Answer
a. 0.350 mC.
b. Zero.
Work Step by Step
Performing the actions described produces an LC circuit with initial current through the inductor of 3.50 A. The current will oscillate over time.
a. Use conservation of energy. The maximum energy stored in the inductor, $\frac{1}{2}Li_{max}^2$, equals the maximum energy stored in the capacitor, $\frac{1}{2}\frac{q_{max}^2}{C}$.
$$\frac{1}{2}Li_{max}^2=\frac{1}{2}\frac{q_{max}^2}{C}$$
$$ q_{max}=(\sqrt{LC})i_{max}$$
$$ q_{max}=(\sqrt{(2.0\times10^{-3}H)(5.0\times10^{-6}F)})(3.50A)=0.350mC$$
b. When the charge q is a maximum, the capacitor has all the energy, and the current is zero.
