A parallel plate capacitor with vacuum between the plates ha

A parallel plate capacitor with vacuum between the plates has a capacitance of 3.547 MuF. A dielectric material with k= 4.617 is placed between the plates, completely filling the volume between them. The capacitor is then connected to a battery that maintains a potential difference of 10.03 v across the plates. Mow much work is required to pull the dielectric material out of the capacitor?

Solution

Capacitance C = 3.547 x10 ^-6 F

Dielectric constant k = 4.617

Potential difference V = 10.03 volt

Capacitance of the capacitor with dielectric is C \' = kC = 4.617 x3.547 x10 ^-6 F

                                                                        = 16.376x10 ^-6 F

Energy stored in capacitor with dielectric U \' = (1/2)C \'V ²

Energy stored in capacitor with out dielectric U = (1/2)CV ²

Work is required to pull the dielectric material out of the capacitor,W = U \' - U

                     W = (1/2)C \'V ² -(1/2)C V ²

                         = (1/2)(C \' -C )V ²

                         = 0.5 (4.617C-C)V ²

                        = 0.5 x 3.617 CV ²

                        = 0.5x3.617 x3.547 x10 ^-6 x10.03 ²

                        = 6.453x10 ^-4 J