Electron in Homogeneous Electric Field

[jakk2]

please help me :(

Electron inserted in a homogeneous electric field to measure 300 N / C, which
directed vertically upwards. The initial velocity of the electron is far
5,00 10^ × 6 m/s and goes to 30 degrees, above the skyline. a) Find the maximum height that
reaches the electron above the original level.

please help i m stuck

 

[rl.bhat]

What is the force on the electron due to electric field?
What is the acceleration and its direction?
What are the vertical and horizontal components of velocity?
Which component remains constant and which component changes?
Which kinematic equation relates initial velocity, final velocity, displacement and acceleration?

 

[tomcue]

I can provide some insights and guidance on this topic. First, let's break down the information provided in the question. We are dealing with an electron in a homogeneous electric field with a strength of 300 N/C, directed vertically upwards. The initial velocity of the electron is 5.00 x 10^6 m/s and it is launched at an angle of 30 degrees above the horizontal.

To find the maximum height reached by the electron, we can use the equations of motion in a constant electric field. Since the electric field is directed upwards, it will cause a force on the electron in the same direction, resulting in its acceleration. The equation for calculating the displacement in this scenario is given by:

y = y0 + v0y t + 1/2 a t^2

Where y is the final displacement, y0 is the initial displacement (in this case, it is 0 as the electron is launched from the ground level), v0y is the initial vertical component of velocity (calculated using the given angle and initial velocity), t is the time, and a is the acceleration due to the electric field (calculated by dividing the electric field strength by the mass of the electron).

To find the time at which the electron reaches its maximum height, we can use the fact that at the maximum height, the vertical component of velocity becomes 0. This gives us the equation:

v = v0 + a t
0 = v0y + a t_max
t_max = -v0y / a

Substituting this value of t_max in the first equation, we get:

y_max = v0y^2 / 2a

Plugging in the values from the question, we get:

y_max = (5.00 x 10^6 sin 30)^2 / (2 x 300 / 9.11 x 10^-31)
= 3.8 x 10^-4 m

Therefore, the maximum height reached by the electron above the original level is 3.8 x 10^-4 meters. I hope this helps you understand the concept better. If you have any further questions, please feel free to ask.