A parallel-plate capacitor stores charge Q.The capacitor is then disconnected from its voltage source,and the space between the plates is filled with a dielectric of constant κ = 2.What is the relationship between the stored energies of the capacitor,where PE_iis the initial stored energy,and PE_f is the stored energy after the dielectric is inserted? A)PEf = 0.25PE_i B)PEf = 0.5PE_i C)PEf = PE_i D)PEf = 2PE_i

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the equation: PE = (1/2)CV^2 where PE is the stored energy, C is the capacitance, and V is the voltage. When the dielectric is inserted, the capacitance increases by a factor of κ, so the new capacitance is given by: C' = κC = 2C Since the charge Q on the capacitor remains the same, the voltage V decreases by a factor of κ: V' = V/κ = V/2 Therefore, the new stored energy is: PEf = (1/2)C'V'^2 = (1/2)(2C)(V/2)^2 = (1/2)CV^2/2 = (1/4)PE_i So, the stored energy is reduced by a factor of 1/4, or 0.5. Therefore, the correct answer is B: PEf = 0.5PE_i.

A Parallel-Plate Capacitor Stores Charge Q