Alright, I have an LC circuit.
Using conservation of energy, I get \frac{dI}{dt}= \frac{-q}{CL} which gives \frac{d^2q}{dt^2}= \frac{-q}{CL}, therefore, q=q_o sin(\omega t +\phi), where \omega= \frac{1}{\sqrt {LC}}, giving q=q_o cos(\omega t) which implies I= -q_o \omega sin(\omega t).
Now...