I am an Indian High School student this Explanation is remarkable. Not many people are able to explain it that perfectly and smoothly. I am very thankful.
My professor made it over compilicated in his slides But know it's much easier I am 1st year Computer Engineering from Egypt and happy I found this video Keep up the good work
Amazing analysis, I have tons of questions related to Earth motion in relation to electron motion and the warbling effect of earth around the sun. This is just a mind blowing phenomenon. I was thinking about the electron speed vector at the right side of the field, and now I can sleep better.
Sir, is there any condition of charge particle could leave the space between two plate that contains the electric field? Some, case charge particle couldn't leave.
@@stepbystepscience If in problem asks, Could charged particle leave the space (uniform electric field) between the plate? How could we convince, it can leave or can not leave.
neglectable, normally for questions for this kind of stuff, the answers don't have to be 100 spot on, as long as they are accurate. If you use Fg in the equation, your answer won't be far off from when you solve w/o Fg, since the weight of the particle is so tiny.
Starting from a = qE/m. The potential difference (voltage) across the plates is U = Ed, rearrange for E gives E = U/d and substitute into the equations acceleration equations for E....finally you have a = qU/dm...Does that help?
While everyone is writing they are studying this in University......I am studying this for grade 12 CBSE boards in India 🤦🙇.... Btw ,Thankyou for the well versed explanation....I am a subscriber now 😄
For part B, you used the time in the x, which is d/v and got 7.2*10^-9s. Shouldn't the y time also be the same though? If I used y=1/2(at^2), I get a different number for my answer. Unless, I am falsely assuming that the x-axis is in the midpoint of the capacitor. Since I used .025m as my y.
No, y value is not 0.025m, instead 0.0137m needs to be applied in the equation, because after leaving the electric field, the electron has only travelled 0.0137m in y-direction
Hey, man thanks for the lesson but the question I was trying to attempt gave me the initial velocity in the x-axis, the length of the plates and magnitude of the electric field. They want me to solve for the final velocity as the electron leaves the uniform field. What equations would I use in this case? Thanks
Very nice and clear. However, you've confused tangent with its inverse in a significant way. The *tangent* of the angle is equal to the ratio of the side lengths given. The inverse tangent is simply used in the *solution* of the equation, to solve for the angle.
I'm studying at university and of all the videos I've covered, THIS... is the most professional and simplest of them all!!
Thank you for the very nice, very positive comment. Means a lot to me!
I have been trying to figure out a theoretical variant of this problem for like 6 hours and NOW I FINALLY KNOW HOW TO DO IT! Bless you man!
That's awesome!!
I am an Indian High School student this Explanation is remarkable. Not many people are able to explain it that perfectly and smoothly. I am very thankful.
Thank you for such a nice comment.
Professors make this concept much more complicated than it needs to be... thank you
You're very welcome
The explanation is clear, his workings are good i have to say this is genuine.
Thanks so much!
These explanations are so clear and simple. Thank you sir, much appreciated!
Glad you like them, you're most welcome!
My professor made it over compilicated in his slides
But know it's much easier
I am 1st year Computer Engineering from Egypt and happy I found this video
Keep up the good work
Thanks for your comment and best wishes for your studies.
I was struggling in physics today with a very similar question. I could not thank you enough for this video! Awesome job :D
Glad to hear the video helped! Thanks for your comment.
Thank you so much for the detailed explanation! You're a savior :)
You are very welcome, thanks for the great comment.
I'm so glad I found this video , Thank You
You are so welcome!
Amazing analysis, I have tons of questions related to Earth motion in relation to electron motion and the warbling effect of earth around the sun. This is just a mind blowing phenomenon. I was thinking about the electron speed vector at the right side of the field, and now I can sleep better.
Great!
thank you for your effort, clear and simple
So nice of you, all the best
The best teacher ever
So nice of you, thanks!
Your videos are so helpful. Thank you so much.😊
You are so welcome!
so helpful!! couldn't survive physics II without :)
Great to hear! Best wishes.
Sir, is there any condition of charge particle could leave the space between two plate that contains the electric field? Some, case charge particle couldn't leave.
if the particle is moving fast enough or if the electric field is not ver strong then it could leave the space between the plates.
@@stepbystepscience If in problem asks, Could charged particle leave the space (uniform electric field) between the plate? How could we convince, it can leave or can not leave.
Thank you so much! My final is in 5 hours you helped me so much
You are very welcome, hope it goes well.
hey, to calculate the acceleration wouldn't we also have to take into account the effects of gravity? thank you for the video
The effect of gravity in these cases is negligible i guess
neglectable, normally for questions for this kind of stuff, the answers don't have to be 100 spot on, as long as they are accurate. If you use Fg in the equation, your answer won't be far off from when you solve w/o Fg, since the weight of the particle is so tiny.
I'm confused where the accell equation a=qV/md come from? I usually see other use common one (v')^2=(v)^2 + 2as (well thts if the velocity were known)
Starting from a = qE/m. The potential difference (voltage) across the plates is U = Ed, rearrange for E gives E = U/d and substitute into the equations acceleration equations for E....finally you have a = qU/dm...Does that help?
@@stepbystepscience oouu thnkss i just know theres other equation U = Ed thnkss :D
explained really well, thank you!
You're very welcome!
Best explanation ever 👌👌
Glad you think so and thanks for the comment.
While everyone is writing they are studying this in University......I am studying this for grade 12 CBSE boards in India 🤦🙇.... Btw ,Thankyou for the well versed explanation....I am a subscriber now 😄
Thanks for the comment and happy to have you as a subscriber!
also grade 12 Boards, but not Indian curriculum
Studying this for the Canadian grade 12 curriculumn too you're not alone!!!
I did. this first term grade 11
Great explanation!💪
Glad you liked it, thanks!
For part B, you used the time in the x, which is d/v and got 7.2*10^-9s. Shouldn't the y time also be the same though? If I used y=1/2(at^2), I get a different number for my answer. Unless, I am falsely assuming that the x-axis is in the midpoint of the capacitor. Since I used .025m as my y.
Yes, the time in the y-direction is the same as the time in the x-direction.
No, y value is not 0.025m, instead 0.0137m needs to be applied in the equation, because after leaving the electric field, the electron has only travelled 0.0137m in y-direction
how does the acceleration of an electron compared to a alpha particle?
Would be much greater because it has a much lower mass.
When the charge exit the plates it has increased its kinetic energy. Where dies this energy comes from?
Thank you! Great job
Thank you too!
Sir... What is the force of electric field on an electron....
Do you mean what is the force on an electron from the electric field? It is F = Eq.
This was so helpful
Glad to hear it, thanks!
Really helped me in our assignment
Thank you!
Welcome!
Well explained Sir💯
Thanks so much!
Thank u☺️
U clear my doubt by just diagram 😄
Glad to hear that
Its very helpful ❤
Glad you think so!
you just saved my grade
That's great. Glad to hear it!
Thanks a gazillion timesssss
You bet, thanks for watching!
Very helpful !!
That's great, glad you think so!
Hey, man thanks for the lesson but the question I was trying to attempt gave me the initial velocity in the x-axis, the length of the plates and magnitude of the electric field. They want me to solve for the final velocity as the electron leaves the uniform field. What equations would I use in this case? Thanks
s=ut + 1/2 at^2
Very nice and clear. However, you've confused tangent with its inverse in a significant way. The *tangent* of the angle is equal to the ratio of the side lengths given. The inverse tangent is simply used in the *solution* of the equation, to solve for the angle.
ok
U really explain very well sir 😃. Thank u soo much 😊
So nice of you
ur a king 👊😎
So nice of you!
i'm a bit confused on the arctan part. why cant it just be regular tan theta
because you need to find the angle and not the tangent of the angle
Glad i find this
Super, I'm glad too!
Great u helped me to solve my quiz
Great, I hope the quiz went well.
@@stepbystepscience yeah thanks
Thank you so much man!!!!!!
No problem!
Thanks
Welcome
Nice positive comment!
Thanks!
wow this is soo good . thanks
Glad you like it! And thanks a lot for commenting.
So damn good!
Thank you
Woyooooo nice one sir,,thanks alot
You are very welcome.
Wow.....thank you sir...
Most welcome and thanks for the comment.
GOATed
Thanks!
nice positive comment
thanks 😊
Is that the Cookie Monster?
/
Bart
I don't get it.....