Question

A ball of mass 100 g and with a charge of 4.9 × 10⁻⁵ C is released from rest in a region where a horizontal electric field of 2.0 × 10⁴ N C⁻¹ exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?

Answer 1

Given:
Charge of the ball, q = 4.9 × 10⁻⁵ C
Electrical field intensity, E = 2 × 10⁴ N/C
Mass of the ball, m = 100 gm
Force of gravity, F_g = mg 
Electrical force, F_e = Eq
The particle moves due to the resultant force of F_g and F_e.  

\[R^2  =  {F_g}^2  +  {F_e}^2 \] 

\[       = (0 . 1 \times 9 . 8 )^2  + (4 . 9 \times  {10}^{- 5}  \times 2 \times  {10}^4  )^2 \] 

\[       = 0 . 9604 + 96 . 04 \times  {10}^{- 2} \] 

\[       = 1 . 9208  N\] 

\[ \Rightarrow R = 1 . 3859  N\] 

 F_g = F_e
⇒ tanθ = 1
⇒ θ = 45°
θ is the angle made by the horizontal with the resultant.
Hence, the path of the ball is straight and is along the resultant force at an angle of 45° with the horizontal
Vertical displacement in t = 2 s,

\[y = \frac{1}{2}g t^2 \] 

\[ \Rightarrow y = \frac{1}{2} \times 9 . 8 \times 2 \times 2 = 19 . 6  \] m

Both the forces are same.
So, vertical displacement in 2 s = Horizontal displacement in 2 s
Net displacement

\[= \sqrt{\left( 19 . 6 \right)^2 + \left( 19 . 6 \right)^2} = \sqrt{768 . 932} = 27 . 7  \] m