Coulomb's Law (7 of 7) Force on Three Charges Arranged in a Right Triangle
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- čas přidán 14. 04. 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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Great to hear and thanks for the comment.
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Hope the mid term goes well, thanks for watching and commenting.
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Great and thank you for subscribing. You can see a listing of al my videos at www.stepbystepscience.com
Thank you so much for this easy to follow break down!
Glad it was helpful!
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Wonderful! Thanks for watching and commenting. Hope AP goes well.
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Glad to hear the videos have been helpful! Thanks for your comment.
Thanks for this informative video. I have recommended that my physics students watch it.
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You are very welcome, thanks for commenting.
This was so clear and I understood it well, thank you!
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Glad to be of help. Best of luck!
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You are most welcome. Glad the series was helpful.
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@@stepbystepscience where can I find more similar examples
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
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
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
thanks man!
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ur welcome
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?
thank you! this helped me in my homework
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Thank you very much for the explanation. I really needed to understand it.. Thank you
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Thank you for a very clear explanation
Glad you liked it, you're welcome.
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You are very welcome and thanks for commenting.
super easy to follow, helped a ton.
Great to hear! Thanks!
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?
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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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Thanks, this is really helpful.
You're welcome! Glad it was helpful.
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
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.
the F31 is poitung downwards so it is the one that should be negative not the F32y
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.
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
i love you, thank you soooo much dude
You're so welcome!
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?
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?
Thank u so much
Wonderful explanation
Cleared my all doubts
Glad to hear that, thanks for watching.
Thank you very much that really helped a lot
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Glad to hear it!
why did we have to find the angle again at 7:39, since we already had it as 60?
Thank u so much sir ❤
U made this soo easy😭
Most welcome, glad it's easy now!
Thankyou very and understandable....
You're most welcome
In the question, there were no angles given, where did you get the angles from
THANK YOU!!
You're welcome!
Why did u put Fx and Fy in the negative direction though they are positive
hi i'm not good at trigo can i ask how did you determine that the triangle is a 30-60-90 triangle? thanks :
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.
Why dis you add 180 to the angle? Arent you suppose to minus it? Im confused
How do you assume the angle of the F32??😊
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+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!
Thank you so much
You're welcome. Thanks for watching.
Indeed it's awesome explanation 👌
Glad you think so!
Life saver, didn't really go to class much and midterm tomorrow:)
Hope the mid-term goes well.
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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
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.
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?
This helped me answer my homework! Thanks 😂
Great and thanks for the comment.
Thanks, you saved me hours of headache
You're welcome!
Thank you sir
Welcome and all the best
The magnitude of the total force acting on q1 is 249 Newtons. but you need to find the charge of Q1?
thankyouuu
thanks
Welcome
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
Great video
Thanks!
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Wonderful, thanks to you too
how to get the 68 degre
🙏 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?
Why using cosine to get the y component instead of sine
you can do that.
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 :
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
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?
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
hi what to do if q3 is - ve
what is -ve?
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
You're the reason my class is passing physics 😭
That's awesome!
What is the net force?
The net force is the sum of all the forces.
🙌🏽
Impressive
Thanks!
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?
nice video
Thank you.
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.
tk!
Welcome!
AMAZINGGGGGGG
Thanks so much!!
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
how come you are using sin for X and cos for Y
Because one of the basic trig functions says: sin(θ)=Opp/Hyp, So: Opp=sin(θ)*Hyp
while Cos(θ)=Adj/Hyp, meaning: Adj=cos(θ)*Hyp
Thank you.