Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectric Field and Potential
A ball having a mass of 100 g and a charge of 4.9 10⁻⁵ C is released from rest into an area containing a horizontal electric field of 2.0 10^4 N C ⁻¹ . Where will the ball be located after 2 s ?
Options
- A28 m from the starting point along its line of motion
- B14 m from the starting point along its line of motion
- C20 m from the starting point along its line of motion
- D56 m from the starting point along its line of motion
Correct answer
A. 28 m from the starting point along its line of motion
Step-by-step solution
Horizontal force acting on the ball: F_x = qE = (4.9 10⁻⁵) (2.0 10^4) = 0.98 N Vertical force acting on the ball (taking g = 10 m s ⁻² ): F_y = mg = 0.1 10 = 1 N Net force acting on the ball: F = F_x^2 + F_y^2 = (0.98)^2 + (1)^2 = 0.9604 + 1 = 1.9604 1.4 N Acceleration of the ball: a = F m = 1.4 0.1 = 14 m s ⁻² Distance traveled in t = 2 s : s = 1 2 at^2 = 1 2 14 (2)^2 = 28 m