Coulomb's Law (7 of 7) Force on Three Charges Arranged in a Right Triangle
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- Опубліковано 14 кві 2016
- How to use Coulomb's law to calculate the net force on one charge from two other charges arranged in a right triangle. Coulomb's law states that the magnitude of the force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. If the sign of the charges is the same, both positive or both negative, then there will be a force of repulsion between the charges. If they have the opposite sign then there will be a force of attraction between the objects. The force is along the straight line joining the two charges. The SI unit for the force is the newton.
The force between the charged bodies at rest is commonly known as the electrostatic force. Coulomb's law is an inverse-square law and is similar to Isaac Newton's inverse-square law of universal gravitation, but gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive.
The law is named after the French physicist Charles-Augustin de Coulomb who first discovered it in 1785. The SI unit for electric charge was named after him.
This video can be shared at: • Coulomb's Law (7 of 7)...
well clear and easy to understand with a short time. Thanks for your effort.
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Life savior. I spent so much time watching useless videos before i came across this gem. Thank you for your very good and clear explanation.
Great to hear and thanks for the comment.
Thank you so much! I have been scrambling before my first Physics 2 midterm this week. my professor would not do any examples of these so this was a lifesaver!
Hope the mid term goes well, thanks for watching and commenting.
GREAT VIDEO!
Helpful and straight to the point!
You clarified so many steps for me in a logical manner. Thank you!!
That's what I try to do, glad it was helpful!
I use your video to show my daughter how to resolve this kind of problems. You make my day! Thanks!
Wonderful! Thank for the comment and thanks for helping your daughter. Parental involvement is very important.
Man have I been struggling with this concept and you just simplified it so well! Thank you sooo much for your videos. I also want to add that I really like that you keep your videos as a series of short videos rather than a long one to focus one concept at a time. Super helpful! :)
Glad to hear the videos have been helpful! Thanks for your comment.
I'm so so thankful you give multiple examples for different topics. Khan Academy only does the simplest sample and I'm left stumped on my homework. Watching your channel has basically saved my AP Physics grade. Thank you!
Wonderful! Thanks for watching and commenting. Hope AP goes well.
Subscribed man! This is the clearest video i've watched of this topic you saved me from failure!! Thank you
Great and thank you for subscribing. You can see a listing of al my videos at www.stepbystepscience.com
Thanks for this informative video. I have recommended that my physics students watch it.
This was great thank you so much!
Thank you so much for this easy to follow break down!
Glad it was helpful!
Thank you for the video, I have a question if I broke the first half of the problem into their X and Y components so F31y+F32y and F32x and then did Pythagoreans theorem I would end up with the same answer right?
Thanks for this informational video. Very systematic. I had one problem regarding the component vector addition. Isn't change in vector x associated with cosine value and change in vector y associated with sine value?
how would you do it if you need to find the magnitude and the direction on q2(75 microcoulomb) and lets just say that the distance between q1 and q2 have not been given.....please help
This was so helpful. The explanation and the visual representations are clear. Thank you!
You're very welcome! Glad it was helpful.
This was so clear and I understood it well, thank you!
You're very welcome! Glad it was helpful.
your videos are helping me with my problem solving skills for physics needed for IB, thank you so much!!!!!
Great to hear and thanks for letting me know.
@@stepbystepscience where can I find more similar examples
This was super helpful thank you so much!
You are very welcome, thanks for commenting.
Dude, I have my physics final tomorrow and this is a lifesaver. Thank you!!!!!
Glad to be of help. Best of luck!
Thank youuu! You the real MVP :)
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thank you! so simple and easy to understand :)
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Neat. I remember using this in chemistry but we never went too in depth in it.
+fakingtrels Great, thanks for comenting.
You can see a listing of all my videos at my website, www.stepbystepscience.com
Thanks alot..... Very clear explanation. Really helpful.
What if the distances aren't given? I have a similar homework problem but neither one distance between any two charges on a 90-degree angle plane does not have a distance?
hello, I'm having trouble figuring out when to use sin and cos when solving for f32 (5:03) you use sin and not cos why
Wow, you explained that so easily. I really did overthink it.
Keep it simple. You can a listing of all my videos from my website at www.stepbystepscience.com
Why can't you use the cosine rule with the angle between F32 and F31 as 120°? is there something I'm not understanding about net forces?
This is so so helpful. Thank you so much!
You're so welcome!
the F31 is poitung downwards so it is the one that should be negative not the F32y
thank you! this helped me in my homework
Thank tou very much
The way you explain step by step is amazing
I have enjoyed in doing that
Thanks
You are very welcome! Glad it was helpful.
hi i'm not good at trigo can i ask how did you determine that the triangle is a 30-60-90 triangle? thanks :
you helped me through out this series thank you
You are most welcome. Glad the series was helpful.
I did subscribe its a life savior
I hope you still read comments on this video. Thank you sir for all the amazing videos that illustrates questions.
Sir, I'm only confused why you didn't consider the Fx as negative? it was on the negative direction
2nd.. Why you didn't make the sign of Fy negative when you counted all the forces in Y direction?
Were you only taking the magnitude of forces ??
I still try to respond to all my comments. Yes, when using the equation you are finding the magnitude of the forces and in the end I gave the magnitude of the resultant force and because force is a vector quantity I also gave the direction of the force in degrees from the x axis.
But it is a good point, I suppose for the component forces I could also have put a minus sign in front of the answers, but then again I did also show their direction on the coordinate system.
And thanks for the positive comment.
Thank you very much for the explanation. I really needed to understand it.. Thank you
You're very welcome. Glad it was helpful!
Hello mr Swarthout . I just wanna thank you for sharing these lessons with us . I was also interested in knowing which software or programme are you using to write and record them . I'm doing the same thing with my students in Albania , but i've recorded videos with my phone only . Wished you could help me :)
Hello from Berlin Germany, I have always wanted to visit Albania. You are ver welcome and thanks for commenting. My vidoes are made very low budget. I have a MacBook Pro and all of the presenations are made with Keynote. Then I just recoed with QuickTime, nothing more. In the past I have used a progam called ScreenFlow which can be used to record what is on your computer as well as from your built in camera.
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why did we have to find the angle again at 7:39, since we already had it as 60?
Yours vids are great. Ty
Thank you very much.
Great lesson step by step!
Yes I try to go over everything step by step, thanks for the comment. You can see a listing of all my videos at my website, www.stepbystepscience.com
Why did u put Fx and Fy in the negative direction though they are positive
thanks man!
great job!
Thank you very much.
ur welcome
In the question, there were no angles given, where did you get the angles from
super easy to follow, helped a ton.
Great to hear! Thanks!
The magnitude of the total force acting on q1 is 249 Newtons. but you need to find the charge of Q1?
Thank you, so much better than my TAs
You are very welcome and thanks for commenting.
Why dis you add 180 to the angle? Arent you suppose to minus it? Im confused
Thank you so much. Short time so much to learn.
You are very welcome!
Thanks, this is really helpful.
You're welcome! Glad it was helpful.
Thank u so much
Wonderful explanation
Cleared my all doubts
Glad to hear that, thanks for watching.
Thank you for a very clear explanation
Glad you liked it, you're welcome.
i love you, thank you soooo much dude
You're so welcome!
I understood! Thanks. ❤️❤️❤️❤️✨
Glad you understood, thanks for watching!
Thanks man, helped a lot
Glad to hear it!
Thank u so much sir ❤
U made this soo easy😭
Most welcome, glad it's easy now!
How do you assume the angle of the F32??😊
Thank you very much that really helped a lot
Awesome!
Your explanation is very good. But you have an error in the place you put theta. Theta should be placed above the negative x-axis according to the prolem.
OK, I will check the video.
Dear Mr. Brian! Do you have a video about "diopter"? I've seen your video about farsightedness and nearsightedness, but i can't find anything about diopter. Sorry for this non - related question.
Btw your videos are awesome, and you are really really great teacher!
+Kiril Blazevski Sorry but I do not, but thanks for the positive comment.
You can see a listing of all my videos at my website, www.stepbystepscience.com
Thank you anyways!
Your direction for q3 on q1 should be opposite because the magnitude of attraction is greater on q3 when you disregard the signs.
Not sure I understand your comment, but I did everything correct....of course.
@@stepbystepscience whattttt
Thank u so much! Got an exam tomorrow so hopefully I’ll ace it ❤️
Good luck!!
@@stepbystepscience Thanks! I passed my test by the way
@@ritajabal Excellent, very happy for you.
i don't understand how you got 214 and 80 N my standard form isn't so great, I've tried calculating it with a calculator but it's not giving me the same values..... it'd be of great help if you could give a detailed explanation
Me too
This helped me answer my homework! Thanks 😂
Great and thanks for the comment.
Thankyou very and understandable....
You're most welcome
made my day
Life saver, didn't really go to class much and midterm tomorrow:)
Hope the mid-term goes well.
lmao same but my midterm is today
@@danjamoth4926 tomorrow it's mine
I GOT A 19/20 BLESS THIS VID
Indeed it's awesome explanation 👌
Glad you think so!
THANK YOU!!
You're welcome!
hello may I ask where did you get the angle 68 degrees??
You are looking for the angle that has a target of 2.52. On your calculator you probably have to push the shift tan key to get tan-1 2.52 = 68 degrees. does that help?
🙏 for the lesson, but why did u use sin in resolving for x-axis and cos for y a-axis then the degree I use 180 + your angle 68 which is = to 248 instead of 180-68🤔
I used the sine because the we went to get the x component which is the opposite side and sine is opposite over the hypothenuse and the I used the cosine because we want to get the y component which is the adjacent side and cosine is the adjacent side over the hypothenuse.
And I used 68 plus 180 to get the angle. Does that answer your question?
Bless you
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You're welcome. Thanks for watching.
Thanks, you saved me hours of headache
You're welcome!
Thank you sir
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Wonderful, thanks to you too
Great video
Thanks!
how to get the 68 degre
thankyouuu
Hello sir, I'm still confuse with why you used sine to find the x-component of F31 and cosine to find the y-component of F31. Could you pls help me out?
sine is for the opposite side of the right triangle, cosine is for the adjacent side
@@stepbystepscience alright..thanks very much
you forgot the negative sign it should be -50 because in the diagram it's written -50 so F31=-214
Why using cosine to get the y component instead of sine
you can do that.
How do u determine the angles? How do u know if it’s 30 or 60 degreesss help ur girl out
In this case you know it is this type of triangle because I told you in the video. But if you know the lengths of the sides or have other information about the triangle then you can calculate the angles.
Please, l have question why The angle (60) with y - axis, I Think it with x - axis
it is a 30, 60 ,90 triangle. 30 degrees with the x-axis and 90 degrees with the y-axis.
!! Nice, Thank you
What if there's no angles given but it's still in a right angle?
Use the distances between the charges and the trig functions to determine the angles????
I also need to know how to solve that
thanks
Welcome
shouldn’t it supposed to towards q3 from q1.From minus towards plus
If I understand your question....the force on q3 from q1 is down. They are opposite charges so they attract each other. q1 pulls q3 down. Does that help?
If I understand you question....the force on q3 from q1 is down. The are opposite charges and they attract each other, than means q1 pulls q3 down. Does that help?
hi what to do if q3 is - ve
what is -ve?
Fx should be cos not sin and Fy should be sin
You're the reason my class is passing physics 😭
That's awesome!
How did you get 248°?
Zero degrees is at the positve x axis and then is counted couter clockwise. The angle I calculated is 68 degree and that is actually 68 degrees past 180 dregrees, so 180 + 68 = 248 degrees. Does that help?
Impressive
Thanks!
nice video
Thank you.
What is the net force?
The net force is the sum of all the forces.
You made some mistakes, but overall helped me understand the concept more. Thank you!
What mistakes? Please elaborate.
@@stepbystepscience I'm pretty sure the y component of the F31 should be negative, rather than the F32 y component.
@@PrometeurYeah the problem clearly states the forces influenced upon Q3. The vector he uses expresses the force of Q3 on Q1.. not the other way around which is what the problem is asking. Otherwise it’s good.
How did you get the 60° at 4:02?
The triangle is a 30 60 90 triangle, the 60 is at the upper left, that means for the second part the opposite angle is also 60 degrees.
Step-by-Step Science Ohh thanks! I understand now, thank you so much!
I was about to ask the same question. Thanks though
Due to a property of angles named _vertical angles_ or _opposite angles_ or _vertically opposite angles._
@@stepbystepscience hi i'm not good at trigo can i ask how did you determine that the triangle is a 30-60-90 triangle? thanks :
For the total of Fy..why minus? Not plus? Hehe please answer
Because it is pointing in the opposite direction?
@@stepbystepscience wait but it's going down, shouldnt it be 40N - 214N im confused
@@datboyandrei You end up with the exact same value that (when it's squared later on in Pythagorean) turns positive anyway. You can do it either way you want since you'll get the same answer :) hope that helped
AMAZINGGGGGGG
Thanks so much!!