Two identical metal plates with small holes are aligned a distance L=6.7 cm apart. They have a potential difference of 131 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 3290 km/s at the negative plate, with what velocity does it exit at the positive plate?
To find the velocity of the electron as it exits the positive plate, we will use the principle of conservation of energy.
The potential difference (V) between the two plates is given as 131 Volts. Since the electron is negatively charged, it will experience a force that will accelerate it from the negative plate to the positive plate.
The work done by the electric field in accelerating the electron is equal to the change in its kinetic energy, assuming there is no loss of energy due to other factors such as collisions or friction.
The work done by the electric field can be calculated using the equation:
Work = Charge × Potential Difference
Since the electron has a charge of -1.6 × 10^(-19) Coulombs, we can substitute these values into the equation:
Work = (-1.6 × 10^(-19) C) × (131 V)
Next, we need to find the change in kinetic energy of the electron. The kinetic energy of an object can be calculated using the equation:
Kinetic Energy = 1/2 × Mass × Velocity^2
Since the electron's mass is approximately 9.11 × 10^(-31) kg, and the initial velocity is given as 3290 km/s, we can substitute these values into the equation:
Initial Kinetic Energy = 1/2 × (9.11 × 10^(-31) kg) × (3290000 m/s)^2
The final velocity of the electron can be found by rearranging the equation for kinetic energy and solving for velocity:
Velocity = √(2 × Final Kinetic Energy / Mass)
Now, we can equate the work done by the electric field to the change in kinetic energy and solve for the final velocity:
(-1.6 × 10^(-19) C) × (131 V) = 1/2 × (9.11 × 10^(-31) kg) × (3290000 m/s)^2 - 1/2 × (9.11 × 10^(-31) kg) × Velocity^2
Simplifying and rearranging the equation, we can solve for the final velocity:
Velocity = √(2 × (-1.6 × 10^(-19) C) × (131 V) / (9.11 × 10^(-31) kg) + (3290000 m/s)^2)
Plugging in the given values and evaluating the expression, we can find the velocity at which the electron exits the positive plate.