A parallel plate capacitor filled with mica having `epsilon_(r)` = 5 is connected to a `10V ` battery . The area of the parallel plate is `6 m^(2)` and separation distance is 6 mm.
(a) Find the capacitance and stored charge . (b) After the capacitor is fully charged the battery is disconnected and the dielectric is removed carefully .
`Calculated the new values of capacitance stored energy and charge.

Explanation

The capacitance of the capacitor in the presence of dielectric is
`C =(epsilon_(r)epsilon_(0)A)/(d)=(5xx8.85xx10^(-12)xx6)/(6xx10^(-3))`
`=44.25xx10^(-9)F=44.25nF`
The stored charge is Q=CV= `44.25xx10^(-9)xx10==44.2nC`
The stored energy is `U=(1)/(2)CV^(2)=(1)/(2) xx44.25xx10^(-9)xx100=2.21xx10^(-6)J=2.21muJ`
(b) After the removal of the dielectric since the battery is already disconnected the total charge will not change . But the potential difference between the the plates increases . As a result the capacitance is decreased .
New capacitance is `C_(0)=(C )/(epsilon_(r))=(44.25xx10^(-9))/(5)=8.85xx10^(-9)F=8.85nF`
The stored charge remains same and 442.5 nC. Hence newly stored energy is
`U_(0)=(Q^(2))/(2C_(0))=(Q^(2)epsilon_(r))/(2C) = epsilon_(r)U`
`=5xx2.21 muJ= 11.05 mu J`
The increased energy is `DeltaU=11.05 muJ-2.21 muJ=8.84muJ`
When the dielectric is removed it experience an inward pulling force due to the plates . To remove the dielectric an external agency has to do work on the dielectric which is stored as additional energy . This is the source for the extra energy `8.84 mu J`