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52 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

In an electrostatic field, if a positive charge is moved along an equipotential surface, what can be said about the work

**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. An equipotential surface has a constant potential at all points. The work done by the electric field when a charge moves along such a surface is zero because the potential difference between any two points on the surface is zero ( W = q Δ V , and Δ V = 0 ). Additionally,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Equipotential Surfaces and Relation Between Field and Potential

A dipole \( p = 7 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**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: W = p E (cos θ₀ - cos θ₁) = 7 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 0°) . W = 7 × 10⁻⁹ × 2 × 10⁵ × (0 - 1) = -1.4 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq =

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

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Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 6 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 150 \, \text{V} \). What

**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 = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴ J = 0.9 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 0.9 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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Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 3 \, \mu\text{C} \) is brought from infinity to a point where the potential is \( 400 \, \text{V} \). Wha

**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 = 3 × 10⁻⁶ × 400 = 1.2 × 10⁻³ J = 1.2 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 1.2 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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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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Electric Potential and Potential Difference

What is the work done in moving a \( 5 \, \mu\text{C} \) charge from infinity to a point where the potential is \( 2000

**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

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Electric Potential and Potential Difference

A charge of \( 9 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 40 \, \text{V} \). What i

**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

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Superposition Principle and Equilibrium of Charges

What characteristic of an electric field ensures that work done in moving a charge along a closed path is zero in electr

**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

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

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Work Heat Distinction and Energy Transfer Modes

What distinguishes work from heat as a mode of energy transfer?

**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 in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**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

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Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically, doing 300 J of work on the system. What is the change in internal energy?

**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

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