A particle (q-4.0 mC.m-50 g) has a velocity of 25 m/s in the positive x direction when it first enters a region where the electric field is uniform (60 N/C in the positive y direction). What is the speed of the particle 5.0 s after it enters this region?

Question

is simple steps
HINT: the answer is 35m\s

Answer 1

To solve this problem, we can use the equation:

F = qE

Where F is the force experienced by the particle, q is its charge, and E is the electric field. Since the force is in the y-direction, we can rewrite the equation as:

Fy = qEy

Using Newton's second law, we know that force equals mass times acceleration:

F = ma

Therefore, we can rewrite the equation as:

ma = qEy

Since we know the mass of the particle is 50 g (or 0.05 kg), the charge is 4.0 mC (or 4.0 * 10^-3 C), and the electric field is 60 N/C, we can substitute these values into the equation:

0.05a = (4.0 * 10^-3)(60)

Simplifying the right side of the equation:

0.05a = 0.24

Now, we can solve for acceleration:

a = 0.24 / 0.05 = 4.8 m/s^2

Acceleration is the rate of change of velocity over time, so we can use the equation:

a = (vf - vi) / t

Rearranging the equation to solve for the final velocity:

vf = a * t + vi

Plugging in the given values:

vf = (4.8)(5.0) + 25

vf = 24 + 25

vf = 49 m/s

However, the question asks for the speed of the particle, which is the magnitude of its velocity. In this case, since the particle is moving in the positive x direction, the velocity in the y direction does not contribute to the speed. Therefore, the speed is simply the magnitude of the x component of the velocity, which is 25 m/s.