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A parallel plate capacitor filled with mica having epsilon_(r) = 5 is connected to a 10Vbattery . 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. |
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Answer» Solution :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` |
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