Off course you would need to have a numerical value for the right side of the equation first. Then you would want to do it numerically. Try a value, see how close you get, then pick a closer value etc.
How can you get the angle theta given those variables. You used sine and cosine functions in the last equation, still u need to simplify further to get a single trigonometric function and do the inverse trig function to get the theta alone :)
This problem would defeat me because I assumed it is asking me to solve for theta in terms of those variables which requires using trigonometric identities and reversing the function to get the angle in pure form.
Not sure what you mean by "both axes". Are you asking why we don't calculate the tension in both strings? If that is the question? If yes, they are perfectly symmetric so the answer is valid for both strings.
If the sum of the forces add up to zero, then there is no net force, which means that the particles will not accelerate. Consequently, if the particles are not accelerating, then there is no net force and therefore the forces will add up to zero.
getting theta by using small angle approximation (as d approaches 0, cos d=1 and sin d = d). Therefore the final answer is θ= (q^2/((epsilon_0)*pi*16*l^2*m*g))^(1/3)
Hey, I just wanted to know why you are able to make the triangle from the forces and also why that theta is equal to the angle we are looking for? Thanks!
Since the two charged objects are in static equilibrium (because they are not moving), the net force on each must be zero. (Newton's second law, F = ma). If there is no net force then the sum of all the forces acting on each charged object must be zero. Hence the triangles. (This is a very typical solution involving vectors adding up to zero).
3 charge particles Q1= -5 micro C, Q2= 5micro C, Q3= -5micro C, assume that Q1 is at the origin and Q3 is right next to Q1, Q2 is at the top between Q1 and Q3. Angle Q2 Q3 Q1 = 20 degrees and angle Q3 Q1 Q2 =40 degrees. HOW MUCH WORK MUST BE DONE TO MOVE Q2 FROM ITS POSITION TO HAFLWAY SITUATED BETWEEN Q2 AND Q3. Sir pls help me with this problem, i can simply get the force for each charge affect by the other charges but i really don't know how to get the WORK.
This video (and the one following this one) may help you figure it out. Physics - Electrical Potential and Electrical Potential Energy (6 of 6) th-cam.com/video/y8vGuA6k6so/w-d-xo.html&index=6&list=PLX2gX-ftPVXXFqBJixIbQcyXZD6kQnAQH
When you place a charge near another set of charges, the work done for each additional charge is equal to the sum of potential energy gained or lost by being close to another set of charges. Change in energy = k * Q1* Q2 / R. Q1 is the charge already there and Q2 is the charge added. You have to calculate that for each Q1 as there could be multiple Q1 charges already present.
sir if i may ask im not good in trigonometry i want to know how can you compute for the angle in your equation sin^3 theta / cos theta... thanks sir...
Ryan Jumar Pantoja no problem. You could write the left side as sin^2 theta * tan theta. The. Divide both sides by sin theta tan theta, then take arc sin of both sides to isolate theta. 😉 I absolutely love trig
Couldn't you also solve this problem by adding the components of the forces like we do in regular static problems instead of making the triangle? I tried it and I got sin(theta)/cos(theta) = (mg*4pi*epsilonR^2)/q^2. So basically I got what you had, but my answer is flipped.
There are usually multiple ways in which problems can be solved. The "triangle method" is a method that is used in many applications and very quick and useful.
Hey, I am somehow not able to solve for thetha...i tried breaking up sin^3 thetha into sin^2 thetha and sin thetha but it didn't work out. Can you plz help me out? Thanks :)
k is an experimentally derived constant of nature. It is related to the properties of the fabric of space, also known as the permittivity of free space
Hi sir, I would like to ask you about how to solve charges in an oblique triangle. Is there another way of solving it without using the cosine law? thank you.
@@MichelvanBiezen yes sir, I followed your solution for a similar problem however in this problem I am to assume that the angle a is small(meaning sin a=a).
Off course you would need to have a numerical value for the right side of the equation first.
Then you would want to do it numerically. Try a value, see how close you get, then pick a closer value etc.
Thanks. From Nigeria 🙃
I'm glad I found your channel
You are welcome. We are glad you found our channel as well. Welcome to the channel.
How can you get the angle theta given those variables. You used sine and cosine functions in the last equation, still u need to simplify further to get a single trigonometric function and do the inverse trig function to get the theta alone :)
You will need to solve it numerically
how did 4 come from k? isn't k supposed to be 9x10^9
5:47
In the denominator: 2^2 = 4
this is way more challenging than physics 104 will call for, and still I'm glad I watched this
Looking at the more challenging problems often helps us with the basic concepts.
Wish my teacher was explaining stuff like you mr. Thanks endlessly
Happy to help! 🙂
I don't understand. where did 1/4piepsilon came from?. I found in the internet it said, "it's equal to coulomb constant." But why is that?
That is correct. 1 /4 pi epsilon is indeed the coulomb constant. epsilon sub knot is the permittivity of free space. It is experimentally determined.
This problem would defeat me because I assumed it is asking me to solve for theta in terms of those variables which requires using trigonometric identities and reversing the function to get the angle in pure form.
Thank you sir, hope you live a long prosperous life. I wish you nothing but the best❤
why tension isn't resolved on both axes against the repulsive force or the gravitational force?
Not sure what you mean by "both axes". Are you asking why we don't calculate the tension in both strings? If that is the question? If yes, they are perfectly symmetric so the answer is valid for both strings.
3:50 sorry to ask but didn't you make an error placing the mg in the denominator?😢
Amazing video Mr. Van Biezen! Thank you so much!
Couldn't you just shift tangent to the right to find theta and so we'll have theta equals arc tangent of the whole thing in the end?
At 2:40, would you please explain why the sum of the three forces is equal to 0?
If the sum of the forces add up to zero, then there is no net force, which means that the particles will not accelerate. Consequently, if the particles are not accelerating, then there is no net force and therefore the forces will add up to zero.
@@MichelvanBiezen Thanks !
getting theta by using small angle approximation (as d approaches 0, cos d=1 and sin d = d). Therefore the final answer is θ= (q^2/((epsilon_0)*pi*16*l^2*m*g))^(1/3)
hi! this might be a silly question but how did end up with 4d^2? isn't 4d only ?? I really don't see the algebra done behind it
Fc = kq^2 / (2d)^2
oh I see it !! thank you! :)
Excellent explanation. Thank you so very much.
You are welcome. Glad you find it helpful. 🙂
What would happen if the charges were opposites of each other? Would the calculations be different?
Then the objects would attract and there would be zero distance between the objects.
@@MichelvanBiezen Got it, Thanks!
Hey, I just wanted to know why you are able to make the triangle from the forces and also why that theta is equal to the angle we are looking for? Thanks!
Since the two charged objects are in static equilibrium (because they are not moving), the net force on each must be zero. (Newton's second law, F = ma). If there is no net force then the sum of all the forces acting on each charged object must be zero. Hence the triangles. (This is a very typical solution involving vectors adding up to zero).
Dear Sir. I would use a small-angle approximation to find the final answer. So the angle = (A/B)^(1/3)
Why? This is a more general answer.
In this case,How to find tension in the rope?
The mass and quantity of charge are independent. The mass represents how much material is in each object.
3 charge particles Q1= -5 micro C, Q2= 5micro C, Q3= -5micro C, assume that Q1 is at the origin and Q3 is right next to Q1, Q2 is at the top between Q1 and Q3. Angle Q2 Q3 Q1 = 20 degrees and angle Q3 Q1 Q2 =40 degrees. HOW MUCH WORK MUST BE DONE TO MOVE Q2 FROM ITS POSITION TO HAFLWAY SITUATED BETWEEN Q2 AND Q3.
Sir pls help me with this problem, i can simply get the force for each charge affect by the other charges but i really don't know how to get the WORK.
This video (and the one following this one) may help you figure it out. Physics - Electrical Potential and Electrical Potential Energy (6 of 6) th-cam.com/video/y8vGuA6k6so/w-d-xo.html&index=6&list=PLX2gX-ftPVXXFqBJixIbQcyXZD6kQnAQH
done watching those videos sir ,over and over,still no idea in solving it. i dont know what formula to use in getting the WORK. pls help me sir.
When you place a charge near another set of charges, the work done for each additional charge is equal to the sum of potential energy gained or lost by being close to another set of charges. Change in energy = k * Q1* Q2 / R. Q1 is the charge already there and Q2 is the charge added. You have to calculate that for each Q1 as there could be multiple Q1 charges already present.
sir if i may ask im not good in trigonometry i want to know how can you compute for the angle in your equation sin^3 theta / cos theta... thanks sir...
Ryan Jumar Pantoja no problem. You could write the left side as sin^2 theta * tan theta. The. Divide both sides by sin theta tan theta, then take arc sin of both sides to isolate theta. 😉 I absolutely love trig
i from india and you are doing a right for student like me and others thank you sir , you are genius
Welcome to the channel!
Please make videos on hard problems on electricity
Thanks for this. Makes so much sense now. You are awesome
I'm from India 🇮🇳
You cleared my very important doubts.
Thanks 😊😊❤
Glad to help. Welcome to the channel!
Thank you so much❤
You're welcome 😊
could you please list problems 4- 8 on ilecture online?
We still need to link a number of videos to our web site. Thanks for the reminder.
Please let me know if you need any help, I am available to help. I am a also web programmer. Thank you for the videos excellent explanation.
Amazing teaching over their sir.thank u sooo much .greetings from India
Couldn't you also solve this problem by adding the components of the forces like we do in regular static problems instead of making the triangle? I tried it and I got sin(theta)/cos(theta) = (mg*4pi*epsilonR^2)/q^2. So basically I got what you had, but my answer is flipped.
There are usually multiple ways in which problems can be solved. The "triangle method" is a method that is used in many applications and very quick and useful.
Hey, I am somehow not able to solve for thetha...i tried breaking up sin^3 thetha into sin^2 thetha and sin thetha but it didn't work out. Can you plz help me out? Thanks :)
What if the is no gravity
Without gravity the strings would be horizontal.
You, kind sir, just earned a subscriber! Thank you so much for your videos!
Love from 🇮🇳 (Indian)
Thank you and welcome to the channel!
I have a question. Why the 16pieEsubnot0????
k = 1/(4 * pi * epsilon sub knot)
+Michel van Biezen Is it like an established principle or law? That K equals that Dr. Biezen?
k is an experimentally derived constant of nature. It is related to the properties of the fabric of space, also known as the permittivity of free space
Hi sir, I would like to ask you about how to solve charges in an oblique triangle. Is there another way of solving it without using the cosine law? thank you.
What is the tension when the system is taken deep into space ??
+ViDdU vidyarth
Then the distance between them would be 2L and you would use Coulomb's law to calculate the force ( = tension) between them.
Because there would be no weight (no gravity) and only the charge forces acting to cause tension in the string. Nice.
Sir please tell me where i can study my maths
on a desk
Thank you so much for the clear, easily understood lecture.
thank u sir.. its amazing
Sir why do we use 2d^2
we use (2d)^2 because its the distance of two charges.
Thank you so much
You're most welcome
Awesome sir
how about the tension ?
Thank you sir
You are welcome.
amazing... nice lecture sir...
Excellent videos, greetings from Chile!
Welcome to the channel.
Such questions are very much easy for 12th students from India
This a very important comment, and certainly contributes much in this context. ;)
Thank you so much!
You're welcome!
Hello, you forgot to add these videos to ilectureonline
Thanks for letting us know. :)
@@MichelvanBiezen Hello sir.I think you don`t added yet.
Any NCERT here????
Yeah
Could u please make some more challenging problems. Thank you.
Thank you so much for this.
Also if you could solve it for theta it would be better
You are right man! Just another angle of consideration
nice sir
thank you soo much you are really great
Sir, I get answer = (q^2/2π £o mgL^2)^1/3 but its different from yours.
Did you work it out the same way?
@@MichelvanBiezen yes sir, I followed your solution for a similar problem however in this problem I am to assume that the angle a is small(meaning sin a=a).
Why don't you plug in some numerical values and see if the 2 answers are close?
You are awesome
Thank u sir
a legend
Glad you found our videos. 🙂
simple problems
I thought we are going to find Q=...... I was afraid
wow i blinked
Wow!
lato lato
? 🙂
👍
In this case,How to find tension in the rope?
Since we know the Coulomb force and mg, we use the Pythagorean theorem (or sin, cos, or tan).
But mass is unknown,Sir!