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.
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?
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1 answer