Practice question

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. In a series combination, capacitors are connected end-to-end, forming a single path for charge flow. When a voltage is applied, the same charge Q accumulates on each capacitor (as charge conservation ensures the same Q passes through each during charging). However, the potential difference across each capacitor is V = Q/C , so V