An electron is accelerated through a uniform electric field of magnitude 2.5x10^2 N/C with an initial speed of 1.2x10^6 m/s parallel to the electric field.

a) Calculate the work done on the electron by the field when the electron has travelled 2.5 cm in the field.
b) Calculate the speed of the electron after it has travelled 2.5 cm in the field.
c) If the direction of the electric field is reversed, how far will the electron move into the field before coming to rest?

Please, please, please, this is the first question of many and I don't get it.

An electron is projected with an initial speed v0 = 1.60 × 106m/s into the uniform field

between two parallel plates as shown below. Assume that the field between the plates is
uniform and directed vertically downward, and that the field outside the plates is zero. The
electron enters the field at a point midway between the plates

To solve this problem, we need to use the formulas for work done, final speed, and distance traveled.

First, let's calculate the work done on the electron.

a) The work done (W) is given by the formula:

W = F * d * cos(theta)

where F is the magnitude of the force, d is the distance traveled, and theta is the angle between the force and the displacement.

In this case, the force and displacement are parallel, so the angle theta is 0 degrees and the cosine of 0 degrees is 1. Therefore, we can simplify the formula to:

W = F * d

Substituting the given values:

W = (2.5x10^2 N/C) * (2.5 cm) = (2.5x10^2 N/C) * (2.5x10^-2 m) = 6.25x10^0 N*m = 6.25 J

So, the work done on the electron by the electric field is 6.25 Joules.

b) To calculate the final speed (v) of the electron after traveling a certain distance, we can use the work-energy theorem:

W = (1/2) * m * (v^2 - u^2)

where W is the work done, m is the mass of the electron (approximately 9.11x10^-31 kg), u is the initial speed (1.2x10^6 m/s), and v is the final speed.

Rearranging the equation and solving for v:

v^2 = 2 * W / m + u^2

v = sqrt(2 * W / m + u^2)

Substituting the given values:

v = sqrt(2 * 6.25 J / (9.11x10^-31 kg) + (1.2x10^6 m/s)^2)
= sqrt(1.373x10^32 m^2/s^2 + 1.44x10^12 m^2/s^2)
= sqrt(1.373x10^32 m^2/s^2 + 0.144x10^14 m^2/s^2)
= sqrt(1.373x10^32 m^2/s^2)
= 3.71x10^16 m/s

So, the speed of the electron after traveling 2.5 cm in the field is approximately 3.71x10^16 m/s.

c) When the direction of the electric field is reversed, the force on the electron will also reverse its direction. The electron will then decelerate until it comes to rest. To find the distance it travels before stopping, we can calculate the distance (d) using the formula:

d = (v^2 - u^2) / (2a)

where v is the final velocity (0 m/s), u is the initial velocity (3.71x10^16 m/s), and a is the acceleration.

Rearranging the equation and solving for d:

d = (0 - (3.71x10^16 m/s)^2) / (2 * a)
= (-1.373x10^33 m^2/s^2) / (2 * a)

The acceleration can be calculated using the formula:

F = m * a

where F is the magnitude of the force and m is the mass of the electron.

Since the electric field magnitude is 2.5x10^2 N/C and the force experienced by the electron is given by F = q * E (where q is the charge of the electron), we can substitute these values into the formula:

2.5x10^2 N/C = (9.11x10^-31 kg) * a

Solving for a:

a = (2.5x10^2 N/C) / (9.11x10^-31 kg)
= 2.746x10^32 m^2/s^2

Substituting this value of acceleration into the equation for distance:

d = (-1.373x10^33 m^2/s^2) / (2 * 2.746x10^32 m^2/s^2)
= -0.5

Since distance cannot be negative, we take the absolute value:

d = 0.5 m

Therefore, the electron will travel 0.5 meters into the field before coming to rest.