Capacitor Energy Simulation
This simulation demonstrates how energy is stored in capacitors and how it changes when capacitors are connected together:
- Capacitance: 900 pF (each capacitor)
- Battery voltage: 100 V
Scenario (a): Single Charged Capacitor
100V
C
Energy: 0 J
Scenario (b): Two Connected Capacitors
Cā
Cā
Energy: 0 J
Physics Explanation:
Scenario (a): Single capacitor charged by battery
Charge: \( Q = CV = 900 \times 10^{-12} \, \text{F} \times 100 \, \text{V} = 9 \times 10^{-8} \, \text{C} \)
Energy: \( U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{1}{2} \times 9 \times 10^{-8} \times 100 = 4.5 \times 10^{-6} \, \text{J} \)
Scenario (b): Two capacitors connected together
When disconnected from battery and connected to another capacitor:
Charge is conserved: \( Q' = \frac{Q}{2} = 4.5 \times 10^{-8} \, \text{C} \) (on each capacitor)
New voltage: \( V' = \frac{Q'}{C} = \frac{4.5 \times 10^{-8}}{900 \times 10^{-12}} = 50 \, \text{V} \)
Total energy: \( U_{total} = 2 \times \frac{1}{2}Q'V' = \frac{1}{4}QV = 2.25 \times 10^{-6} \, \text{J} \)
(a) Initial energy stored: \( 4.5 \times 10^{-6} \, \text{J} \)
(b) Final energy stored: \( 2.25 \times 10^{-6} \, \text{J} \)
The "missing" energy is dissipated as heat and electromagnetic radiation during the charge transfer.
