Practice question

**Capacitance depends on geometry** not charge, C = Q/V constant for given arrangement. Parallel plate C ∝ A/d, so larger area and smaller separation increase capacitance, principle used to increase storage by using large area foils with thin dielectric. For r = 0.15 m > R = 0.1 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (2 × 10⁻⁷/(0.15)²) = 9 × 10⁹ × (2 × 10⁻⁷/0.0225) = 8 × 10⁴ N/C . 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