Final answer:
Inserting a dielectric into a capacitor reduces the electric field within the capacitor due to the induced charge on the dielectric material, which generates an opposing field. This results in a lower voltage for the same charge and an increased capacitance as the energy polarizes the dielectric material.
Step-by-step explanation:
When a dielectric material is inserted into a capacitor, the electric field inside the capacitor is altered. In an empty capacitor connected to a battery, the electric field Eo is determined by the voltage Vo supplied by the battery and the separation of the plates. With the introduction of a dielectric of dielectric constant κ (kappa), the capacitance increases because a dielectric reduces the electric field strength within the capacitor.
This reduction in electric field, E, can be understood by considering the induced charge on the dielectric material, which generates its own electric field that opposes the original field Eo. There are fewer electric field lines extending from one side of the capacitor to the other because they end on charges in the dielectric. Consequently, for the same charge Q, the voltage V across the capacitor plates is reduced (V = Ed), which means the capacitance C (C = Q/V) is greater. The energy from the electric field goes into polarizing the dielectric material, effectively stretching it like tiny springs, and reduces the electric field in the process.
Overall, the electric field within a capacitor containing a dielectric is weaker than in a vacuum, resulting in a lower voltage needed to maintain the same amount of charge on the plates. Therefore, a dielectric increases the overall capacitance of a capacitor while reducing the electric field.