A small signal voltage V(t) = V_o sin $ωt$ is applied across an ideal capacitor C:

over a full cycle the capacitor C does not consume any energy from the voltage source

current I(t) is in phase with voltage V(t)

Current I(t) leads voltage V(t) by 180^o

Current I(t) lags voltage V(t) by 90^o

Solution

over a full cycle the capacitor C does not consume any energy from the voltage source

For an ac circuit containing capacitor only, the phase difference between current and voltage will be $straight pi over 2$ (i.e., 90^o).
In this case, the current is ahead of voltage by $straight pi over 2$
Therefore, power is given by,
P = VI cos $straight ϕ$
where, $straight ϕ$ is the phase difference between voltage and current.
P = VI cos 90^o = 0