Use the following values for mass and charge: an electron has mass me = 9.11×10-31 kg and charge -e, a proton has mass mp = 1.67×10-27 kg and charge +e, an alpha particle has mass malpha = 6.65×10-27 kg and charge +2e, where e = 1.60×10-19 C. An electron is released from rest in a vacuum between two flat, parallel metal plates that are 13.0 cm apart and are maintained at a constant electric potential difference of 780 Volts. If the electron is released at the negative plate, what is its speed just before it strikes the positive plate?
To determine the speed of the electron just before it strikes the positive plate, we need to apply the principles of electrostatics and motion in a uniform electric field.
First, let's determine the electric field strength between the plates. We can use the formula:
Electric field strength (E) = Potential difference (V) / Distance between the plates (d)
Given the potential difference (V) = 780 Volts and the distance (d) = 13.0 cm = 0.13 m, we can calculate the electric field strength:
E = 780 Volts / 0.13 m = 6000 Volts/m
Now, let's determine the force experienced by the electron in the electric field. Using the equation:
Force (F) = Electric field strength (E) * Charge (q)
Since the charge of an electron is -e, where e = 1.60×10^-19 C, the force experienced by the electron is:
F = E * (-e) = 6000 Volts/m * (-1.60×10^-19 C)
Next, we can use Newton's second law of motion to find the acceleration of the electron. The force acting on the electron is provided by the electric field, and the mass of the electron is me = 9.11×10^-31 kg. The equation is:
F = mass (m) * acceleration (a)
Solving for acceleration (a):
a = F / m = (6000 Volts/m * (-1.60×10^-19 C)) /9.11×10^-31 kg
Once we have the acceleration, we can determine the final velocity (v) of the electron just before it strikes the positive plate using the equations of motion for uniform acceleration:
v^2 = u^2 + 2 a d
Since the electron starts from rest, the initial velocity (u) is 0 m/s. The distance between the plates (d) is given as 13.0 cm = 0.13 m.
Plugging in the values, we can solve for the final velocity (v):
v^2 = (0 m/s)^2 + 2(a)(0.13 m)
v^2 = 0 + 2(a)(0.13 m)
v^2 = 2(a)(0.13 m)
Finally, we can substitute the value of acceleration (a) from the previous calculation and solve for the final velocity (v).
To find the speed of the electron just before it strikes the positive plate, we can use the concepts of electric potential energy and kinetic energy.
Step 1: Determine the electric potential difference (V) between the plates.
Given: V = 780 Volts
Step 2: Convert the distance between the plates (d) to meters.
Given: d = 13.0 cm = 0.13 m
Step 3: Calculate the electric field (E) between the plates using the formula E = V/d.
E = V/d = 780 V / 0.13 m = 6000 V/m
Step 4: Determine the force (F) experienced by the electron inside the electric field using the formula F = E * magnitude of charge.
Given: Charge of an electron = -e = -1.60×10^-19 C
F = E * charge = 6000 V/m * (-1.60×10^-19 C) = -9.6 × 10^-16 N
Step 5: Calculate the acceleration (a) experienced by the electron using Newton's second law F = ma, where m is the mass of the electron.
Given: Mass of an electron (me) = 9.11×10^-31 kg
a = F / m = (-9.6 × 10^-16 N) / (9.11×10^-31 kg) = -1.055 × 10^15 m/s^2
Step 6: Use the kinematic equation v^2 = u^2 + 2as to find the final velocity v, where u is the initial velocity (0 m/s), s is the displacement between the plates (0.13 m), and a is the acceleration (-1.055 × 10^15 m/s^2).
v^2 = (0 m/s)^2 + 2 * (-1.055 × 10^15 m/s^2) * (0.13 m)
v^2 = -2.746 × 10^14 m^2/s^2
v = sqrt(-2.746 × 10^14 m^2/s^2)
NOTE: The negative sign indicates that the electron will be moving in the opposite direction of the electric field.
Step 7: Calculate the speed (v) of the electron using the equation v = |v|.
v = |v| = |-16573 m/s|
v ≈ 16573 m/s
Therefore, the speed of the electron just before it strikes the positive plate is approximately 16573 m/s.