A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1.0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.

Given -


Area of the plates of the capacitor, A = 100 cm2 = 10-2 m2


Separation between the plates, d = 1 cm = 10-2 m


Emf of battery, V = 24 V


We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


Therefore,



Capacitance,


=




Energy stored by the capacitor





Now, force of attraction between the plates,



where


E = energy stored and d is the separation between the plates





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