Where Does the Energy Go in a Capacitor Without Resistance?

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SUMMARY

The discussion centers on the energy dynamics in a capacitor with capacitance C charged by a battery with electromotive force (emf) E, under the assumption of no resistance. The final charge on the capacitor is Q = C*E, and the work done by the battery is W = C*E^2. However, the energy stored in the capacitor is U = (C*E^2)/2, leading to a discrepancy where the work done does not equal the stored energy. The conclusion drawn is that the model without resistance is not physically realistic, as introducing resistance resolves the energy loss issue.

PREREQUISITES
  • Understanding of capacitor fundamentals and capacitance (C)
  • Knowledge of electromotive force (emf) and its role in charging
  • Familiarity with energy equations related to capacitors (U = (C*E^2)/2)
  • Basic principles of electrical resistance and its effects on circuit behavior
NEXT STEPS
  • Research the impact of resistance on capacitor charging and energy storage
  • Explore the concept of energy dissipation in electrical circuits
  • Learn about the role of kinetic energy in charge movement within capacitors
  • Investigate real-world applications of capacitors in circuits with resistance
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Students of electrical engineering, physicists studying circuit theory, and anyone interested in the principles of energy storage in capacitors.

Yegor
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The capacitor of capacitance C is charged by battery (emf = E) (assume that there is no resistance). In the end charge of the capacitor Q = C*E; Work done by the battery W = Q*E=C*E^2. But The energy of charged capacitor is U = (C*E^2)/2. Work doesn't equals to stored energy. Where we lost energy?? Can we forget about kinetic energy of the charges it this case?? Or it is exactly that "lost" energy?
 
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Thank you very much. As i understood, the model without resistance isn't really physical. With nonzero resistance everything is ok.
 

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