Dielectric in an Isolated Capacitor
For an isolated ideal capacitor with no leakage, free plate charge Q is conserved; complete dielectric insertion gives C=KC_0, V=Q/C, and U=Q^2/(2C).
Why this shows up in the exam
battery-removed insertion · isolated dielectric replacement · energy decrease and mechanical work
Learn the idea
With fixed free charge, inserting dielectric lowers voltage and stored energy. After battery removal, charge has nowhere to go. A dielectric increases capacitance, so the same charge produces a smaller voltage and less field energy.
🧠 Memory hook: Battery removed means Q fixed.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Q = constant — isolated capacitor, negligible leakage
- V_new = V_old/K; U_new = U_old/K — complete insertion into initially vacuum capacitor
How to approach it
- 1Circle the fixed quantity Q
- 2Update C
- 3Use Q/C and Q squared/(2C)
Common slip-ups that cost marks
- •Keeping voltage fixed after disconnection
- •Concluding energy disappears without mechanical transfer
- •Confusing free charge with bound polarization charge
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
Two point charges 1 microC and 2 microC are 1 m apart in vacuum. Take k = 9 x 10^9 SI. Find the force magnitude.
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