An electron of mass me and charge q=-e is injected horizontally midway between two very large oppositely charged plates. The upper plate has a uniform positive charge per unit area +σ and the lower plate has a uniform negative charge per unit area -σ . You may ignore all edge effects. The particle has an initial velocity vo=vo x^ . What is the y-component of the position of the particle when the particle reaches the plane defined by x=L ?

What is the magnitude of the electric field between the plates? Express your answer in terms of the following variables and constants, if relevant, me,σ, Vo ,L ,e and epsilon_0

What is the magnitude of the acceleration of the electron when it is between the plates? Express your answer in terms of the following variables and constants, if relevantme,σ, Vo ,L ,e and epsilon_0

What is the y-component of the position of the particle when it reaches the plane defined by ? Express your answer in terms of the following variables and constants, if relevant me,σ, Vo ,L ,e and epsilon_0

Please some one answer.

Please I also have to submit the assignment by today...

To find the y-component of the position of the electron when it reaches the plane defined by x = L, we need to analyze the motion of the electron under the influence of the electric field between the plates.

First, let's consider the forces acting on the electron. There are two forces at play: the electric force and the gravitational force.

The electric force experienced by the electron is given by the equation F = qE, where q is the charge of the electron and E is the electric field between the plates.

The gravitational force experienced by the electron is given by the equation F = mg, where m is the mass of the electron and g is the acceleration due to gravity.

Since the electron is injected horizontally and the motion is in the y-direction, there is no initial velocity in the y-direction. Therefore, the motion of the electron is purely vertical.

Using Newton's second law, F = ma, we can equate the electric force and the gravitational force to find the acceleration of the electron.

qE = mg

Now, let's solve for the acceleration:

a = (qE) / m

We have the formula for the acceleration. Now, we need to find the electric field E between the plates.

To find the electric field, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).

In this case, the electric field between the plates is uniform, and the symmetry allows us to use a Gaussian surface parallel to the plates.

We can assume a Gaussian surface in the shape of a rectangular box with the upper and lower faces parallel to the plates.

Since the electric field is parallel to the plates, it will have the same magnitude and direction on each face of the Gaussian surface. Therefore, the electric flux through each face will be the same.

Next, we need to calculate the electric flux through each face. The area of each face is equal to the area of the plate, A.

The electric flux through each face of the Gaussian surface is given by the equation Φ = E * A.

Since the electric flux is the same through each face, the total electric flux through the Gaussian surface is 2Φ.

By Gauss's law, the total electric flux through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space.

Therefore, we can write the equation 2Φ = (σA + (-σA)) / ε₀.

Since the charge enclosed is zero, the equation simplifies to 2Φ = 0, and we can conclude that the electric flux through the Gaussian surface is zero.

Since the electric flux is zero, the electric field through the Gaussian surface is also zero.

Now, we have determined that the electric field is zero between the plates. Therefore, the magnitude of the electric field E is zero.

Substituting E = 0 in the equation for the acceleration, we get:

a = (q * 0) / m

a = 0

Hence, the magnitude of the acceleration of the electron when it is between the plates is zero.

Since the acceleration is zero, the y-component of the position of the particle does not change. Therefore, the y-component of the position of the particle when it reaches the plane defined by x = L is the same as its initial y-component, Vo.