Final answer:
When a dielectric material with dielectric constant k is inserted into an isolated, charged air capacitor, the stored energy decreases. This is because the capacitance of the capacitor increases by a factor of k, leading to a decrease in energy stored by the same factor, as long as the charge on the capacitor remains constant.
Step-by-step explanation:
Effect of Dielectric on Stored Energy in a Capacitor
When a dielectric material with a dielectric constant k is inserted into an air capacitor that is isolated and charged, the energy stored in the capacitor decreases. This is because the dielectric increases the capacitance of the capacitor by a factor of k while the charge remains constant, consequently decreasing the potential difference across the capacitor plates. The energy stored in the capacitor is given by the equation U = Q^2 / (2C), where U is the stored energy, Q is the charge, and C is the capacitance. Since the charge Q remains the same and the capacitance C increases, the stored energy U decreases by a factor of k. So, with the introduction of the dielectric, the energy stored in the capacitor decreases.
As an example, if a Teflon™ dielectric with a dielectric constant of 2.1 is inserted between the plates of an isolated capacitor, the capacitance would increase by a factor of 2.1, and the energy stored would decrease by the same factor, assuming the charge on the capacitor remains unchanged.