ch 23 physics serway "electric field

Question

a particle (q=4.0 mc , m=50g) has a velocity of 25m/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's the speed of particle 5.0 s after it enters this region ?

 

a.49 m/s

b.35 m/s

c.32 m/s

d.44 m/s

e.24 m/s

 

Answer 1

Velocity is a vector quantity,

            _{∧           ∧}

v = v_x  x + v_y  y

             _{∧                                                                              ∧}

Where x is the unit vector in the x direction and y is the unit vector in the y direction.  Since the force is in the +y direction, the particle's speed in the x-direction remains unchanged at:

 

v_x = 25 m/s. 

 

Initially, the particle has no velocity in the y-direction.  When it enters the region with the electrostatic field, it experiences a constant force of F = qE in the y-direction, where q = 4.0 mc = 0.004 c and E = 60 N/c.  Since the force is a constant, the particle undergoes a constant acceleration, a:

 

F = ma = qE

a = qE/m

 

The particle's velocity in the y-direction, v_y, will increase in time at a constant rate:

 

v_y = a*t = (qE/m)*t

• q = the particle's charge = 4 mc = 0.004 c
• E = the electric field strength in the y-direction = 60 N/c
• m = the particle's mass = 50 g = 0.05 kg
• t = time in seconds

 
To compute speed at time t = 5 s, set t = 5 and compute the magnitude of the velocity vector:

 

speed = √(v_x²+v_y²)