Final answer:
Insertion of a dielectric into a fully charged capacitor that remains connected to a battery results in an increase in capacitance and stored charge, while the voltage remains constant and the energy stored in the capacitor increases.
Step-by-step explanation:
Impact of Inserting a Dielectric into a Parallel-Plate Capacitor
When a dielectric is inserted into a fully charged parallel-plate capacitor that remains connected to a battery, the following changes occur:
The capacitance (C) of the capacitor increases because the dielectric increases the capacitor's ability to store charge.
The charge (Q) on the capacitor increases as the capacitance increases for a given voltage, because Q = CV, where V is the voltage.
The voltage (ΔV) across the capacitor stays the same because the capacitor is still connected to the battery, which maintains constant voltage.
The energy stored in the capacitor increases since the energy is given by the formula ½ CV2, and both capacitance and voltage have remained constant or increased.
Note that if the capacitor were disconnected from the battery before the introduction of the dielectric, the voltage across the capacitor would decrease because the charge would remain constant while the capacitance increased.