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 ( (3d/4) ) + (E₀/K) ( (d/4) ) = E₀ d ( (3/4) + (1/4 × 4) ) . V = E₀ d ( (3/4) + (1/16) ) = E₀ d × (13/16) . C = (Q/V) = (Q/(13/16) V₀) = (16/13) × 120 ≈ 147.69 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =