#potential energy

Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of an electron in the \( n = 3 \) state is -1.51 eV. What is its potential energy?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = K + U , K = -E = 1.51 eV . U = -2K = -2 × 1.51 = -3.02 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields -3.02

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of an electron in the ground state is -13.6 eV. What is the magnitude of its potent

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = K + U , K = 13.6 eV , U = -2K = -27.2 eV . Magnitude = |U| = 27.2 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the ratio of the kinetic energy to the potential energy of an electron in the ground state of a hydrogen atom? (

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. E = K + U , K = 13.6 eV , U = -27.2 eV . Ratio = (K/|U|) = (13.6/27.2) = 0.5 . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 0.5, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the potential energy of an electron in the \( n = 5 \) state of a hydrogen atom? (Use \( E_n = -\frac{13.6}{n^2}

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. E₅ = -0.544 eV , K = 0.544 eV . U = -2K = -2 × 0.544 = -1.088 eV ≈ -1.09 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields -1.09 eV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Energy Stored in Capacitor and Energy Density

Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (-6, 0, 0) \) and \( (6, 0, 0) \, \text{cm} \)

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Mutual energy: U₁₂ = 9 × 10⁹ × (8 × 10⁻⁶ × (-4 × 10⁻⁶)/0.12) = -2.4 J . External potential: V(r) = (10⁵/r) , at r = 0.06 m , V = (10⁵/0.06) = 1.67 × 10⁶ V . External energy: 8 × 10⁻⁶ × 1.67 × 10⁶ + (-4 × 10⁻⁶) ×

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

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Energy Stored in Capacitor and Energy Density

Two charges \( 32 \, \mu\text{C} \) and \( -16 \, \mu\text{C} \) are placed 32 cm apart. What is the potential energy of

**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. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (32 × 10⁻⁶ × (-16 × 10⁻⁶)/0.32) . U = 9 × 10⁹ × (-512 × 10⁻¹²/0.32) = -14.4 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 -14.4 J follows, reflecting potential-capacitance relations.

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

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Combination of Capacitors - Series and Parallel

A dipole with \( p = 5 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 3 \t

**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. U = -p E cos θ = -5 × 10⁻⁹ × 3 × 10⁴ × cos 30° . cos 30° = (√(3)/2) ≈ 0.866 , so U = -5 × 10⁻⁹ × 3 × 10⁴ × 0.866 = -1.3 × 10⁻⁴ 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

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 with \( p = 4 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 5 \t

**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. U = -p E cos θ = -4 × 10⁻⁹ × 5 × 10⁴ × cos 30° . cos 30° = (√(3)/2) ≈ 0.866 , so U = -4 × 10⁻⁹ × 5 × 10⁴ × 0.866 = -1.732 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

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Conductors, Electrostatic Shielding and Dielectrics

Why does the potential energy of a system of two charges depend only on their separation and not on their orientation in

**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. The potential energy between two point charges is U = (1/4 π ε₀) (q₁ q₂/r) , where r is the distance between them. This expression depends only on the magnitude of the separation r , not on the direction or orientation of the line joining the charges in space, because the Coulomb force is isotr

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

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 12 cm apart. What is the potential energy of the syst

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (6 × 10⁻⁶ × (-3 × 10⁻⁶)/0.12) = 9 × 10⁹ × (-18 × 10⁻¹²/0.12) = -1.35 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 -1.35 J 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 dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) makes an angle of \( 60^\circ \) with a uniform field \( E = 1 \t

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. U = -p E cos θ = -8 × 10⁻⁹ × 1 × 10⁵ × cos 60° . cos 60° = 0.5 , so U = -8 × 10⁻⁹ × 1 × 10⁵ × 0.5 = -4 × 10⁻⁴ 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

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

Two charges \( 18 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are placed 18 cm apart. What is the potential energy of

**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. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (18 × 10⁻⁶ × (-6 × 10⁻⁶)/0.18) . U = 9 × 10⁹ × (-108 × 10⁻¹²/0.18) = -5.4 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

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

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