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Not only due to resistance in realistic cases.In an ideal cap2cap transfer (without intermediating resistance and inductance) infinite current would flow initially and the entire energy loss would be due to EM radiation.

Dose it take twice the energy to charge a cap,than what you have in the cap after the charge is complete?.

What i mean to say is,if i use 10 joules of energy to charge a cap to say 10 volt's,how much of that 10 joules i used to charge the cap will be actually stored in the cap?--will it be close to the 10 joules used to charge it,or will there only be 5 joules of energy stored in that cap?. Did we loose half of the energy when charging the cap?.

Yes.You will lose 5 Joules, so 5 Joules will be left in the capacitor.

According to the the electron count exercise we did recently, no EM loss is necessary to account for half the loss.

That's not to say the loss has been accounted for in that scenario though, it is still apparently a mystery as to where the missing energy goes.

Most caps i have pulled apart,have the terminal wires fixed to the center of each capacitor plate strip.This would mean that the current flow would see a CW wound coil for half of the turns,and a CCW coil for the other half of the turns of each of the two plates that make up the capacitor. To me,this would mean a cancellation,or more accurately,a confinement of the EM field produced by current flowing through the plates-

Does it take twice the energy to charge a cap,than what you have in the cap after the charge is complete?

If I have 2 parallel plates separated by 10 distance units and I apply a 10 coulomb charge,, so I pull 10 coulombs of electrons from one plate and push 10 coulombs of electrons into the other plate.I now have X energy stored within the cap, now I move the plates closer together so that they are only at 1 distance unit of separation.

Ask yourself the question, whether the energy used to polarize the dielectric slug can be recovered, once the slug becomes attracted all the way between the plates ?

Thanks Poynt,i was not sure on that one.So the question remain's--how or why was it lost?It would seem that this loss of energy is not just through the cap to cap transfer,but also from any energy source that is used to charge the cap.But why is it then that we can reduce this loss by way of inductive charging?--what is it in the inducting charging method that reduces this loss?.Brad