/sci/ - Science & Math » Thread #11542373

3MiB, 1920x1080, ScreenShot14.png

View Same Google iqdb SauceNAO

help with physics

Anonymous Tue 07 Apr 18:51:29 2020 No.11542373 View Reply Original Report

Quoted By: >>11542660 >>11542671 >>11542688

Just tell me what formula i should use pls. I have no clue how speed factors in. I am rarted

An electron (me = 9.11 · 10?31 kg) in the homogeneous electric field of a cathode ray tube is supposed to
achieve a velocity of 6 · 107 m/s after an accelerating distance of 10 cm.
a) What potential difference is necessary for this?
b) What is the corresponding electric field strength?

Anonymous

Anonymous Tue 07 Apr 2020 20:16:52 No.11542660 Report

Quoted By:

>>11542373
1/2mv^2=qV
qEd=1/2mv^2

Anonymous

Anonymous Tue 07 Apr 2020 20:19:58 No.11542671 Report

Quoted By:

>>11542373
How do you even know an electron exists? Have you seen one? Why are you going over this mental masturbation about things youre not even sure if are real

Anonymous

Anonymous Tue 07 Apr 2020 20:24:01 No.11542688 Report

Quoted By:

>>11542373
In this case, the change in potential energy is equal to the change in kinetic energy for the electron, or $U=\Delta K = K = (\gamma -1)mc^2$ , (or you can use the classical expression for $K$ , they're pretty close but the relativistic effects are non-negligible) and of course $U=qV$ , so $V = \frac{(\gamma -1)mc^2}{q}$ . Then, in the vase of a homogeneous electric field, $V = Es$ , so $E = \frac{V}{s}$ , where $s$ is the separation between the electrodes.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches