relation Between Field and Potential

When a charged conductor is placed in contact with an uncharged conductor of smaller size, why does the smaller conducto

**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. When two conductors reach equilibrium, they share charge and attain the same potential. Potential on a conductor's surface is V = (Q/4 π ε₀ R) for a sphere (or similar for other shapes). For equal V , (Q₁/R₁) = (Q₂/R₂) , so Q ∝ R . Surface charge density sigma = (Q/4 π R²) , so sigma ∝ (Q/R²)

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

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Capacitor with Dielectric and Effect of Inserting Slab

In a system where a positively charged sphere is enclosed by an uncharged conducting shell, why does the outer surface o

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. The positively charged sphere induces a negative charge on the inner surface of the shell and a positive charge on its outer surface to maintain zero field inside the conductor. When a negative charge is brought near the shell, it repels negative charges to the far side of the shell and attracts positive

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a conductor vanish in electrostatic equilibrium, even if the conductor is irregularly

**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. In electrostatic equilibrium, free charges in a conductor redistribute to cancel any internal electric field. If there were a field inside, it would exert a force on the free charges, causing them to move until the field becomes zero everywhere inside. This principle holds regardless of shape because conductors allow charge mobility, and equilibrium requires E = 0 inside, with all excess charge residing

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

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a uniformly charged spherical shell remain zero even if the shell is placed in an ext

**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. For a uniformly charged spherical shell in electrostatic equilibrium, the electric field inside is zero regardless of external fields due to electrostatic shielding. Applying Gauss’s law inside the shell (no charge enclosed within the cavity), E = 0 . The external field induces charge redistribution on the shell's outer surface, but this does not affect the interior, as the induced charges ensure the internal field

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

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Electric Potential and Potential Difference

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

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (4 × 10⁻⁸/0.08) = 9 × 10⁹ × 5 × 10⁻⁷ = 4500 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

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

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Gauss's Theorem and Total Flux

A thin spherical shell of radius 7 cm has \( q = 3 \, \mu\text{C} \). What is the electric field at 4 cm from the center

**Closed-surface flux** depends solely on net charge inside, not external charges. This principle allows flux calculation without detailed field integration and forms cornerstone for symmetric charge distributions. Inside shell ( r < R ): E = 0 (Gauss’s law). Substituting values gives 0 N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.

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

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Gauss's Theorem and Total Flux

What explains why the electric field inside a charged non-conducting sphere is non-zero and varies with position?

**Closed-surface flux** depends solely on net charge inside, not external charges. This principle allows flux calculation without detailed field integration and forms cornerstone for symmetric charge distributions. In a non-conductor, charges are fixed and distributed throughout the volume. Gauss’s law shows the field inside depends on the enclosed charge, which increases with radius, leading to a non-zero, position-dependent field. Substituting values gives Volume charge distribution, 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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Electric Dipole - Moment, Field and Torque

Why can’t the electric field inside a charged insulator be zero, unlike in a conductor?

**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°. In insulators, charges are fixed and cannot move to cancel an internal field. If charges are present inside, they generate a field that persists, as there are no free charges to redistribute and neutralize it, unlike in conductors. Substituting values gives Lack of free charges, 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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Gauss's Theorem and Total Flux

Why does Gauss’s law fail to determine the electric field for a finite charged object without symmetry?

**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. Gauss’s law requires a Gaussian surface with symmetry matching the charge distribution to simplify field calculation. For finite, asymmetric objects, the field varies in complex ways, making symmetry-based simplification impossible without additional methods. Substituting values gives Lack of symmetry, 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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Field Due to Special Charge Configurations

A thin spherical shell of radius 12 cm has \( q = 10 \, \mu\text{C} \). What is the electric field at 14 cm from the cen

**Symmetric configurations** from Gauss's law produce characteristic fields. Uniformly charged infinite plane gives uniform field E = σ/(2ε₀) independent of distance due to planar symmetry, spherical shell acts as point charge outside E = kq/r² and zero inside, reflecting zero enclosed charge interior. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (10 × 10⁻⁶/(0.14)²) = 9 × 10⁹ × (10 × 10⁻⁶/0.0196) = 4.59 × 10⁶ N/C . Substituting values gives 4.59 × 10⁶ N/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 > Field Due to Special Charge Configurations

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Superposition Principle and Equilibrium of Charges

What characteristic of conductors allows charges to distribute uniformly over their surface when placed in an external e

**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. In conductors, charges (free electrons) can move freely. In an external field, they redistribute until the internal field cancels the external field, achieving equilibrium. This results in charges residing only on the surface, distributed uniformly for a spherical conductor due to symmetry. Substituting values gives Mobility of charges, 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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