#work done

Energy Stored in Capacitor and Energy Density

A charge of \( 7 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 100 \, \text{V} \). What

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². Work done = Potential energy = q V . W = 7 × 10⁻⁶ × 100 = 7 × 10⁻⁴ J = 0.7 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 0.7 mJ follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 5 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 5 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 180°) . W = 5 × 10⁻⁹ × 2 × 10⁵ × (0 - (-1)) = 5 × 10⁻⁹ × 2 × 10⁵ × 1 = 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 10 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 10 × 10⁻⁹ × 4 × 10⁵ × (cos 0° - cos 90°) . W = 10 × 10⁻⁹ × 4 × 10⁵ × (1 - 0) = 4 × 10⁻³ J . 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 > Conductors, Electrostatic Shielding and Dielectrics

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Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 2 \, \mu\text{C} \) is moved from infinity to a point with potential \( 500 \, \text{V} \). What is the w

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. Work done = Potential energy = q V . W = 2 × 10⁻⁶ × 500 = 10⁻³ J = 1 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 1 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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Potential Energy of System of Charges

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻⁹ × 1 × 10⁵ × (cos 0° - cos 180°) . W = 8 × 10⁻⁹ × 1 × 10⁵ × (1 - (-1)) = 8 × 10⁻⁹ × 1 × 10⁵ × 2 = 1.6 × 10⁻³ J . Using V = kQ/r,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Electric Potential and Potential Difference

A charge of \( 8 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 50 \, \text{V} \). What i

**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. Work done = Potential energy = q V . W = 8 × 10⁻⁶ × 50 = 4 × 10⁻⁴ J = 0.4 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,

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

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Electric Potential and Potential Difference

A dipole \( p = 6 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \(

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 6 × 10⁻⁹ × 3 × 10⁵ × (cos 0° - cos 90°) . W = 6 × 10⁻⁹ × 3 × 10⁵ × (1 - 0) = 1.8 × 10⁻³ J . 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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Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically, doing 360 J of work. What is the change in its internal energy?

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Heat Transfer Work Distinction and Internal Energy Change

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. First Law: Δ Q = Δ U + Δ W . Δ Q = 850 J , Δ W = -300 J (work on system). 850 = Δ U - 300 ⇒ Δ U = 850 + 300 = 1150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Isochoric Processes and Pressure-Temperature

A system absorbs 600 J of heat and does 150 J of work. What is the change in internal energy?

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. First Law: Δ Q = Δ U + Δ W . Given Δ Q = 600 J , Δ W = 150 J (work by system). 600 = Δ U + 150 ⇒ Δ U = 600 - 150 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

A system releases 600 J of heat and performs 250 J of work. What is the change in internal energy?

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. First Law: Δ Q = Δ U + Δ W . Δ Q = -600 J (heat released), Δ W = 250 J (work by system). -600 = Δ U + 250 ⇒ Δ U = -600 - 250 = -850 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 980 J of heat and performs 420 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 980 - Q_reject = 420 ⇒ Q_reject = 980 - 420 = 560 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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