Q45 of 94 Page 166

A capacitor having a capacitance of 100 μF is charged to a potential difference of 24V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12V with the positive plate of the capacitor joined with the positive terminal of the battery.

a) Find the charges on the capacitor before and after the reconnection.


b) Find the charge flown through the 12V battery.


c) Is work done by the battery or is it done on the battery? Find its magnitude.


d) Find the decrease in electrostatic field energy.


e) Find the heat developed during the flow of charge after reconnection.


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.


More from this chapter

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43

A large conducting plane has a surface charge density 1.0 × 10–4 C m–2. Find the electrostatic energy stored in a cubical volume of edge 1.0 cm in front of the plane.

 44

A parallel-plate capacitor having plate area 20 cm2 and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm.

a) Calculate the charge flown through the circuit during the process.


b) How much energy is absorbed by the battery during the process?


c) Calculate the stored energy in the electric field before and after the process.


d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart.


e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.


 46

Consider the situation shown in figure. The switch S is open for a long time and then closed.

a) Find the charge flown through the battery when the switch S is closed.


b) Find the work done by the battery.


c) Find the change in energy stored in the capacitors.


d) Find the heat developed in the system.



 47

A capacitor of capacitance 5.00 μF is charged to 24.0V and another capacitor of capacitance 6.0 μF is charged to 12.0V.

a) Find the energy stored in each capacitor.


b) The positive plate of the first capacitor is now connected to the negative plate of the second nd vice versa. Find the new charges on the capacitors.


c) Find the loss of electrostatic energy during the process.


d) Where does this energy go?