Physics
Physics
Problem
A parallel-plate air capacitor of capacitance is connected to a battery. A dielectric slab of relative permittivity is then completely inserted between the plates while the capacitor remains connected to the battery. a. (i) Calculate the charge on the capacitor and the energy stored in it before the slab is inserted. b. (ii) Calculate the charge and the energy stored after the slab is fully inserted. c. (iii) Determine the ratio of the final energy to the initial energy.Intuitive explanation
The key insight is that keeping the battery connected fixes the voltage, so the capacitor's behavior is completely different from the isolated-charge case. When you insert the dielectric, the capacitance increases by a factor of , which means more charge flows from the battery onto the plates to maintain that constant voltage. Focus on how capacitance, charge, and energy all change—but voltage stays put.
Intuitive explanation
The key insight is that keeping the battery connected fixes the voltage, so the capacitor's behavior is completely different from the isolated-charge case. When you insert the dielectric, the capacitance increases by a factor of , which means more charge flows from the battery onto the plates to maintain that constant voltage. Focus on how capacitance, charge, and energy all change—but voltage stays put.