An electron enters the region of a uniform electric field between two oppositely charged parallel plates as shown, with an initial velocity of 3 x 106 m/s. The electric field between the plates is measured to be E = 200N/C. The horizontal length of the plates is 10cm.
a. Find the acceleration of the electron while it is in the electric field.
b. How long does it take the electron to travel the length of the plates and
leave the field?
a. Acceleration is
a = e*E/m
e is the electron charge and m is the electron mass
b. Solve the equation
Since the electron's velocity component parallel to the plates remains the same as when it entered,
t = 0.10 m/(3*10^6 m/s)
how do we find out the electron charge and mass if we don't know what type it is?
a. To find the acceleration of the electron in the electric field, we can use the equation F = qE, where F is the force experienced by the electron, q is the charge of the electron, and E is the electric field strength.
Given that the charge of an electron is -1.6 x 10^-19 C and the electric field strength is 200 N/C, we can substitute these values into the equation to find the force experienced by the electron:
F = (-1.6 x 10^-19 C) * (200 N/C)
F = -3.2 x 10^-17 N
Since the force experienced by the electron is in the opposite direction of its initial velocity, the acceleration can be found using Newton's second law: F = ma, where m is the mass of the electron and a is its acceleration.
The mass of an electron is 9.1 x 10^-31 kg, so we can rearrange the equation to solve for acceleration:
a = F / m
a = (-3.2 x 10^-17 N) / (9.1 x 10^-31 kg)
a ≈ -3.516 x 10^13 m/s²
Therefore, the acceleration of the electron in the electric field is approximately -3.516 x 10^13 m/s² (directed opposite to the initial velocity).
b. To calculate the time it takes for the electron to travel the length of the plates and leave the field, we can use the equation of motion for constant acceleration:
s = ut + (1/2)at²
where s is the distance traveled, u is the initial velocity, t is the time taken, and a is the acceleration.
Given that the initial velocity is 3 x 10^6 m/s and the acceleration is -3.516 x 10^13 m/s², and the horizontal length of the plates is 10 cm (0.1 m), we can substitute these values into the equation:
0.1 = (3 x 10^6) t + (1/2) (-3.516 x 10^13) t²
Simplifying the equation:
(1/2) (-3.516 x 10^13) t² + (3 x 10^6) t - 0.1 = 0
We can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
In our case, a = (1/2) (-3.516 x 10^13), b = 3 x 10^6, and c = -0.1. Plugging these values into the quadratic formula, we get:
t = [-(3 x 10^6) ± √((3 x 10^6)² - 4(1/2) (-3.516 x 10^13) (-0.1))] / [2(1/2) (-3.516 x 10^13)]
Simplifying further:
t ≈ [-(3 x 10^6) ± √(9 x 10^12 + 7.032 x 10^12)] / (-1.758 x 10^13)
t ≈ [-(3 x 10^6) ± √(16.032 x 10^12)] / (-1.758 x 10^13)
t ≈ [-(3 x 10^6) ± 4 x 10^6] / (-1.758 x 10^13)
We take the positive value for t since time cannot be negative:
t ≈ (7 x 10^6) / (1.758 x 10^13)
t ≈ 3.979 x 10^-7 s
Therefore, it takes approximately 3.979 x 10^-7 seconds for the electron to travel the length of the plates and leave the electric field.
To find the answers to these questions, we can use the equations of motion and the relationship between force, electric field, and charge.
a. To find the acceleration of the electron while it is in the electric field, we can use the equation:
F = m * a
where F is the force acting on the electron, m is the mass of the electron, and a is its acceleration.
The force experienced by the electron in an electric field is given by the equation:
F = q * E
where q is the charge of the electron and E is the electric field.
Since the electron has a negative charge (q = -1.6 x 10^(-19) C), the force it experiences will be in the opposite direction to the electric field. Therefore, we can rewrite the equation as:
F = -q * E
Substituting this in our equation for force, we have:
-m * a = -q * E
Rearranging the equation, we can solve for the acceleration:
a = (q * E) / m
Substituting the values given in the problem, where q = -1.6 x 10^(-19) C, E = 200 N/C, and m = mass of an electron = 9.11 x 10^(-31) kg, we can calculate the acceleration.
a = (-1.6 x 10^(-19) C * 200 N/C) / (9.11 x 10^(-31) kg)
b. To find the time it takes for the electron to travel the length of the plates and leave the field, we can use the equation:
t = d / v
where t is the time, d is the distance traveled, and v is the velocity of the electron.
The distance traveled by the electron is given as the horizontal length of the plates, which is 10 cm or 0.1 m.
We can assume that the acceleration of the electron remains constant while it is in the electric field. Therefore, we can use the equation of motion:
v^2 = v₀^2 + 2 * a * d
where v₀ is the initial velocity, v is the final velocity, a is the acceleration, and d is the distance traveled.
Rearranging the equation, we can solve for the final velocity:
v = sqrt(v₀^2 + 2 * a * d)
Substituting the values given in the problem, where v₀ = 3 x 10^6 m/s, a is the acceleration we calculated in part (a), and d = 0.1 m, we can calculate the final velocity.
Finally, substituting the values of d and v into the equation for time (t = d / v), we can find the time it takes for the electron to travel the length of the plates and leave the field.