Two parallel metal plates P and …

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search-icon-header Search for questions & chapters search-icon-image Class 12 >>Physics >>Electrostatic Potential and Capacitance >>Effects of Dielectrics in Capacitors >>Two identified parallel plate capacitors Question Bookmark Two identified parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. 623081 expand Medium Updated on : 2022-09-05 Solution verified Verified by Toppr Initially charge on either capacitor is Q A ​ =Q B ​ =CV . After dielectric is introduced, new capacitance of either capacitor =KC After opening switch potential across capacitor A is V volts. Let potential across capacitor B be V 1 ​ Therefore Q B ​ =CV=C 1 ​ V 1 ​ =KCV 1 ​ V 1 ​ = K V ​ Initial energy in both capacitors = 2 CV 2 ​ + 2 CV 2 ​ =CV 2 Final energy of capacitor A = 2 KCV 2 ​ Final energy of capacitor B = 2K 2 KCV 2 ​ = 2K CV 2 ​ Total final energy of both capacitors = 2 KCV 2 ​ + 2K CV 2 ​ =( 2K K 2 +1 ​ )CV 2 Ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric= ( 2K K 2 +1 ​ )CV 2 CV 2 ​ = K 2 +1 2K ​