#Coulomb's constant

Energy Stored in Capacitor and Energy Density

A point charge \( Q = 6 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 2 m aw

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (6 × 10⁻⁹/2) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 V . 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 > Energy Stored in Capacitor and Energy Density

View Details

Combination of Capacitors - Series and Parallel

A spherical conductor of radius 20 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the electric field at

**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. For r = 0.5 m > R = 0.2 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (8 × 10⁻⁸/(0.5)²) = 9 × 10⁹ × (8 × 10⁻⁸/0.25) = 2.88 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

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

View Details

Equipotential Surfaces and Relation Between Field and Potential

An electric dipole with moment \( p = 2 \times 10^{-10} \, \text{C m} \) is at the origin, aligned along the x-axis. 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. Position vector r = (0, 2, 0) , r = 2 m , hatr = (0, 1, 0) . Dipole moment p = (2 × 10⁻¹⁰, 0, 0) . V = (1/4 π ε₀) (p · hatr/r²) = 9 × 10⁹ × ((2 × 10⁻¹⁰) · (0)/2²) = 0 V (since cos θ = 0 , equatorial plane). Using V = kQ/r, U

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

View Details

Potential Energy of System of Charges

Two charges \( 12 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (5, 0, 0) \) and \( (-5, 0, 0) \, \text{cm} \)

**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 = 0.05 m. V = 9 × 10⁹ ( (12 × 10⁻⁶/0.05) + (-3 × 10⁻⁶/0.05) ) = 9 × 10⁹ × (9 × 10⁻⁶/0.05) . V = 9 × 10⁹ × (9 × 10⁻⁶/0.05) = 1.62 × 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

View Details

Potential Energy of System of Charges

A point charge \( Q = 24 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 8 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (24 × 10⁻⁹/8) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 C

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

View Details

Potential Energy of System of Charges

A point charge \( Q = 8 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 4 m aw

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (8 × 10⁻⁹/4) = 9 × 10⁹ × 2 × 10⁻⁹ = 18 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 C V² and common potential V = Q_total/C_total, result

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

View Details

Electric Potential and Potential Difference

Three charges \( +8 \, \mu\text{C} \), \( -5 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (

**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. Distances: r₁ = √(8² + 8²) = 8√(2) m , r₂ = 8 m , r₃ = 8 m . V = 9 × 10⁹ ( (8 × 10⁻⁶/8√(2)) + (-5 × 10⁻⁶/8) + (3 × 10⁻⁶/8) ) . V = 9 × 10⁹ ( (8 × 10⁻⁶/11.314) - (5 × 10⁻⁶/8) + (3 × 10⁻⁶/8)

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

View Details