AC Circuit Having Only Capacitor
Current in the capacitor is proportional the rate of voltage change across it (proportional to how quickly the voltage across the capacitor is changing).
The faster the voltage change (frequency of an AC signal is high) the large the current flow through the capacitor.

$$I = C*\frac{dV}{dt} $$

This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand, if the voltage is kept constant no current will flow no matter how large the voltage.
Likewise, if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.


The special case is when the voltage input voltage is a sine-wave.
For the sine-wave, the rate of voltage change (\$\frac{dV}{dt}\$) reaches the maximum when voltage signal crossing zero. And the rate of voltage change reaches minimum when the voltage is at a peak value. 





This is why current reaches \$0A\$ because at peak the rate of change is \$\frac{dV}{dt} = 0\$ and when a input sine-wave crossing 0, the rate of change is at max \$\frac{dV}{dt} = A_{peak}2\pi f \$. 

For example for \$A=1V\$ and \$f = 50Hz;C=1\mu F\$  

The highest slope of the signal is \$\frac{dV}{dt} = A_{peak}2\pi f = 1V*6.28*50Hz = 314V/s\$ therefore the capacitor peak current values is 

\$I = C*\frac{dV}{dt} = 1\mu F * 314 \frac{V}{s} = 0.314mA\$

And this diagram is trying to shows the charging and discharging phases for a sine-wave input voltage. 



All this analysis assumes steady-state situation (The circuit is ON for a very long time).