Question

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through.

Calculate the horizontal component of the acceleration of the electron if the field strength is 2.55*10^4 N/C. Express your answer in m/s^2 and assume the electric field is pointing in the negative x-direction.

Answer 1

The horizontal component of acceleration of an electron in a uniform electric field of 2.55*10^4 N/C is approximately 4.48 × 10^15 m/s^2. This acceleration is in the positive x-direction due to the force acting on the negatively charged electron being in the opposite direction of the field.

Step-by-step explanation:

To calculate the horizontal component of the acceleration of an electron in a uniform electric field, we can use the formula a = F/m, where a is the acceleration, F is the force on the electron, and m is the mass of the electron. Considering that the electron carries a charge of e = -1.60 × 10^-19 C and the mass of an electron is approximately m = 9.11 × 10^-31 kg, we can find the force exerted by the electric field using F = qE, where q is the charge of the electron and E is the electric field strength.

In this example, if the electric field strength is E = 2.55 × 10^4 N/C, the force on the electron would be F = eE = (-1.60 × 10^-19 C) × (2.55 × 10^4 N/C), which results in F = -4.08 × 10^-15 N. Notice the negative sign indicates the force direction is opposite to the field direction (force direction is positive x-direction due to the negative charge of the electron). Then, the acceleration of the electron would be a = F/m = (-4.08 × 10^-15 N) / (9.11 × 10^-31 kg), which equals to approximately a = 4.48 × 10^15 m/s^2 in the positive x-direction.

Answer 2

Answer :
`4.483* 10^(15)\ m/s^2`.

Step-by-step explanation:

It is given that,

Electric field strength,
`E=2.55* 10^(4)\ N/C`

We know that,

Charge of electron,
`q=1.6* 10^(-19)\ C`

Mass of electron,
`m=9.1* 10^(-31)\ kg`

From the definition of electric field,
`F=qE`...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

`ma=qE`

`a=(qE)/(m)`

`a=(1.6* 10^(-19)\ C* 2.55* 10^(4)\ N/C )/(9.1* 10^(-31)\ kg)`

`a=0.4483* 10^(16)\ m/s^2`

or

`a=4.483* 10^(15)\ m/s^2`

So, the horizontal component of acceleration of an electron is
`4.483* 10^(15)\ m/s^2`.

Hence, it is the required solution.