Kinetic energy of charged particle and Electron-volt

 # Kinetic energy of charged particle:

Let potential difference between two parallel charge plates, V1-V2 = V

Distance between two plates = d

Hence, electric field intensity,E = ∆V/∆X= V/d

A positively charged particle,P experience an electric force
F = q.E
F = q.(V/d)

By the Newton's law of motion
F= m.a

Hence, m.a = q.(V/d)

Or,.   a = q.V/m.d

By the third eqn of motion
v² = u² + 2as

Putting the values

v² = O +2(q.V/m.d)×d
v² = 2q.V/m
mv² = 2q.V
1/2mv² = q.V
(∆K = q.V)

Increases in k.E = charge×potential diff.

#) Electron-volt :
                   It is the small unit of energy.
It can be defined as the energy gained by an electron when it is accelerated by potential difference 1 volt .

we know that
Increase in k.E        = charge × potential diff. of charged particle                  

∆K = q × ∆V

If q = e and ∆V = 1 volt
Then, ∆K = 1 electron-volt

Hence, 1electron-volt = e × 1volt
                  = (1.6×10^-19 C)× 1volt

1eV = 1.6×10^-19   Joule