Practice question

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. In series, the charge Q on each capacitor is the same. Capacitance is C = (K ε₀ A/d) , where K is the dielectric constant. A higher K increases C . Since V = (Q/C) , a larger C (due to higher K ) results in a smaller V . Thus, capacitors with higher dielectric constants