Charged capacitor as output
You seem to be a bit confused between energy and power, which is the rate of energy transfer. If no current is going into or out of the capacitor, the energy it stores is not changing (\$E = \dfrac{1}{2} \cdot \dfrac{Q^2}{C}\$, and \$Q\$ isn't changing). The power you describe that is being input into the circuit must therefore be dissipated somewhere else in the circuit, for example by heating up a resistor.