Answer
See the detailed answer below.
Work Step by Step
The mentioned equation is
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
And for only the capacitor circuit, $R=0$ and $X_L=0$
$$\phi=\tan^{-1}\left[\dfrac{0-X_C}{0}\right]$$
where we know that $\dfrac{x}{0}=\infty$, thus,
$$\phi=\tan^{-1}\left[-\infty\right]=\dfrac{-\pi}{2}$$
This means that the voltage lags the current by an angle of $\pi/2$ which is the case for a capacitor.
