Answer
The maximum charge on the capacitor is $~~5.56\times 10^{-6}~C$
Work Step by Step
In part (a), we found that the total energy in the circuit is $1.98\times 10^{-6}~J$
We can find the maximum charge $q$ on the capacitor:
$U_E = 1.98\times 10^{-6}~J$
$\frac{q^2}{2C} = 1.98\times 10^{-6}~J$
$q^2 = (1.98\times 10^{-6}~J)(2C)$
$q = \sqrt{(1.98\times 10^{-6}~J)(2)(7.80\times 10^{-6}~C)}$
$q = 5.56\times 10^{-6}~C$
The maximum charge on the capacitor is $~~5.56\times 10^{-6}~C$
