A charged particle of mass 20 mg and charge 5 C is projected in the direction opposite to a uniform electric field. It comes to rest momentarily after travelling a distance of 0.4 m in 0.02 s . The magnitude of the electric field is:

Step-by-step solution

Let the initial velocity be u and the constant deceleration be a . Since the final velocity is zero, using the equations of motion backward from rest, the distance s covered in time t is given by: s = 1 2 a t^2 a = 2s t^2 = 2 0.4 (0.02)^2 = 0.8 4 10⁻⁴ = 2000 m s ⁻² The deceleration is provided by the electric force, so a = qE m . E = ma q = 20 10⁻⁶ 2000 5 10⁻⁶ = 4 2000 = 8000 V m ⁻¹ Answer: 8000 V m ⁻¹