Answer :


Given –


Plate area 20 cm2 = 0.002m2


Separation between the plates 1.00 mm = 0.001m


Battery Voltage = 12.0 V


We know capacitance, C


1)



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space


= 8.854 × 10-12 m-3 kg-1 s4 A2


From1),


Capacitance when distance d = 0.001m, C1



Substituting values,


= 2)


When The plates are pulled apart to increase the separation to –


2.0 mm = 0.002m, then capacitance C2 becomes,



Substituting values


= 3)


From 1) and 2)


4)


We know capacitance in terms of voltage is given by –


Q = CV 5)


Where


Q= charge stored on the capacitor


C= capacitance of the capacitor


V = applied voltage


Given applied v = 12V


From 2) and 3) and 5)



=


= 6)


And



7)


a) The charge flown through the circuit during the process



From 6) and 7)



=


=


= 8)


b)Energy absorbed by the battery during the process-


We know Energy E is given by -


E=QV 9)


Where


Q = charged present on the surface


V= Applied voltage


From 8),



Applied voltage V = 12V


From 9),


Energy absorbed,



=


c)Stored energy in the electric field before and after the process


We know that stored energy in the electric field,



Before process, the energy stored -



From 2),



10)


From 3), After process, the energy stored will become



11)


d) The work done by the person pulling the plates apart.


We know, work done, W


12)


Where, f = force


d = displacement


We, know in parallel plate capacitor, the force between the plates is given by.


13)


Where, Q = charge enclosed,


σ = surface charge density,


σ, surface charge density is given by ,



From 12) and 13)


Work done,


Given, Plate area 20 cm2 = 0.002m2


Separation between the plates changed to, 2.00 mm = 0.002 m by


pulling apart.


Substituting values, from 7),






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