Electrons projected through an electron field problem

[glitterjewels]

Electrons projected through an electron field problem...please help!

Here's the problem:
Electrons with initial speed of 5 x 10^6 m/s are projected into a region where a constant electric field of 1 x 10^5 V/m exists. The field is directed in a way that causes the electrons to DECELERATE. How far do the electrons travel before they turn around and move in the opposite direction?

My problem is I'm not sure how to find the distance or "how far". I may be missing an equation. Here's what I do have

F/m=acceleration=Fg=mg
The electric field=qE(volts/m)= 1 x 10^5 V/m
charge of electron qe= 1.06 x 10^-19c
acceleration upward= qE/m
Your help would be greatly appreciated! Thank you!

 

[xanthym]

Here's the problem:
Electrons with initial speed of 5 x 10^6 m/s are projected into a region where a constant electric field of 1 x 10^5 V/m exists. The field is directed in a way that causes the electrons to DECELERATE. How far do the electrons travel before they turn around and move in the opposite direction?

My problem is I'm not sure how to find the distance or "how far". I may be missing an equation. Here's what I do have

F/m=acceleration=Fg=mg
The electric field=qE(volts/m)= 1 x 10^5 V/m
charge of electron qe= 1.06 x 10^-19c
acceleration upward= qE/m
Your help would be greatly appreciated! Thank you!

Since motion direction isn't specified, it's presumed gravitational effects can be ignored. For these conditions, the following kinematic equation applies:

$$1: \ \ \ \ v_{final}^{2} \, - \, v_{initial}^{2} \, \ = \, \ 2 a d$$

where "v"s are the indicated velocities (here: v_final=0, & v_initial=5e6 m/s), "d" the distance traveled, and "a" the constant acceleration (here: a=qE/m). Solve for "d".


~~

 

[thepatientmental]

Sure, I'd be happy to help! Let's break down the problem and see what equations we can use to solve it.

First, we know that the electrons are projected with an initial speed of 5 x 10^6 m/s. This means that they have an initial kinetic energy, which we can calculate using the equation KE = 1/2 * mv^2.

Next, we know that the electrons are in a constant electric field of 1 x 10^5 V/m, which is causing them to decelerate. This means that the electrons are experiencing a force in the opposite direction of their motion, which we can calculate using the equation F = qE.

Now, we can use Newton's second law, F = ma, to relate the force to the acceleration. We know that the mass of an electron is 9.11 x 10^-31 kg, so we can plug in our values and solve for the acceleration (a = F/m).

Since we now have the acceleration, we can use the kinematic equation vf^2 = vi^2 + 2ad to find the final velocity (vf) of the electrons when they turn around. We know that the final velocity will be 0 m/s, since the electrons are turning around, so we can solve for the distance (d) that they travel before coming to a stop.

Therefore, the distance that the electrons travel before turning around is d = vf^2 / 2a. I hope this helps! Let me know if you have any further questions.