Answer :


Given:


C1=5 μF


V1=24 V


To calculate the charge present on the capacitor, we use the formula



where,


c = capacitance of the capacitor and


v = voltage across the capacitor


For first capacitor, the stored charge q1 is given by





Similarly for second capacitor, the stored charge q2 is given by-



Given, C2=6 μF and V2=12




a) Energy stored in each capacitor-


Energy stored in a capacitor is given by



Where, v = applied voltage


C =capacitance


For capacitor C1, energy stored is given by



=



Similarly, for capacitor C2, energy stored is given by





b) New charges on the capacitors when the positive plate of the first capacitor is now connected to the negative plate of the second nd vice versa



The capacitors are connected as shown on the right hand side.


The positive of first capacitor is connected to the negative of the second capacitor.


So charge flows from positive of first capacitor to the negative of the second capacitor.


Then, the net charge for connected capacitors becomes





Now, let V be the common potential of the two capacitors


Since, charge is conserved, we know that electric charge can neither be created nor be destroyed, hence net charge is always conserved.



From the conservation of charge before and after connecting, we get, common voltage V





We know,



where v = applied voltage and C is the capacitance


Using above relation, the new charges becomes-








c) Loss of electrostatic energy during the process


Energy stored in a capacitor is given by


1)


Where, v = applied voltage


C =capacitance


For capacitor C1, energy stored is given by



=


Similarly, for capacitor C2 , energy stored is given by





=


2)


But given


for c1, actual V1 = 24V


and c2, actualV2 = 12V


Using 1)



And



Now, change in energy,



3)


From 2) and3)


Loss of electrostatic energy =



d) This energy, which is lost as electrostatic energy gets converted and dissipated from the capacitor in the from of heat energy.


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