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
⇒