Question

A particle (mass = 4.0 g, charge = 80 mC) moves in a region of space where the electric field is uniform and is given by Ex = -2.5 N/C, Ey = Ez = 0. If the velocity of the particle at t = 0 is given by vx = 80 m/s, vy = vz = 0, what is the speed of the particle at t = 2.0 s?

Answer 1

Answer:

20 m/s

Explanation:

The force experienced by a charged particle in an electric field is given by

$F=qE$

where, in this problem:

$q=80 mC=0.080 C$ is the charge of the particle

E is the electric field

The electric field here has components:

$E_x=-2.5 N/C\\E_y=0\\E_z=0$

So the components of the force experienced by the particle are:

$F_x=qE_x=(0.080)(-2.5)=-0.2 N\\F_y=0\\F_z=0$

Now we can find the components of the acceleration experienced by the particle, using Newton's second law of motion:

$a=\frac{F}{m}$

where

m = 4.0 g = 0.004 kg is the mass of the particle

The 3 components of the acceleration are:

$a_x=\frac{F_x}{m}=\frac{-0.2}{0.004}=-50 m/s^2\\a_y=0\\a_z=0$

Now we can find the components of the velocity of the particle at time t using the suvat equation:

$v=u+at$

where:

$u_x=80 m/s\\u_y=0\\u_z=0$

are the initial components of the velocity

Therefore, at t = 2.0 s, we have:

$v_x=u_x+a_xt=80+(-50)(2.0)=-20 m/s\\v_y=u_y+a_yt=0+0=0\\v_z=u_z+a_zt=0+0=0$

And so, the speed of the particle is the magnitude of the final velocity:

$v=\sqrt{v_x^2+v_y^2+v_z^2}=\sqrt{(-20)^2+0+0}=20 m/s$