#charge transfer

Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 10 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 30 \, \text{V} \). What

**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. Work done = Potential energy = q V . W = 10 × 10⁻⁶ × 30 = 3 × 10⁻⁴ J = 0.3 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 0.3 mJ follows, reflecting potential-capacitance relations.

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

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Coulomb's Law and Force Between Point Charges

A plastic rod gains a charge of \( -1.28 \times 10^{-7} \, \text{C} \) when rubbed. How many electrons were transferred

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Negative charge means electrons gained. q = n e , e = -1.6 × 10⁻¹⁹ C . n = (q/|e|) = (1.28 × 10⁻⁷/1.6 × 10⁻¹⁹) = 8 × 10¹¹ . Substituting values gives 8 × 10¹¹, 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 body acquires a charge of \( -9.6 \times 10^{-8} \, \text{C} \) when rubbed. How many electrons were transferred to it

**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. Negative charge indicates electrons gained. q = n e , e = -1.6 × 10⁻¹⁹ C . n = (q/|e|) = (9.6 × 10⁻⁸/1.6 × 10⁻¹⁹) = 6 × 10¹¹ . Substituting values gives 6 × 10¹¹, 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 plastic rod gains \( 4.8 \times 10^{-8} \, \text{C} \) of negative charge when rubbed. How many electrons were transfe

**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. Negative charge means electrons are gained. q = n e , e = -1.6 × 10⁻¹⁹ C . n = (q/|e|) = (4.8 × 10⁻⁸/1.6 × 10⁻¹⁹) = 3 × 10¹¹ . Substituting values gives 3 × 10¹¹, 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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Electric Charge, Quantization and Conservation

Which principle explains why rubbing two neutral objects can result in one becoming positively charged and the other neg

**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. The conservation of charge ensures that the total charge remains zero. Rubbing transfers electrons from one object to another due to differing electron affinities, leaving one with a net positive charge (electron loss) and the other negative (electron gain). Substituting values gives Conservation of charge, which matches expected magnitude for this ele

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

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Electric Charge, Quantization and Conservation

A glass rod loses \( 6.4 \times 10^{-8} \, \text{C} \) of charge when rubbed with silk. How many electrons were transfer

**Quantization of charge** states observable charge is integer multiple of elementary charge e = 1.6×10⁻¹⁹ C, q = n·e, and total charge is conserved in isolated systems. Loss of electrons produces positive charge, and number of transferred electrons follows n = q/e, linking macroscopic charge measurement to microscopic carriers. Losing charge means electrons are removed, so charge is positive. q = n e , e = 1.6 × 10⁻¹⁹ C . n = (q/|e|) = (6.4 × 10⁻⁸/1.6 × 10⁻¹⁹) = 4 × 10¹¹ . Substituting values gives 4 × 10¹¹, which matches expected magnitude for this electrostatic configuration, confirming Cou

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

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