A dielectric slab is inserted into a charged capacitor. How do I know whether stored energy rises or falls?
Do not start from a memorised result; identify the connection.
Do not start from a memorised result; identify the connection. If the capacitor is isolated, Q stays fixed. Since C becomes KC, U = Q^2/(2C) falls by factor K. If the battery remains connected, V stays fixed and the battery supplies extra charge; U = CV^2/2 rises by factor K. The mechanical work and battery energy account for the difference.
Key point
Choose the energy formula using the quantity that remains fixed.
Common mistake
using U = CV^2/2 while silently treating V as fixed for an isolated capacitor.
Memory tip
disconnected: Q fixed; connected: V fixed.
Tap to watch or see it right here — the app stays open (it only opens a quick search if nothing embeddable is found).