Answer
The maximum charge that will appear on the capacitor is $~~1.8\times 10^{-4}~C$
Work Step by Step
We can use conservation of energy to find the maximum charge that will appear on the capacitor:
$U_{E,max} = U_{B,max}$
$\frac{q^2}{2C} = \frac{L~i^2}{2}$
$q^2 = C~L~i^2$
$q = \sqrt{C~L~i^2}$
$q = \sqrt{(2.70\times 10^{-6}~F)(3.00\times 10^{-3}~H)(2.00~A)^2}$
$q = 1.8\times 10^{-4}~C$
The maximum charge that will appear on the capacitor is $~~1.8\times 10^{-4}~C$
