Worked example: A parallel-plate capacitor C0 = 100 pF is charged to V0 = 50 V, then disconnected. A glass slab (κ = 5) is slid fully between the plates. Find the new voltage, capacitance, and energy change. Step 1: Q is fixed (isolated): Q = C0·V0 = (10⁻¹⁰)(50) = 5 nC. Step 2: New C = κC0 = 500 pF. New V = Q/C = 5 nC/500 pF = 10 V — the voltage drops fivefold. Step 3: Initial energy U_i = (1/2)C0V0² = 0.5 × 10⁻¹⁰ × 2500 = 1.25 × 10⁻⁷ J. Final U_f = (1/2)CV² = 0.5 × 5 × 10⁻¹⁰ × 100 = 2.5 × 10⁻⁸ J — one-fifth. Step 4: The lost 1.0 × 10⁻⁷ J became the work done pulling the slab in (plus a little heat). If instead the battery had stayed connected at 50 V, C would still quintuple but Q would rise to 25 nC and energy would increase fivefold — the battery supplies the extra.