Question

A particle (m = 20 mg, q = -5.0 μC) moves in a uniform electric field of 60 N/C in the positive x direction. At t = 0, the particle is moving 30 m/s m the positive x direction and is passing through the origin. Determine the maximum distance beyond x = 0 the particle travels in the positive x direction. a. 25 m b. 20 m c. 15 m d. 30 m e. 60 m

Answer 1

The correct answer is d d= 30 m

Step-by-step explanation:

The electric force is given by

Fe = q E

Since the particle charge is negative, the force and acceleration oppose the movement, let's use Newton's second law

F = m a

q E = m a

a = q E / m

Let's reduce

q = -5.0 mC = -5.0 10⁻⁶ C

m = 20 mg = 20 10⁻³ g (1kg / 1000 g) = 20 10⁻⁶ kg

a = - 5.0 10⁻⁶ 60/20 10⁻⁶

a = -15 m / s²

Now we can use kinematics

v² = v₀² + 2 a d

At the point where stop (v=0)

0 = v₀² + 2 a d

d = - v₀² / 2 a

d = 30² / (2 -(15))

d = 30 m

The correct answer is d