#rotation

Combination of Capacitors - Series and Parallel

A dipole \( p = 8 \times 10^{-10} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻¹⁰ × 10⁶ × (cos 0° - cos 90°) = 8 × 10⁻⁴ × (1 - 0) = 8 × 10⁻⁴ 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 > Combination of Capacitors - Series and Parallel

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Combination of Capacitors - Series and Parallel

A dipole \( p = 2 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. Work done: W = p E (cos θ₀ - cos θ₁) = 2 × 10⁻⁹ × 5 × 10⁵ × (cos 90° - cos 0°) = 2 × 10⁻⁹ × 5 × 10⁵ × (0 - 1) = -10⁻³ 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

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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Potential Energy of System of Charges

Why does the potential energy of a dipole in a uniform field increase when rotated from alignment (\( \theta = 0^\circ \

**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. The potential energy of a dipole is U = -p E cos θ . At θ = 0° , cos 0 = 1 , U = -p E , the minimum. At θ = 90° , cos 90 = 0 , U = 0 . From θ = 0° to 90° , U increases from -p E to 0

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

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PHYSICS

A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the n

Given: A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the nter? These values define the system as per NCERT data. Formula: L = m v r. This is standard NCERT relation. Substitution & Calculation: m = 4 kg, v = 2 m/s, r = 0.25 m . L = 4 × 2 × 0.25 = 2 kg m²/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: System of Particles and Rotational Motion, Topic: Angular momentum L = m v r, rotating mass and moment of momentum. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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Moment of Inertia and Radius of Gyration

In a rigid body undergoing both translation and rotation, what describes the motion of any point?

The motion of any point is a combination of the translational velocity of the body (same for all points) and the rotational velocity about the axis, varying with distance from the axis.