Energy in Capacitors
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
A charged capacitor stores energy U = ½QΔV = ½C(ΔV)² = Q²/2C. The ½ is there because the voltage builds up from zero while you charge it. The energy lives in the electric field, with density u = ½ε0E².
1. Where the ½ comes from
2. Which form to use
Isolated (disconnected): Q is fixed, so use U = Q²/2C. Pulling the plates apart lowers C and raises U: you do work against their attraction.
Connected to a battery: ΔV is fixed, so use U = ½C(ΔV)². Pulling the plates apart lowers C and lowers U (charge flows back to the battery).
Energy density check: for parallel plates, u × volume = ½ε0E² × Ad = ½(ε0A/d)(Ed)² = ½C(ΔV)² ✓.
3. Example, then try it
A camera flash charges a 50 µF capacitor to 200 V. How much charge and energy does it store?
Q = CΔV = (50 × 10⁻⁶)(200) = 0.010 C.
U = ½C(ΔV)² = ½(50 × 10⁻⁶)(200)² = 1.0 J, released in a millisecond flash.
A 100 pF air capacitor is charged to 50 V, disconnected, and then its plates are pulled apart to three times the separation. Find Q, the new C and ΔV, and the energy before and after. Where does the extra energy come from?
Charge (fixed): Q = (100 × 10⁻¹²)(50) = 5.0 nC.
After: C = 100/3 ≈ 33 pF, so ΔV = Q/C = 150 V.
Energy: before ½(5.0 × 10⁻⁹)(50) = 125 nJ; after ½(5.0 × 10⁻⁹)(150) = 375 nJ.
The extra 250 nJ is the work you did pulling the attracting plates apart.
Answer: 5.0 nC; 33 pF and 150 V; energy triples from 125 nJ to 375 nJ.
4. Common mistakes
A capacitor stays connected to a battery while its plates are pulled farther apart. What happens to its stored energy?
ΔV is fixed by the battery and C decreases, so U = ½C(ΔV)² decreases.
Ready to practice?
4 questions on Energy in Capacitors, each with a hint and an explanation for every wrong answer.