A parallel-plate

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

⇒