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. Potential difference: V = E₀ ( (5d/6) ) + (E₀/K) ( (d/6) ) = E₀ d ( (5/6) + (1/6 × 5) ) . V = E₀ d ( (5/6) + (1/30) ) = E₀ d × (26/30) = E₀ d × (13/15) . C = (Q/V) = (Q/(13/15) V₀) = (15/13) × 200 ≈ 230.77 pF . Using V = kQ/r, U = k q₁q₂/r,