A particle (mass = 6.7 *10-27 kg, charge = 3.2 * 10-19 C) moves along the positive x axis with a speed of 4.8 *105 m/s.It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field.What is the magnitude of the electric field?

Question

Quizplus · Accepted Answer

We can use the equation of motion in uniform acceleration to find the time taken by the particle to come to rest:

v^2 = u^2 + 2as

where v = 0 (final velocity), u = 4.8 x 10^5 m/s (initial velocity), s = 2 m (distance travelled in the field). Solving for time (t), we get:

t = s/u = 4.17 x 10^-6 s

Now, we can use the equation for electric force:

F = qE

where F is the force on the particle, q is the charge on the particle, and E is the magnitude of the electric field. Since the particle comes to rest, the electric force must balance the force due to motion:

F = ma = qE

where m is the mass of the particle and a is the acceleration. Solving for E, we get:

E = F/q = ma/q = (6.7 x 10^-27 kg)(4.8 x 10^5 m/s)/(3.2 x 10^-19 C) = 1.2 x 10^3 N/C

Therefore, the magnitude of the electric field is 1.2 kN/C, which is answer choice C.

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