Answer
$\phi = 46.9^{\circ}$
Work Step by Step
In part (b) we found that the maximum charge on the capacitor is $~~Q = 5.56\times 10^{-6}~C$
It is given that at time $t = 0$, the charge on the capacitor is $q = 3.80\times 10^{-6}~C$
We can find $\phi$:
$q = Q~cos~(\omega t+\phi)$
$cos~(\omega t+\phi) = \frac{q}{Q}$
$cos~[\omega (0)+\phi] = \frac{3.80\times 10^{-6}~C}{5.56\times 10^{-6}~C}$
$cos~\phi = \frac{3.80\times 10^{-6}~C}{5.56\times 10^{-6}~C}$
$\phi = cos^{-1}~(\frac{3.80\times 10^{-6}~C}{5.56\times 10^{-6}~C})$
$\phi = 46.9^{\circ}~~$ or $~~\phi = -46.9^{\circ}$
Note that the capacitor is discharging at $t = 0$, so the value of $~~cos~(\omega t+\phi)~~$ must be decreasing as $t$ increases.
Therefore, $~~\phi = 46.9^{\circ}$
