#### Question

A parallel-plate capacitor with the plate area 100 cm^{2} 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.

#### Solution

Area of the plates of the capacitor, *A* = 100 cm^{2} = `10^-2 "m"^2`

Separation between the plates, *d* = 1 cm = `10^-2 "m"`

Emf of battery, *V* = 24 V

Therefore,

Capacitance , `C = (∈_0A)/d = ((8.85 xx 10^-12) xx (10^-2))/(10^-2) = 8.85 xx 10^-12 V`

Energy stored in the capacitor,

`E = 1/2 CV^2 = 1/2 xx (8.85 xx 10^-12) xx (24)^2`

= `2548.8 xx 10^-12 "J"`

Force of attraction between the plates, `F = E/d = (2548.8 xx 10^-12)/(10^-2) = 2548.8 xx 10^-10 "N"`

Is there an error in this question or solution?

Solution 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. Concept: Capacitors and Capacitance.