Skip to content

Question

A \( 12 \, \mu\text{F} \) capacitor charged to \( 30 \, \text{V} \) is connected to an uncharged \( 12
\, \mu\text{F} \) capacitor. What is the final potential difference?

Options

Choose one · Correct answer highlighted

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 = 12 × 10⁻⁶ × 30 = 3.6 × 10⁻⁴ C . Total capacitance: 12 + 12 = 24 μF . Final voltage: V = (Q/C) = (3.6 × 10⁻⁴/24 × 10⁻⁶) = 15 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.