Say we have a battery at a fixed voltage of $V$ charging a capacitor with capacitance $C$.  When the capacitor is fully charged, the energy stored in it will be $\frac{1}{2}QV$, where $V$ is the final voltage of the capacitor (and the voltage of the battery).  However, I'm a little confused as to the work the battery does...  To me, it seems that if the battery pushed $Q$ onto the capacitor, and the voltage of the battery is $V$, it must've done $QV$ of work, which means that the energy stored in the capacitor only accounts for half of it.  Where did the rest of the work the battery did go?
For a simple circuit with a battery supplying a voltage $V$ to a capacitor, let us assume that the charge on the capacitor is $Q$. Now, the work done by the battery or the energy supplied is given by the relation:  $$W=QV$$  But the energy stored in the capacitor is given by:  $$U = \tfrac12 QV$$  The value of $Q$ as well as that of $V$ should be the same in both the equations.  Now my question is, where is the other half of the energy that the battery supplied?
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EDIT:  When I run:  "INSERT INTO INVOICE (INVOICE_DATE, BOOK_FEE, ADMIN_FEE, TOTAL_FEE, CUSTOMER_ID) VALUES ('2009:01:01 10:21:12', 50, 7, 57, 2134);last_insert_id();"   I get:  {"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'last_insert_id()' at line 1"}