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#superposition

12 public questions tagged with this topic.

Two charges \( 20 \, \mu\text{C} \) and \( -10 \, \mu\text{C} \) are at \( (1, 0, 0) \) and \( (-1, 0, 0) \, \text{cm} \

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. Distance to midpoint = 0.01 m. V = 9 × 10⁹ ( (20 × 10⁻⁶/0.01) + (-10 × 10⁻⁶/0.01) ) = 9 × 10⁹ × (10 × 10⁻⁶/0.01) . V = 9 × 10⁹ × (10 × 10⁻⁶/0.01) = 9 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½

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

Two charges \( 18 \, \mu\text{C} \) and \( -9 \, \mu\text{C} \) are at \( (2, 0, 0) \) and \( (-2, 0, 0) \, \text{cm} \)

**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. Distance to midpoint = 0.02 m. V = 9 × 10⁹ ( (18 × 10⁻⁶/0.02) + (-9 × 10⁻⁶/0.02) ) = 9 × 10⁹ × (9 × 10⁻⁶/0.02) . V = 9 × 10⁹ × (9 × 10⁻⁶/0.02) = 4.05 × 10⁶ V . 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 > Potential Energy of System of Charges

Four charges \( +q, -q, +q, -q \) are at the corners of a square of side \( 2 \, \text{m} \) in order. What is the poten

**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. Distance from center to each corner = √(2) m . V = 9 × 10⁹ ( (1 × 10⁻⁶/√(2)) + (-1 × 10⁻⁶/√(2)) + (1 × 10⁻⁶/√(2)) + (-1 × 10⁻⁶/√(2)) ) = 0 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel,

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

In a system of two identical charged spheres brought close together, why does the potential at the midpoint differ from

**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. The potential at a point due to multiple charges is the algebraic sum of the potentials due to each charge (principle of superposition). However, when two charged spheres are close, they influence each other’s charge distribution (mutual interaction), polarizing each other. This alters the effective potential at the midpoint compared to the sum of their individual

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

What underlies the periodic nature of SHM when expressed as a superposition of sine and cosine functions?

**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. The periodicity arises from the repeating nature of sine and cosine functions, which have a fixed period ( 2π/ω ), ensuring the motion repeats consistently. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The periodicity of trigonometric functions follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

Two waves \( y_1 = 4 \sin (5x - 10t) \) and \( y_2 = 4 \sin (5x - 10t + \frac{\pi}{6}) \) interfere. What is the amplitu

**Relative motion** changes effective wavelength encountered. For moving source approaching stationary observer, wavelength ahead λ' = (v - v_s)/f, so f' = v/λ' = f·v/(v - v_s) > f, basis for calculating apparent pitch shift in sound. Amplitude: A = 2a cos (Φ/2) , a = 4 m , Φ = (π/6) . A = 2 × 4 cos (π/12) ≈ 8 × 0.966 = 7.73 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 7.7 m, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Doppler Effect

Two waves \( y_1 = 3 \sin (4x - 10t) \) and \( y_2 = 3 \sin (4x - 10t + \pi/2) \) interfere. What is the amplitude of th

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. Amplitude: A = 2a cos (Φ/2) , where a = 3 m , Φ = π/2 . A = 2 × 3 cos π/4 = 6 × (1/√(2)) = 3√(2) ≈ 4.24 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 4.2 m, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

What fundamental principle allows the electric field due to multiple charges to be calculated as the vector sum of the f

**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. The superposition principle states that the electric field produced by multiple charges is the vector sum of the fields produced by each charge independently. This principle holds because electric forces and fields are linear and unaffected by the presence of other charges. Substituting values gives Superposition, 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

Two charges \( +11 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are 80 cm apart. What is the electric field magnitude a

**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. Midpoint distance = 40 cm = 0.4 m. E₁ = 9 × 10⁹ × (11 × 10⁻⁶/(0.4)²) = 6.1875 × 10⁵ N/C (towards -5 μC ). E₂ = 9 × 10⁹ × (5 × 10⁻⁶/(0.4)²) = 2.8125 × 10⁵ N/C (towards -5 μC ). Net E = 6.1875 × 10⁵ + 2.8125 × 10⁵ = 9 × 10⁵ N/C . Substituting values gives 9.0 × 10⁵ N/C, which matches expected magnitude for this electrostatic configuration,

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

Two charges \( +4 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are 30 cm apart. What is the electric field at a point 1

**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. Distance from -6 μC = 15 cm = 0.15 m . E₁ = 9 × 10⁹ × (4 × 10⁻⁶/(0.15)²) = 1.6 × 10⁶ N/C (away). E₂ = 9 × 10⁹ × (6 × 10⁻⁶/(0.15)²) = 2.4 × 10⁶ N/C (towards). Net E = 2.4 × 10⁶ - 1.6 × 10⁶ = 8 × 10⁵ N/C

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

Three equal charges \( q = 3 \, \mu\text{C} \) are at the vertices of an equilateral triangle of side 1 m. What is the e

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. Distance from vertex to centroid: r = (l/√(3)) = (1/√(3)) m . E = (k q/r²) = 9 × 10⁹ × (3 × 10⁻⁶/(1/√(3))²) = 81 × 10⁴ N/C per charge. By symmetry (all +q ), vectors cancel, so Eₙₑt = 0 . Substituting values gives 0 N/C, 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

Which principle underlies the fact that the electric field due to a system of charges can be analyzed by considering eac

**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 superposition principle allows the total electric field to be the vector sum of fields from individual charges, assuming each field is unaffected by others. This linearity stems from the nature of electrostatic forces, enabling independent analysis. Substituting values gives Superposition, 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