Practice question

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. The capacitance of a parallel plate capacitor is C = (K ε₀ A/d) , where d is the separation between plates, A is the area, and K is the dielectric constant. When the plates are moved closer, d decreases, and since C ∝ (1/d) , the capacitance increases. A smaller d means a stronger field for the same charge ( E = (Q/ε₀ A) ), allowing more charge