Two identical metal plates with small holes are aligned a distance L=7.8 cm apart. They have a potential difference of 127 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 4630 km/s at the negative plate, with what velocity does it exit at the positive plate?

To determine the velocity with which an electron exits the positive plate, we need to use the principles of electrostatics.

The potential difference between the plates causes an electric field to be established between them. This electric field exerts an electric force on the electron, causing it to accelerate as it moves from the negative plate to the positive plate.

The formula that relates the electric force on a charged particle to its charge and the electric field is given by:
F = q * E, where F is the force, q is the charge, and E is the electric field.

In this case, the charge of the electron is e = 1.6 * 10^-19 C, and the electric field strength can be calculated using the formula:
E = V / d, where V is the potential difference and d is the distance between the plates.

Given:
V = 127 V
d = 7.8 cm = 0.078 m

Using the formula, the electric field strength is:
E = V / d = 127 V / 0.078 m = 1628.21 V/m

Now, we can calculate the electric force on the electron using:
F = q * E = (1.6 * 10^-19 C) * (1628.21 V/m) = 2.60 * 10^-16 N

The electric force is responsible for the acceleration of the electron using Newton's second law, F = m * a, where m is the mass of the electron and a is its acceleration.

The mass of an electron is m = 9.11 * 10^-31 kg.

Rearranging the formula, we can find the acceleration:
a = F / m = (2.60 * 10^-16 N) / (9.11 * 10^-31 kg) ≈ 2.85 * 10^14 m/s²

Knowing the initial velocity at the negative plate (v₀ = 4630 km/s) and the acceleration, we can use the equation of motion to find the velocity at the positive plate.

v = v₀ + a * t

Since we know the distance between the plates (d = 0.078 m) and the initial velocity (v₀ = 4630 km/s = 4630 * 10^3 m/s), we can solve for time (t) using the equation:
d = v₀ * t + (1/2) * a * t²

Rearranging and substituting the values, we get:
0.078 = (4630 * 10^3) * t + (1/2) * (2.85 * 10^14) * t²

This is a quadratic equation that can be solved, resulting in two possible values for time (one of which is negative). The positive value of time will give us the time it takes for the electron to travel from the negative plate to the positive plate.

Once we have the time, we can substitute it back into the equation v = v₀ + a * t to find the velocity with which the electron exits the positive plate.