Practice question
Question
A \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 6
\, \mu\text{F} \) capacitor. What is the final potential difference?
Explanation
**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial charge: Q = 6 × 10⁻⁶ × 100 = 6 × 10⁻⁴ C . Total capacitance: 6 + 6 = 12 μF . Final voltage: V = (Q/C) = (6 × 10⁻⁴/12 × 10⁻⁶) = 50 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq
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