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.
$$\begin{gathered} R^2={F_g}^2+{F_e}^2 \\ =(0.1\times 9.8{)}^2+(4.9\times {10}^{-5}\times 2\times {10}^4{)}^2 \\ =0.9604+96.04\times {10}^{-2} \\ =1.9208\mathrm{N} \\ \Rightarrow R=1.3859\mathrm{N} \end{gathered}$$

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,
$$\begin{gathered} y=\frac{1}{2}g{t}^2 \\ \Rightarrow y=\frac{1}{2}\times 9.8\times 2\times 2=19.6\mathrm{m} \end{gathered}$$

Both the forces are same.
So, vertical displacement in 2 s = Horizontal displacement in 2 s
Net displacement$=\sqrt{{\left(19.6\right)}^2+{\left(19.6\right)}^2}=\sqrt{768.932}=27.7\mathrm{m}$