Where is has the other half of the energy gone ?

  • Thread starter Thread starter MrPhy
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
When two capacitors are connected in parallel, the initial energy stored in one capacitor is reduced due to charge sharing, leading to a lower total energy after the switch is closed. The energy loss, which can be attributed to radiative losses and heat dissipation, raises the question of where the missing energy goes. Even in an ideal scenario with no resistance, energy is still lost as electromagnetic radiation or heat, preventing perpetual oscillation of charges. This phenomenon parallels a water tank scenario, where energy dissipates through friction and other losses until equilibrium is reached. Ultimately, energy is conserved, but it can transform into other forms during the process.
MrPhy
Messages
3
Reaction score
0
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
 
Last edited:
Physics news on Phys.org
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 ?
 
Last edited:
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.
 
Back
Top