Aparticle (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

The correct answer is d     d= 30 m

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

it's a b and

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