Question

A particle (mass = 5.0 g, charge = 40 mC) moves in a region of space where the electric field is uniform and is given by Ex = -5.5 N/C, Ey = Ez = 0. If the position and velocity of the particle at t = 0 are given by x = y = z = 0 and vx = 50 m/s, vy = vz = 0, what is the distance from the origin to the particle at t = 2.0 s?

Answer 1

First, we must find the acceleration of the particle in the electric field.

The electric force on the particle is given by:

F = qE

where
F - force on the particle
E - electric field
q - charge of the particle

The electric field is only in the x-direction:

Ex = -5.5 N/C
Ey = Ez = 0

The charge on the particle is:

q = 40 mC = 40 x 10^(-3) C

Now we can find the force on the particle in the x-direction:

Fx = q * Ex = (40 x 10^(-3) C) * (-5.5 N/C) = -0.22 N

Next, we find the acceleration using Newton's second law:

F = m * a

where
F - force on the particle
m - mass of the particle
a - acceleration of the particle

The mass of the particle is:

m = 5 g = 5 x 10^(-3) kg

Now we can find the acceleration in the x-direction:

ax = Fx / m = (-0.22 N) / (5 x 10^(-3) kg) = -44 m/s^2

There is no acceleration in the y and z directions:

ay = az = 0

Now we know the particle's acceleration, initial position, and initial velocity, so we can find its position at t = 2.0 s using the kinematic equations for each coordinate:

x(t) = x0 + vx * t + 0.5 * ax * t^2
y(t) = y0 + vy * t + 0.5 * ay * t^2
z(t) = z0 + vz * t + 0.5 * az * t^2

Initial position and velocity are given:

x0 = y0 = z0 = 0
vx = 50 m/s
vy = vz = 0

At t = 2.0 s, we get:

x(2) = 0 + (50 m/s) * (2 s) + 0.5 * (-44 m/s^2) * (2s)^2 = 100 m - 88 m = 12 m
y(2) = 0 + 0 * (2 s) + 0.5 * 0 * (2 s)^2 = 0
z(2) = 0 + 0 * (2 s) + 0.5 * 0 * (2 s)^2 = 0

Finally, we can find the distance from the origin to the particle at t = 2 s using the Pythagorean theorem:

distance = √(x^2 + y^2 + z^2) = √((12 m)^2 + (0 m)^2 + (0 m)^2) = √144 = 12 m

So the distance from the origin to the particle at t = 2.0 s is 12 m.