When a capacitor is applied with an AC Wave in a circuit containing only capacitor, it charges to its peak value between the time period of 0 to 90 degrees. My doubt is that what happens between the time period 90 to 180 degrees i.e. when the voltage starts decreasing from peak to zero. I understand that there is no dissipating element in the circuit therefore charge will not decay as time constant RC is zero. But still I was eager to know what happens at that time interval i.e. 90 to 180 degrees.
An AC generator is considered low (or theoretically, zero) impedance, so it will "force" the capacitor connected to it to be at the same voltage at every point in time.
An important aspect of this circuit is what happens to the current in the capacitor during the different portions of the curve. For the first 90 degrees the voltage is rising, so current will go into the capacitor, and stop at the 90 degree point where the sine wave flattens out. From 90 to 180 (actually, to 270) degrees, current will go out of the capacitor and into the source of the wave. Presuming the diagram is of voltage, you can draw the current through the capacitor with a cosine wave. It starts at maximum at 0 degrees, then goes to zero at 90 degrees, and then goes negative for the next half-cycle from 90 to 270 degrees. From there it goes positive again to 360 degrees and the cycle repeats at 0 degrees.
What you may be thinking of is when a diode is in series with the AC source and a capacitor. In this case, the capacitor will charge up to the peak value in the first 90 degrees (minus the usual 0.6 volt drop of most silicon diodes) due to the AC voltage being greater than the capacitor voltage, and thus the diode being forward biased. After 90 degrees AC generator voltage drops, so the diode will be reversed biased, and no current will flow into or from the capacitor. It will stay charged at the peak voltage (again, minus the diode drop) of AC generator.