Electrostatic Potential and Capacitance

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Electric potential at any point in an electric field is equal to the work done against the electric force in moving a unit positive charge from infinity to that point. It is a scalar quantity measured

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Electric field is the negative gradient of electric potential: E = -dV/dr. For uniform field: E = (V₁ - V₂)/d, where d is distance between points. The electric field points in the direction of decreas

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For point charge q at distance r: Work done = ∫(kqq₀/r²)dr from ∞ to r = kqq₀(1/r - 1/∞) = kqq₀/r. Therefore, V = W/q₀ = kq/r = q/(4πε₀r), where k = 1/(4πε₀) = 9×10⁹ N⋅m²/C².

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Given: q = 20 μC = 20×10⁻⁶ C, r = 30 cm = 0.3 m Formula: V = q/(4πε₀r) = kq/r Solution: V = (9×10⁹ × 20×10⁻⁶)/0.3 = 180/0.3 = 600,000 V = 6×10⁵ V Answer: 600 kV

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For electric dipole with moment p = q×2l, potential at axial point at distance r: V = p/(4πε₀r²) (positive side), V = -p/(4πε₀r²) (negative side). On equatorial line: V = 0. Note: Potential ∝ 1/r² (un

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Capacitance is the ratio of charge stored on a conductor to the potential difference created: C = Q/V. It measures the ability to store charge. Unit: Farad (F) = Coulomb/Volt. 1 F is very large, so we

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For parallel plates with area A, separation d, charge ±Q: Electric field: E = σ/ε₀ = Q/(ε₀A) Potential difference: V = Ed = Qd/(ε₀A) Capacitance: C₀ = Q/V = ε₀A/d Capacitance ∝ A and C ∝ 1/d

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Dielectric with constant K increases capacitance by factor K: C = KC₀ = Kε₀A/d. This happens because dielectric polarization reduces effective electric field, decreasing potential difference for same