googly from ATGS 13

Okay subho. As I promised here's a detailed discussion. (However, you need to read a bit of inductance). The point is that we generally tend to compartmentalize a real situation. For example, since we are dealing with capacitors we generally ignore some other characteristic property that all physical situations will have. As in this case, if we try to charge the capacitor via superconductor wires connected to an ideal battery, we still need to take into account the physical property of self-inductance (don't confuse this with self-indulgance [6])which any conductor possesses.
Basically, the inductance is something like an (electrical) inertia. The system does not want to change its state. If a current is flowing in a conductor, it will try its best even if you cut it (that's the reason why you see a spark when you switch something off --- the self inductance tries its best to maintain it even if it means jumping across a gap). Similarly, if you want to establish a current, the self-inductance tries to oppose it by inducing an emf in the opposite direction.

Suppose the wires have an inductance L. Let's look at the transients when the capacitor is being charged up. Suppose at any instant a charge q has been accumulated on the capacitor's plates and at that instant a current I is there in the circuit. In the next dt interval of time, a charge dq = I dt will flow through the battery, which will do a work
δW = I dt V₀
on it. (V₀ is the emf of the battery). Part of this work will go to increase the magnetic energy (there always a magnetic field around a current, and the magnetic energy is 12 LI²) and the rest of the energy goes to increase the capacitors electrical energy. Using the fact that the increment in the magnetic energy is LI dI and the increment in the capacitor's energy is qC dq, and using the work energy principle, we get
I V_0\,\mathrm dt = LI\,\mathrm dI + \frac{q}{C}\,\mathrm dq
Dividing through out by dt, we get
I V_0 = LI\dfrac{\mathrm dI}{\mathrm dt} + \frac{q}{C}\dfrac{\mathrm dq}{\mathrm dt}
Using the fact that I = dq/dt, we obtain finally the equation
\dfrac{d^2q}{\mathrm dt^2}+\dfrac{1}{LC}\,q=\dfrac{V_0}{L}
Define 1LC =ω₀²
Then the previous equation becomes
\dfrac{d^2q}{\mathrm dt^2}+\omega_0^2q=\dfrac{V_0}{L}
Take V₀/L to the left and pull outω₀² common out to get
\dfrac{d^2q}{\mathrm dt^2}+\omega_0^2\left(q-\dfrac{V_0}{\omega_0^2L}\right)=0
Using \xi =q-\dfrac{V_0}{\omega_0^2L}\qquad \Rightarrow \dfrac{\mathrm d^2\xi}{\mathrm dt^2}=\dfrac{\mathrm d^2q}{\mathrm dt^2}
the above equation finally transforms to SHM equation
\dfrac{\mathrm d^2\xi}{\mathrm dt^2}+\omega_0^2\xi=0
So the variable ξ (and hence q) oscillates simple harmonically with an angular frequency ω₀. We can write out the solution as
\xi =A\cos(\omega_0t+\varphi)\quad \Rightarrow \ q=\dfrac{V_0}{\omega_0^2L}+A\cos(\omega_0t+\varphi)
However, ω₀² L = 1/C, hence we ultimately get
\boxed{q=CV_0+A\cos(\omega_0t+\varphi)}
The term CV₀ is easily seen to be the equilibrium charge. So then in this case the charge will actually oscillate about its usual equilibrium value. The same thing remains true for the energy of the capacitor as well. At any moment, the energy will in general not same as CV₀²/2, however the average value will turn out to be that much. So that means while charging via resistance-less wires, the extra work done by the battery will be in the magnetic field of the current in the wire.

P.S. Though I am pretending no loss in the above discussion, actually while the charge surges to and fro in the capacitor, it will behave like a dipole antenna and electromagnetic radiations will continuously come out taking away with them some energy. Once the battery has ran out of its emf source, the oscillations will start to damp out and finally all the energy will be gone. In this case, it will seem that the capacitor will discharge through a "characteristic" resistance given by √μ₀/ε₀ ≈ 376.7 Ω. This effective resistance is called the impedance of free space.

To help you visualize the situation I will give you a mechanical analog. Imagine that you have a vertical spring (which follows the Hook's law) whose one end is fixed from the ceiling. Let the spring constant be k. If you now hook a mass m from the free end and let go, the system will oscillate. If there is some internal resistance present in the spring, the oscillations will ultimately die out and system will come to rest with the spring extending by an amount x = mg/k.
If you look at the decrease in the potential energy of the mass, its mgx = (mg)²/k. On the other hand, what is the increase in the spring potential energy? That is, simply
12kx² = (mg)²/2k
which is only half the loss in p.e. of the mass. What about the other half? That's gone as heat (the spring will be a little warmer).

Now think of the situation when the spring has no internal resistance. What will happen in this case?

http://www.physics.tyrannosaur.eu/Accelerating_electrons_don_t_r/accelerating_electrons_don_t_r.html
You might find this interesting!
It's quite thoroughly covered in this. And you'll be a little shocked to see the conclusion, because we're fed textbook crap which is too outdated. Especially this "theory" which says that electrons accelerating in an electromagnetic field lose energy by radiation. It was part of Planck's initial quantum theory. It's just used as a buffer of continuity from Rutherford's model to Bohr's model in textbooks, with no explanation on the validity of the statement.

@Pritish,
I would rather be careful in passing any verdict. The realm which you are pointing to are deep waters and one should tread carefully. The question whether an accelerating charge radiates or no is an open one. You may take a look at this view as well:
http://www.mathpages.com/home/kmath528/kmath528.htm

"....differentiable manifold is locally Euclidean." intuitively means that a small part of a smooth curved surface seems almost flat --- like a differentiable function in the vicinity of any point could be approximated by the tangent at that point. Equivalence principle being a spacetime version of that fact means that the only in flat space time the equivalence principle holds.

yaar i wrote how it is fromed in my post #2 only if u could read it :P :P ....waise i don't have nay more time to waste on this question now anyways!

abey tht pink thing is a comedy...i don't even care if i get a 1000 or 0 pinks here!

so is my reasoning incorect ?
its not i hope !
source:NCERT

this energy is lost during transient period in the form of HEAT and ELECTROMAGNETIC RADIATIONS!

@subhomoy,
What you have said above is not correct. If you take the potential difference across the terminals of a battery to be same as that across the capacitor plates, why should the capacitor charge at all. And probably, you forgot that the battery has the task of maintaining a constant potential difference across its terminals. Furthermore, the total work done by the battery is definitely CV² and not half of it as you have written in the last line.

The reason for this apparent loss of the energy if R=0 is a bit different. Try to see if you can find the actual reason.