A parallel-plate capacitor of capacitance 50 F is fully charged using a 100 V battery. The battery is then disconnected. A dielectric material of dielectric constant K = 5 is subsequently inserted to completely fill the space between the plates. The magnitude of the decrease in potential difference and the magnitude of the loss in electrostatic energy, respectively, are

Step-by-step solution

Initial charge on the capacitor is Q = CV = 50 10⁻⁶ 100 = 5 10⁻³ C . Since the battery is disconnected, the charge Q remains constant. The new capacitance after inserting the dielectric is C' = KC = 5 50 F = 250 F . The new potential difference is V' = Q C' = 100 5 = 20 V . The magnitude of the decrease in potential difference is V = 100 - 20 = 80 V . The initial electrostatic energy is U_i = 1 2 CV^2 = 1 2 50 10⁻⁶ (100)^2 = 0.25 J . The final electrostatic energy is U_f = Q^2 2C' = U_i K = 0.25 5 = 0.05 J . The ma