An electron (mass m = 9.11 * 10^-31 kg) is accelerated in a uniform field vector E ( E =1.45 * 10^4 N/C) between two parallel charged plates. The seperation of the plates is 1.10 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

a) With what speed does it leave the hole?
b) Show that the gravitational force can be ignored.

a) e E d = (1/2) mV^2

d is the plate separation
e amd m are the electron charge and mass, respiectively.
Solve for V
b) Compare Ee, (the electric force) to mg (the gravity force). The ratio may surprise you.

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a) To find the speed with which the electron leaves the hole, we can use the equation for the acceleration of a charged particle in an electric field:

a = eE/m

where a is the acceleration, e is the charge of the electron (1.6 * 10^-19 C), E is the electric field, and m is the mass of the electron.

Given:
e = 1.6 * 10^-19 C
E = 1.45 * 10^4 N/C
m = 9.11 * 10^-31 kg

We can plug these values into the equation to find the acceleration:

a = (1.6 * 10^-19 C)(1.45 * 10^4 N/C) / (9.11 * 10^-31 kg)

Now we can solve for the acceleration:

a ≈ 2.56 * 10^14 m/s^2

Next, we can use the kinematic equation relating acceleration, initial velocity, final velocity, and displacement:

v^2 = u^2 + 2as

where v is the final velocity (unknown), u is the initial velocity (0 m/s since it starts from rest), a is the acceleration, and s is the displacement.

The electron accelerates across the gap between the plates over a distance equal to the separation of the plates, s = 1.10 cm = 0.011 m.

Plugging in the values, the equation becomes:

v^2 = 0 + 2(2.56 * 10^14 m/s^2)(0.011 m)

Now we can solve for v:

v^2 ≈ 5.632 m^2/s^2

Finally, taking the square root of both sides, we find:

v ≈ 2.374 m/s

Therefore, the electron leaves the hole with a speed of approximately 2.374 m/s.

b) To show that the gravitational force can be ignored, we need to compare the magnitudes of the electric force and the gravitational force on the electron.

The electric force on the electron is given by:

Fe = eE

where Fe is the electric force, e is the charge of the electron, and E is the electric field.

The gravitational force on the electron is given by:

Fg = mg

where Fg is the gravitational force, m is the mass of the electron, and g is the acceleration due to gravity.

Given:
e = 1.6 * 10^-19 C
E = 1.45 * 10^4 N/C
m = 9.11 * 10^-31 kg
g = 9.8 m/s^2

We can calculate the magnitudes of the electric force and the gravitational force:

Fe = (1.6 * 10^-19 C)(1.45 * 10^4 N/C) ≈ 2.32 * 10^-15 N

Fg = (9.11 * 10^-31 kg)(9.8 m/s^2) ≈ 8.94 * 10^-30 N

Comparing the magnitudes, we can see that the electric force is much stronger than the gravitational force:

Fe ≫ Fg

Therefore, we can ignore the gravitational force in this scenario.

To determine the speed at which the electron leaves the hole, we can use the principle of energy conservation. We can assume that the work done by the electric field in accelerating the electron is equal to the change in its kinetic energy.

a) To find the speed at which it leaves the hole, we can start by finding the work done by the electric field.

The work done by an electric field on a charged particle is given by the equation:

Work = charge of the particle * electric potential difference

In this case, the charge of the electron is -1.6 * 10^-19 C (since it is negatively charged), and the electric potential difference is the product of the electric field strength and the separation distance between the plates.

Electric potential difference = E * d

where E is the electric field strength (1.45 * 10^4 N/C) and d is the separation distance between the plates (1.10 cm = 0.011 m).

Now, we can calculate the work done by the electric field:

Work = (-1.6 * 10^-19 C) * (1.45 * 10^4 N/C) * (0.011 m)

Next, we equate the work done by the electric field with the change in kinetic energy of the electron:

Work = (1/2) * mass * final velocity^2

Plugging in the values, we get:

(-1.6 * 10^-19 C) * (1.45 * 10^4 N/C) * (0.011 m) = (1/2) * (9.11 * 10^-31 kg) * final velocity^2

Simplifying the equation, we can solve for the final velocity:

final velocity^2 = [(-1.6 * 10^-19 C) * (1.45 * 10^4 N/C) * (0.011 m)] / [0.5 * (9.11 * 10^-31 kg)]

b) To show that the gravitational force can be ignored, we need to compare its magnitude to the force exerted by the electric field.

The force exerted by gravity is given by the equation:

Force_gravity = mass * gravitational acceleration

where the gravitational acceleration is approximately 9.8 m/s^2.

Force_gravity = (9.11 * 10^-31 kg) * (9.8 m/s^2)

Comparing this to the force exerted by the electric field, we can find the ratio:

Ratio = Force_gravity / Electric force

Electric force = charge of the electron * electric field strength

Electric force = (-1.6 * 10^-19 C) * (1.45 * 10^4 N/C)

Calculating the ratio:

Ratio = [(9.11 * 10^-31 kg) * (9.8 m/s^2)] / [(-1.6 * 10^-19 C) * (1.45 * 10^4 N/C)]

We can then evaluate the ratio and compare it to 1. If the ratio is much smaller than 1, then the gravitational force can be ignored.