A particle (mass = 4.0 g, charge = 50 mC) moves in a region of space where the electric field is uniform and is given by Ex=3.0N/C. If the velocity of the particle at t=0 is given by vy=40 m/s, what is the speed of the particle at t=2.0 seconds?
Jajaj
To solve this problem, we need to use the equations of motion for a particle in a uniform electric field.
The equation we will use is:
v = u + at
where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Given:
u = vy = 40 m/s
a = qE/m, where:
q is the charge of the particle (50 mC)
E is the electric field (3.0 N/C)
m is the mass of the particle (4.0 g = 0.004 kg)
t = 2.0 seconds
To find a, we can use the formula:
a = qE/m
Substituting the given values, we get:
a = (50 x 10^-3 C)(3.0 N/C) / (0.004 kg)
Simplifying this expression, we get:
a = 37.5 m/s²
Now, we can plug in the values of u, a, and t into the equation:
v = u + at
v = 40 m/s + (37.5 m/s²)(2.0 s)
Calculating this expression, we get:
v ≈ 40 m/s + 75 m/s
v ≈ 115 m/s
Therefore, the speed of the particle at t=2.0 seconds is approximately 115 m/s.