An electron is accelerated horizontally from rest in a television picture tube by a potential difference
of 1.00 × 104 V.
(a) Find the final velocity after the original acceleration period.
(b) It then passes between two horizontal plates 6.0 cm long and 1.0 cm apart which have a
potential difference of 200 V. At what angle will the electron be traveling after it passes
between the plates
To find the final velocity of the electron after the original acceleration period, we can use the formula for the kinetic energy of a moving particle:
K = (1/2)mv^2
where K is the kinetic energy, m is the mass of the electron, and v is its final velocity.
(a) First, we need to find the mass of the electron. The mass of an electron is approximately 9.11 x 10^-31 kg.
Next, we need to calculate the initial kinetic energy of the electron after being accelerated by the potential difference of 1.00 x 10^4 V. The electrical potential energy gained by the electron is equal to its kinetic energy. Therefore, we can write:
K_initial = qV
where q is the charge of the electron and V is the potential difference applied. The charge of an electron is approximately -1.6 x 10^-19 C. Plugging in these values, we have:
K_initial = (-1.6 x 10^-19 C)(1.00 x 10^4 V)
Solving for K_initial, we get:
K_initial = -1.6 x 10^-15 J
Now, we can equate the initial kinetic energy to the final kinetic energy and solve for the final velocity:
K_initial = K_final
(-1.6 x 10^-15 J) = (1/2)(9.11 x 10^-31 kg)(v_final)^2
Solving for v_final, we get:
v_final = sqrt((-1.6 x 10^-15 J) / (0.5)(9.11 x 10^-31 kg))
Calculating the final velocity using a calculator, we find:
v_final ≈ -5.11 x 10^6 m/s
Note: The negative sign indicates that the electron is moving in the opposite direction of the applied electric field.
(b) To find the angle at which the electron will be traveling after passing between the plates, we need to consider the electric field created by the potential difference between the plates. The electric field will exert a force on the electron, causing it to be deflected.
The force experienced by the electron due to the electric field is given by:
F = Eq
where F is the force, E is the electric field strength, and q is the charge of the electron.
The electric field strength between the plates is given by:
E = V / d
where V is the potential difference applied to the plates, and d is the distance between the plates.
Plugging in the values for V and d, we have:
E = (200 V) / (0.01 m)
Simplifying, we find:
E = 2 x 10^4 N/C
Now, let's calculate the force acting on the electron:
F = (2 x 10^4 N/C) * (-1.6 x 10^-19 C)
Solving for F, we have:
F ≈ -3.2 x 10^-15 N
The electron experiences a force in the vertical direction due to the electric field. This force can be decomposed into a vertical component and a horizontal component.
Since the electron is already moving horizontally, the only force that can cause a change in its direction is the vertical component of the force. This force will act perpendicularly to the velocity and cause the electron to change its direction.
To determine the angle at which the electron will be traveling, we can use trigonometry. Let's define the angle between the electron's initial velocity and the horizontal direction as θ.
Now, we can write the equation for the net force experienced by the electron in the direction perpendicular to its initial velocity:
F_vertical = F * sin(θ)
Since the force is negative, we can ignore the negative sign and write:
|F_vertical| = |F| * sin(θ)
Solving for sin(θ), we find:
sin(θ) = |F_vertical| / |F|
Substituting the values for |F_vertical| and |F|, we get:
sin(θ) = (3.2 x 10^-15 N) / (3.2 x 10^-15 N)
Simplifying, we find:
sin(θ) = 1
Therefore, the angle at which the electron will be traveling after passing between the plates is 90 degrees or perpendicular to its initial velocity.
To solve this problem, we'll utilize the principles of electric potential and energy conservation.
(a) Finding the final velocity after the original acceleration period:
Given:
Potential difference (V) = 1.00 × 10^4 V
We know that the work done on the electron is equal to its change in kinetic energy (KE):
W = ΔKE
The work done is given by:
W = qV
where q is the charge of the electron and V is the potential difference.
The change in kinetic energy is given by:
ΔKE = 1/2 mv^2
where m is the mass of the electron and v is its velocity.
Since qV = ΔKE, we can equate the two equations:
qV = 1/2 mv^2
To solve for the final velocity (v), we need to know the charge (q) and mass (m) of the electron. The charge of an electron is -1.6 × 10^-19 C, and the mass of an electron is 9.1 × 10^-31 kg.
Substituting these values into the equation, we have:
-1.6 × 10^-19 C × 1.00 × 10^4 V = 1/2 × 9.1 × 10^-31 kg × v^2
Simplifying the equation, we get:
v^2 = (2 × -1.6 × 10^-19 C × 1.00 × 10^4 V) / (9.1 × 10^-31 kg)
v^2 = -3.52 × 10^-4 m/s
Taking the square root of both sides, we find:
v ≈ -0.019 m/s
The negative sign indicates that the electron is traveling in the opposite direction to the initial acceleration. Hence, the final velocity after the original acceleration period is approximately -0.019 m/s.
(b) Finding the angle at which the electron will be traveling after passing between the plates:
Given:
Potential difference between the plates (V) = 200 V
Length of the plates (d) = 6.0 cm = 0.06 m
The electric potential energy gained or lost by the electron passing through the plates is given by:
ΔPE = qV
The change in potential energy is equal to the work done by the electric field, and this energy is converted into kinetic energy.
Therefore, we can write:
ΔPE = ΔKE
Since the electron is not changing elevation, the change in kinetic energy is due to the change in its horizontal velocity component. This means the net force on the electron is acting in the horizontal direction, resulting in a change in horizontal velocity.
From the conservation of energy, we can write:
1/2 mv^2 = qV
Substituting the given values, we have:
1/2 × 9.1 × 10^-31 kg × v^2 = -1.6 × 10^-19 C × 200 V
Simplifying the equation, we find:
v^2 = (2 × -1.6 × 10^-19 C × 200 V) / (9.1 × 10^-31 kg)
v^2 = -7.62 × 10^-10 m^2/s^2
Taking the square root of both sides, we get:
v ≈ ± 8.73 × 10^-5 m/s
The positive sign indicates that the electron is traveling in the same direction as its original acceleration. Therefore, the electron will be traveling at an angle of:
θ = atan(v / (d/2))
θ ≈ atan(8.73 × 10^-5 m/s / (0.06 m / 2))
θ ≈ atan(8.73 × 10^-5 m/s / 0.03 m)
θ ≈ atan(2.91 × 10^-3)
Using a calculator, the angle θ is approximately:
θ ≈ 0.166 degrees
So, after passing between the plates, the electron will be traveling at an angle of approximately 0.166 degrees.