Answer :

Given



Capacitance =100 μF


Initial battery voltage used = 24V


Second voltage used = 12V


a) Charges on the capacitor before and after the reconnection.


Before reconnection, the battery used is 24V, hence




We know charge present on a capacitor is given by


q = CV 1)


Where q = charge


C=capacitance


V=voltage


Substituting in 1)






Similarly, after connection of 12V battery –


When C = 100μf


V = 12V


Using equation 1)





b) Charge flown through the 12V battery.


C = 100, V = 12V


From 1),



Substituting the values,





c) Work is done by the battery, and its magnitude is as follows


We know



Where V = applied voltage across the capacitor


W = work done and


q = charge on the surface of the parallel plate capacitor


Which gives,





is the amount of work done on the battery.


d) Decrease in electrostatic field energy


Electrostatic field energy stored is given by –



, c = capacitance


v= voltage across capacitor


Initially, electrostatic field energy stored is given by -



Final Electrostatic field energy



Decrease in Electrostatic field energy


=


=


=


Substituting the values


Capacitance =100 μF


Initial battery voltage used = 24V


Second voltage used = 12V


=



e) Heat developed during the flow of charge after reconnection


After reconnection


C = 100μc, V = 12v


We know the energy stored, E in capacitor is given by



Where c is the capacitance and v is the applied voltage


Substituting values –




This is the amount of energy developed as heat when the charge flows through the capacitor.


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