An electron with a speed of 3.00X10^6 m/s moves into a uniform electric field of magnitude 1.00X10^3 N/C. The
field lines are parallel to the electron’s velocity and pointing in the same direction as the velocity. How far does the electron travel before it is brought to rest?
To find the distance traveled by the electron before it is brought to rest, we need to determine its acceleration first. We can use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
The net force acting on the electron is the electric force, and it can be calculated using the formula:
F = q * E
where F is the force, q is the charge of the electron (1.6 × 10^-19 C), and E is the electric field strength (1.00 × 10^3 N/C).
Let's calculate the force:
F = (1.6 × 10^-19 C) * (1.00 × 10^3 N/C)
F = 1.6 × 10^-16 N
Now, using Newton's second law, we can calculate the acceleration:
F = m * a
where m is the mass of the electron (9.11 × 10^-31 kg).
a = F / m
a = (1.6 × 10^-16 N) / (9.11 × 10^-31 kg)
a ≈ 1.76 × 10^14 m/s^2
Since acceleration is the rate of change of velocity, we can find the time it takes for the electron to come to rest once we have the acceleration. The formula to calculate the time is:
v_final = v_initial + a * t
where v_final is the final velocity (0 m/s), v_initial is the initial velocity (3.00 × 10^6 m/s), a is acceleration (-1.76 × 10^14 m/s^2), and t is the time.
0 = 3.00 × 10^6 m/s + (-1.76 × 10^14 m/s^2) * t
Solving for t:
t = - (3.00 × 10^6 m/s) / (-1.76 × 10^14 m/s^2)
t ≈ 1.70 × 10^-8 s
Finally, to find the distance traveled, we can use the equation:
d = v_initial * t + (1/2) * a * t^2
d = (3.00 × 10^6 m/s) * (1.70 × 10^-8 s) + (1/2) * (-1.76 × 10^14 m/s^2) * (1.70 × 10^-8 s)^2
d ≈ 2.59 × 10^-2 m
Therefore, the electron travels approximately 2.59 × 10^-2 meters before it comes to rest.
To find the distance the electron travels before it is brought to rest, we need to use the equations that relate the electric force to the motion of a charged particle.
First, we need to calculate the electric force acting on the electron. The electric force (F) on a charged particle in an electric field is given by the equation:
F = q * E
Where:
- F is the electric force
- q is the charge of the particle
- E is the electric field strength
In this case, the charge of the electron (q) is -1.6 × 10^-19 C, and the electric field strength (E) is 1.00 × 10^3 N/C. Therefore, the electric force (F) acting on the electron is:
F = (-1.6 × 10^-19 C) * (1.00 × 10^3 N/C)
Next, we use the equation for the force acting on an object to calculate the distance traveled (d) before the electron is brought to rest. The equation is:
F = m * a
Where:
- F is the force
- m is the mass of the electron
- a is the acceleration
The mass of an electron (m) is approximately 9.1 × 10^-31 kg.
Since the electron is brought to rest, its final velocity (vf) is 0, and the initial velocity (vi) is 3.00 × 10^6 m/s.
We can use the equation for acceleration (a) to solve for the distance traveled (d):
vf^2 = vi^2 + 2ad
Rearranging the equation, we get:
d = (vf^2 - vi^2) / (2a)
Substituting the values given, we have:
d = (0 - (3.00 × 10^6 m/s)^2) / (2 * a)
Finally, we substitute the known values into the equation and solve for d:
d = (0 - (3.00 × 10^6 m/s)^2) / (2 * [(F) / m])
Calculating this expression should give us the distance the electron travels before it is brought to rest.