A parallel plate capacitor is made of two 3.5 cm x 3.5 cm square plates that are separated by 1.6 x10 -5 m. The plates are charged to 4.8 nC each (positive and negative). An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.

a. What will be the speed of the electron when it reaches the positive plate?

b. What will be the speed of the proton when it reaches the negative plate?

To find the speed of the electron and the proton, we can use the principle of conservation of energy. We need to calculate the potential energy and the kinetic energy for both particles at their respective starting and ending points.

a. Let's begin by calculating the potential energy of the electron at the negative plate. The potential energy of a charged particle in an electric field is given by the formula:

PE = q * V

Where PE is the potential energy, q is the charge of the particle, and V is the voltage between the plates.

Since the electron is at the negatively charged plate, the voltage V is negative. The potential energy of the electron at the negative plate is:

PE_electron_start = (-4.8 nC) * V

Next, let's calculate the potential energy of the electron at the positive plate. The voltage V between the plates is the same, but since the electron is moving against the electric field, the potential energy is positive:

PE_electron_end = (4.8 nC) * V

Now, let's calculate the kinetic energy of the electron at the positive plate. Initially, the electron is at rest, so its initial kinetic energy is zero. At the positive plate, the electron has gained kinetic energy:

KE_electron_end = (1/2) * m * v_electron^2

where m is the mass of the electron and v_electron is its final velocity.

According to the conservation of energy, the change in potential energy is equal to the change in kinetic energy:

PE_electron_end - PE_electron_start = KE_electron_end - 0

Simplifying the equation:

(4.8 nC) * V - (-4.8 nC) * V = (1/2) * m * v_electron^2

Canceling out the terms:

9.6 nC * V = (1/2) * m * v_electron^2

We can now substitute the given values into the equation and solve for v_electron.

b. Similarly, we can calculate the potential energy and kinetic energy of the proton.

The potential energy of the proton at the positive plate is:

PE_proton_start = (4.8 nC) * V

The potential energy of the proton at the negative plate is:

PE_proton_end = (-4.8 nC) * V

Since the proton is moving in the opposite direction of the electric field, its potential energy at the negative plate is positive.

The kinetic energy of the proton at the negative plate is:

KE_proton_end = (1/2) * m * v_proton^2

Applying the conservation of energy:

PE_proton_end - PE_proton_start = KE_proton_end - 0

- (4.8 nC) * V - (4.8 nC) * V = (1/2) * m * v_proton^2

-9.6 nC * V = (1/2) * m * v_proton^2

Substituting the given values will allow us to solve for v_proton.