A ball of mass 100 g and having a charge of  4⋅9 x 10 − 5 C is released from test in a legion where a horizontal electric field of 2⋅0 x 10 4 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? 

⚡ Motion of a Charged Ball in an Electric Field

Problem: A ball of mass 100 g carrying a charge of 4.9 × 10⁻⁵ C is released from rest in a horizontal electric field of 2.0 × 10⁴ N/C.

(a) Resultant Force on the Ball

F = qE
F = (4.9 × 10⁻⁵) × (2.0 × 10⁴) = 0.98 N

(b) Path of the Ball

The ball moves in a parabolic path. It accelerates:

• Horizontally due to the electric field
• Vertically under gravity

Similar to projectile motion.

(c) Position After 2 Seconds

• Horizontal Acceleration: aₓ = F/m = 0.98 / 0.1 = 9.8 m/s²
• x = ½ aₓ t² = ½ × 9.8 × 4 = 19.6 m
• y = ½ g t² = ½ × 9.8 × 4 = 19.6 m

✅ Final Answer:

Resultant Force: 0.98 N
Path: Parabola
Position after 2 s: (19.6 m, 19.6 m)