An -particle of mass 6.4 10⁻²⁷ ~kg and charge 3.2 10⁻¹⁹ C is situated in a uniform electric field of 1.6 10⁵ Vm ⁻¹ . The velocity of the particle at the end of 2 10⁻² ~m path when it starts from rest is

Step-by-step solution

Given, m_ =6.4 10⁻²⁷ ~kg , q _ =3.2 10⁻¹⁹ C , E =1.6 10⁵ Vm ⁻¹ Force on -particle gathered F = q _ E =3.2 10⁻¹⁹ 1.6 10⁵ =51.2 10⁻¹⁵ ~N gathered Now, acceleration of the particle = F m _ = 51.2 10⁻¹⁵ 6.4 10⁻²⁷ =0.8 10¹³ ~ms ⁻² Initial velocity, u =0 v ²=2 S =2 8 10¹² 2 10⁻²=32 10¹⁰ or v =4 2 10⁵ ~ms ⁻¹