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Capacitor charging & energy stored

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Consider a capacitor (capacitance = C), which is wired to a battery, is charged to potential V volts, and has Q coulombs charge. Now the battery is removed, and a 2nd identical capacitor (uncharged) is connected to the 1st capacitor.

We find that (due to charge conservation in parallel config), the final potential across the system is (V/2) volts, and system capacitance is (2C).

This means that the system's energy is (1/2) * (2C) * (V/2)² = (1/4) CV².

However, the energy stored by the initial capacitor is (1/2) CV².

Where did the remaining energy go ???
 

Hey,

Its not this way, its like the energy in each capacitor is given by 1/2C(V/2)^2.
That is to say Ec=1/8CV^2, Etotal=1/4CV^2
Why?
When you put the capacitors in parallel causing C to be 2C, ie Etotal=1/2(2C)(V/2)^2, i.e. Etotal=1/4CV2, some energy was consumed to move the charges from the plates of the first capacitor and make potential difference via electric field in the second capacitor. Electrons are mass having particles, they need energy to be moved.
:)
 

Most of the energy goes into heating up the wires and capacitors. The circuit is basically two capacitors and a small resistor all in series. Ordinary RC discharge.

If you use perfect capacitors and zero resistance wire, then the energy will oscillate back and forth between the two capacitors at the loop's resonant frequency (the wire has inductance). The loop is an antenna that gradually radiates the energy into space.
 

But Why exactly half of energy? Can anyone prove this?

leomecma
 

I disagree with echo47 and elmolla. This problem is just a trick and it's important to know what that trick is.
Maybe in reality, it is as you said but this is just theory and every thing is ideal. We can see it is very logic but the result is not what we expected. We use the logic A true but lead to logic B wrong.
I think of it but not yet have an answer. Maybe we do approve some thing wrong about his assumtion.
Remember this:
V = Q/C.
we have the same Q but double the C so the V is now just equal to half of it when we parallel another C. I'll check my book then come back later
 

The same current that discharges the first capacitor also charges the second capacitor. They reach equilibrium at V/2.

You can simulate the RC discharge in Spice and see the final V/2 result (or the LC oscillation).

Here's another way of looking at it:
**broken link removed**
 

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The main property of a capacitor is to oppose the change of voltage and the 2nd capacitor would charge till V volts and then 1st starts getting charged and hence it oscillates forming an AC signal with rms value V/√2 and hence conserving the energy ( try reading about inductive kickback which its dual (the art of electronics by horowitz))
 

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