47 public questions tagged with this topic.
**Sign of potential energy** indicates nature of interaction, like charges U>0, opposite U
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges
**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 = Potential energy = q V . W = 5 × 10⁻⁶ × 2000 = 10⁻² J = 0.01 J . 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 0.01 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference
**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 = Potential energy = q V . W = 9 × 10⁻⁶ × 40 = 3.6 × 10⁻⁴ J = 0.36 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 0.36 mJ follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference
**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. The conservative nature of the electrostatic field means the work done depends only on the potential difference between points, not the path. For a closed path, the start and end points are the same, so the net work is zero. Substituting values gives Conservative nature, 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
**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
**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. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 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 > Heat Transfer Work Distinction and Internal Energy Change
**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For adiabatic ( Δ Q = 0 ), First Law: Δ U = -Δ W . Work on system: Δ W = -300 J (negative by convention). Δ U = -(-300) = 300 J . Using first law ΔU = Q - W,
Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change
**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.8 , R = 8.3 , T₁ = 760 , T₂ = 380 , γ = 1.4 . W = (0.8 × 8.3 × (760 - 380))/(1.4 - 1) = (6.64 × 380)/(0.4) = 6312 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P
Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change
**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. In an adiabatic process ( Δ Q = 0 ), Δ U = -Δ W (First Law). If work is done on the system ( Δ W < 0 ), Δ U becomes positive, increasing the internal energy. 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 > Cyclic Processes and Reversibility Concepts
**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. First Law: Δ Q = Δ U + Δ W . Δ Q = -700 J (heat released), Δ W = 250 J (work by system). -700 = Δ U + 250 ⇒ Δ U = -700 - 250 = -950 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 =
Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts
**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. First Law: Δ Q = Δ U + Δ W . Δ Q = -670 (heat released), Δ W = -230 (work on system). -670 = Δ U - 230 ⇒ Δ U = -670 + 230 = -440 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. First Law: Δ Q = Δ U + Δ W . Δ Q = -760 (heat released), Δ W = 240 (work by system). -760 = Δ U + 240 ⇒ Δ U = -760 - 240 = -1000 J . Using first law ΔU = Q - W, W = ∫ P
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
