Two identical metal plates with small holes are aligned a distance L=7.2 cm apart. They have a potential difference of 141 Volts. An electron can enter through the hole in the negative plate and exit through the hole in the positive plate. If it enters with a velocity of 3680 km/s at the negative plate, with what velocity does it exit at the positive plate?
To determine the velocity with which the electron exits the positive plate, we can use the principles of electrostatics and conservation of energy.
1. Calculate the electric field between the plates:
- The potential difference (V) between the two plates is given as 141 Volts.
- The distance (L) separating the plates is given as 7.2 cm, which is equal to 0.072 m.
- The electric field (E) between the plates can be calculated using the formula E = V / L.
- Plugging in the given values, we get E = 141 V / 0.072 m = 1958.33 N/C.
2. Determine the force on the electron:
- The force on an electron in an electric field is given by the equation F = q * E, where q is the charge.
- The charge on an electron is -1.6 x 10^-19 C (Coulombs).
- Plugging in the values, we get F = -1.6 x 10^-19 C * 1958.33 N/C = -3.13 x 10^-16 N.
- Note that the force acts in the opposite direction of the electric field, so it is negative.
3. Calculate the work done by the electric field:
- The work done by an electric field on a charged particle is given by the equation W = F * d * cos(theta), where d is the distance traveled by the electron and theta is the angle between the force and displacement vectors.
- Since the plates are aligned, theta is 0 degrees, and cos(theta) = 1.
- The distance traveled by the electron is the distance between the plates, which is 0.072 m.
- Plugging in the values, we get W = (-3.13 x 10^-16 N) * (0.072 m) * (1) = -2.26 x 10^-17 J.
4. Apply the principle of conservation of energy:
- The work done by the electric field is equal to the change in kinetic energy of the electron.
- The initial kinetic energy of the electron is given by KE_initial = (1/2) * m * v1^2, where m is the mass of the electron and v1 is the initial velocity.
- The final kinetic energy of the electron is given by KE_final = (1/2) * m * v2^2, where v2 is the final velocity.
- Since the mass of the electron is a constant, we can equate the two expressions: KE_initial = KE_final.
- Plugging in the given values, we have (1/2) * (9.11 x 10^-31 kg) * (3680 km/s)^2 = (1/2) * (9.11 x 10^-31 kg) * v2^2.
- Solving for v2, we get v2 = sqrt((3680 km/s)^2 * (v2 / 3680 km/s)^2) = 3680 km/s * (v2 / 3680 km/s).
Thus, the electron exits the positive plate with a velocity of 3680 km/s.