Answer
The maximum energy stored on the capacitor is $~~6.88\times 10^{-9}~J$
Work Step by Step
We can find the maximum charge $Q$ on the capacitor:
$Q = V~C = (0.250~V)(220\times 10^{-9}~F) = 55.0~nC$
We can find the maximum energy stored on the capacitor:
$U_E = \frac{Q^2}{2C}$
$U_E = \frac{(55.0\times 10^{-9}~C)^2}{(2)(220\times 10^{-9}~F)}$
$U_E = 6.88\times 10^{-9}~J$
The maximum energy stored on the capacitor is $~~6.88\times 10^{-9}~J$
