Where is has the other half of the energy gone ?

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Discussion Overview

The discussion revolves around the energy loss observed when two capacitors are connected in parallel, one initially charged and the other uncharged. Participants explore the question of where the apparent loss of energy goes, considering ideal conditions without resistance in the circuit.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant notes that the initial energy in the charged capacitor is 1/2 CV^2, while after connecting the capacitors, the total energy is 1/4 CV^2, questioning where the other half of the energy has gone.
  • Another participant suggests that energy likely dissipates into the resistance of the switch or wire, which is a common explanation for energy loss.
  • Some participants propose that energy is radiated away, raising questions about the role of connecting wires as antennas and the dependency of radiation on wire length.
  • It is mentioned that even with low resistance, energy will still be radiated away, and that oscillation of charge could occur until energy is lost through various mechanisms.
  • A comparison is made to a scenario with two water tanks, where energy is lost to friction and oscillation until equilibrium is reached, paralleling the energy dynamics in the capacitor scenario.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the mechanisms of energy loss, including resistance, radiation, and oscillation. The discussion remains unresolved, with no consensus on the primary cause of energy loss.

Contextual Notes

Participants acknowledge the ideal conditions of the problem but explore the implications of real-world factors such as resistance and radiation, which complicate the analysis.

MrPhy
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Where is has the other half of the energy gone ??

this is quite a common question asked in many interviews.There are two capacitors C.Intially one capacitor is charged to V volts while the other is at zero volts.At t=0,a switch connects both of them in parallel.Due to charge sharing the voltage at the node connecting both of the capacitors become V/2 volts.Now the question is the energy before the switch was closed was 1/2 CV^2 and the total energy after the switch was closed is 1/4 CV^2.WHere is has the other half of the energy gone ?

Note : this problem it is in the ideal case , where no resistance for any switch or wire
 
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Just a guess based on previous threads, but probably into the resistance of the switch/wire. That seems to be where energy usually goes.
 
Energy is radiated away.
 
umm

Doc Al said:
Energy is radiated away.


we never anyone of US while studying charging & discarging of Cap we study radiated

and if so , you mean that the connecting wires are antennas

but our problem is valid whatever was the wire length (so short , so long ...etc)

so the raditation depends a lot on the antenna length ?
 
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Something's got to give somewhere. :smile:

If you ignore resistance and radiative losses, why wouldn't the charge keep oscillating back and forth? Even with very low resistance, you'll have energy radiated away. (Even the capacitor plates will radiate.) Through some mechanism, energy will be lost until the charges have done their thing and settled into the new configuration.
 
MrPhy said:
we never anyone of US while studying charging & discarging of Cap we study radiated

and if so
That's why its a decent interview question. It makes you think a little beyond what you actually studied. Bottom line is that energy is conserved. If it isn't in the capacitors then it must be radiated away as EM or lost to heat.
 
Doc Al said:
If you ignore resistance and radiative losses, why wouldn't the charge keep oscillating back and forth?
Similar scenario: Consider two cylindrical water tanks connected by a pipe at the bottom, with a valve in the pipe. One water tank is full, the other empty. Open the valve. At some point in the future, the water level will be equal in both tanks and the potential energy 1/2 what it was before. Where did the energy go?

The answer is, the water will oscillate back and forth between the two tanks until the friction in the pipe and viscous friction in the water absorbs (and reradiates) the kinetic energy.
 

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