Question

A particle of mass 1 g and charge 2.5 × 10⁻⁴ C is released from rest in an electric field of 1.2 × 10 ⁴ N C⁻¹. What will be the speed of the particle after travelling this distance? 

Answer 1

Given:
Charge of the particle, q = 2.5 × 10⁻⁴ C
Initial velocity, u = 0 
Electric field intensity, E = 1.2 × 10⁴ N/C
Mass of the particle, m = 1 g = 10⁻³ kg
Distance travelled, s = 40 cm = 4 × 10⁻¹ m

 Using the third equation of motion, we get

\[v^2  =  u^2  + 2as\] 

\[ \Rightarrow  v^2  = 0 + 2 \times 3 \times  {10}^3  \times 4 \times  {10}^{- 1} \] 

\[ \Rightarrow v = 4 . 9 \times 10 = 49  \text{ m/s }\]