Final answer:
Inserting a dielectric into a capacitor increases the capacitance by a factor equal to the dielectric constant K, reduces the electric field and voltage across the capacitor plates, and changes the energy stored in the system.
Step-by-step explanation:
Impact of a Dielectric on Capacitance
When a dielectric material is inserted into a capacitor, the capacitance of the system is affected. The dielectric constant, denoted as K, is a measure of how much the dielectric can reduce the electric field within the capacitor. According to the principles of physics, if a dielectric with a dielectric constant K is inserted into an empty capacitor, the new capacitance C becomes C = K × Co, where Co is the original capacitance without the dielectric. Therefore, the capacitance of the capacitor is increased by a factor of K.
The insertion of a dielectric also affects the electric field and the voltage across the plates. The electric field strength is reduced because the dielectric material causes a portion of the electric field lines to end on induced charges within the dielectric rather than crossing from one plate to another. This reduction in field strength means that the voltage across the plates, given by V = Ed (where E is the electric field and d is the distance between the plates), is also reduced. This results in increasing the capacitance since the same charge Q now sees a smaller voltage V, with capacitance being defined as C = Q/V.
In summary, inserting a dielectric into a capacitor increases the capacitance of the capacitor while reducing the electric field and voltage across the plates. The energy of the system will change as a result of the work done in inserting the dielectric, and this energy can be calculated using the formula involving the newly adjusted capacitance and voltage.