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?
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.
To find the distance from the origin to the particle at t = 2.0 s, we can use the equations of motion under constant acceleration and the electric field provided.
1. First, let's find the acceleration experienced by the particle due to the electric field.
The electric force experienced by a charged particle is given by F = qE, where F is the force, q is the charge, and E is the electric field.
The acceleration can be calculated using the formula F = ma, where m is the mass of the particle.
So, a = F/m = (qE)/m.
Given:
Charge (q) = 40 mC = 40 × 10^(-3) C
Electric field (E) = -5.5 N/C (Note: Neglect the negative sign as it only indicates direction)
Mass (m) = 5.0 g = 5.0 × 10^(-3) kg
Therefore, the acceleration (a) experienced by the particle due to the electric field is:
a = (qE) / m = (40 × 10^(-3) C) × (-5.5 N/C) / (5.0 × 10^(-3) kg)
a = -220 N / kg
2. Next, we can use the equations of motion to find the displacement of the particle at t = 2.0 s.
x = x0 + v0xt + (1/2)at^2
y = y0 + v0yt + (1/2)at^2
z = z0 + v0zt + (1/2)at^2
Given:
x0 = y0 = z0 = 0 (Initial positions)
v0x = 50 m/s (Initial velocity in the x-direction)
v0y = v0z = 0 (Initial velocities in the y and z-directions)
t = 2.0 s (Time)
Using the given information, we can calculate the displacements in each coordinate:
x = (1/2)at^2 = (1/2)(-220 N / kg)(2.0 s)^2
y = (1/2)at^2 = (1/2)(-220 N / kg)(2.0 s)^2
z = (1/2)at^2 = (1/2)(-220 N / kg)(2.0 s)^2
3. Finally, the distance from the origin to the particle at t = 2.0 s is given by the magnitude of the displacement vector:
distance = √(x^2 + y^2 + z^2) = √[x^2 + y^2 + z^2] = √[(1/2)(-220 N / kg)(2.0 s)^2]^2 + [(1/2)(-220 N / kg)(2.0 s)^2]^2 + [(1/2)(-220 N / kg)(2.0 s)^2]^2
Evaluate the equation to find the final answer.
Please note that in this solution, I have neglected any other forces acting on the particle and assumed that the initial velocity is in the x-direction only.
To find the distance from the origin to the particle at t = 2.0 s, we need to determine the particle's position at that time.
We can start by finding the acceleration of the particle using Newton's second law:
F = ma
In this case, the force experienced by the particle is given by the electric force, which is given by:
F = qE
where q is the charge of the particle and E is the electric field. Plugging in the values, we have:
F = (40 × 10^-3 C)(-5.5 N/C) = -0.22 N
Since F = ma, we can rearrange the equation to solve for acceleration:
a = F / m = -0.22 N / 5.0 × 10^-3 kg = -44 m/s^2
The acceleration is constant and directed opposite to the x-axis since the electric field only has an x-component.
Next, we can use the kinematic equations to find the position of the particle at t = 2.0 s.
The equation for position (x) as a function of time (t), initial position (x0), initial velocity (v0), and acceleration (a) is given by:
x = x0 + v0t + (1/2)at^2
Since the initial position and velocity are both zero in this case, the equation simplifies to:
x = (1/2)at^2
Plugging in the values:
x = (1/2)(-44 m/s^2)(2.0 s)^2 = -176 m
The negative sign indicates that the particle moves in the negative x direction from the origin.
Finally, we can find the distance from the origin to the particle by taking the absolute value of the position:
distance = |x| = |-176 m| = 176 m
Therefore, the distance from the origin to the particle at t = 2.0 s is 176 meters.