#Coulomb constant

Conductors, Electrostatic Shielding and Dielectrics

A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 3 m aw

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (9 × 10⁻⁹/3) = 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 V² and common potential V =

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Conductors, Electrostatic Shielding and Dielectrics

A point charge \( Q = 18 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (18 × 10⁻⁹/6) = 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 V² and common potential V

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Electric Potential and Potential Difference

Three charges \( +5 \, \mu\text{C} \), \( -2 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (

**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. Distances: r₁ = √(5² + 5²) = 5√(2) m , r₂ = 5 m , r₃ = 5 m . V = 9 × 10⁹ ( (5 × 10⁻⁶/5√(2)) + (-2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V = 9 × 10⁹ ( (5 × 10⁻⁶/7.07) - (2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V =

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Electric Potential and Potential Difference

A spherical conductor of radius 6 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Spherical conductor** behaves as point charge outside, potential V = k Q/R at surface, field E = k Q/r² for r>R, zero inside for rR, so E = k Q/r² =9×10⁹×6×10⁻⁸/0.16=3375 N/C. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (6 × 10⁻⁸/0.06) = 9 × 10⁹ × 10⁻⁶ = 9000 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

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Electric Potential and Potential Difference

A spherical conductor of radius 5 cm is charged with \( 3 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (3 × 10⁻⁸/0.05) = 9 × 10⁹ × 6 × 10⁻⁷ = 5400 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

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

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