Thanks a lot for those very concise and didactally awesome lessons. Although I studied electrical engineering those lessons made contents even more clearer to me. I am retired but a lifelong learner and even sat down trying to solve the challenging tests. Due to your godd lessons I was able to see things more clearly. Keep on the good work, Uwe Janssen
Uwe,Great to see you are staying active with this intellectual stimulation. I find that as I am getting older, I am still making new discoveries on material I never fully comprehended. Enjoy.Michel
Very cool video. I'm excited that I am at my point in my engineering journey such that all of this made perfect sense and I could probably replicate a similar problem pretty easily. Thank you for the video sir, enjoy your week!
First Thx for all your videos. I really did enjoy watching it. I wanna know that are there any textbooks or sources that have the challenging problems like this, because I love doing difficult problems
Sir Michel van Beizen, I really don't understand why you put a -1 (5:13 - 5:15). Is it part of the rule? What I know is Work done = negative Kinetic Energy. Thank you sir. BTW, a good example tho.
I'm not sure if it is going to help you right now, but that -1 comes from the U substitution. He didn't show the U substitution. u= (a-x) du= -1dx . As a result, dx= (-1) du.
Why is force not constant that we had to use calclus? When we calculate attraction force between two particles we never do use calclus so I am curious.
If the particles are at a fixed distance and you want to calculate the force between them you don't need calculus. But if you want to calculate the work or energy, or something like that which depends on the force as the particles move, the force will be constantly changing.
hey, would the work done by the electron be negative.. in that the +ve to -ve is an attractive force, and therefore for the electron to go there to those alpha particles would be negative work. conversely if it were the same like charges it would take work to go to each other and therefore the work is positive
Since the electron will be in a lower energy state, the work done moving the electron closer to the positive charge is indeed negative. W = change in energy
the particle is gonna move to b he called that distance x and since x starts at 0 means that (a-0) = a and should end at a-b means a-(a-b)=b and thats the whole work done from (a) distance down to (b) distance. he just described a function
It's a method that allows anyone to express a variable distance. In this case, X becomes the variable distance. If the distance between the 2 charges was 1 meter, we can describe the distance of one particle moving across that space as (1 - x). He substituted that for r, which he doesn't need. (1 - x) allows us to then apply differentiation and integration techniques.
Hi sir, I did not understand the first step of the exercise or rather see it as illogical Because the change in work is impossible to equal the change in the distance times the force For example between two distances of 5 meters(f=5n) and 6meters(f=6n) the Change in work will be equal 11 But with dw = f * dx the change in work will be 5.5 between the two points so i think that the change in the work is dw = f(x)*dx + df(x)*x =(5.5n)*1m + 1n * 5.5m please sir could you correct my way of thinking if i am wong
Whenever you use differentials, they will give you the instantaneous work. Since the work is not linear, you cannot compare it to a linear approximation. That is the key to differentiation and differentials. The concept of that can be found here: CALCULUS 1 CH 1 LIMITS & DERIVATIVES and here: CALCULUS 1 CH 6 DIFFERENTIALS
@@jworldtrading6490 they are talking about finding the strength of the electric field or voltage and calculate how much energy it takes to move the charge and then relate it to 1/2 mv^2
why we did integration for this question? I thoght , we just need to think W=F.dx in this question we know dx=(a-b) F=k.q.Q/R^2 What is R for force equation? R is changing continuosly.So we need to integrate. (That comment for viewers who do not understand integration)
There are always multiple ways in which you can set up a problem like this. It this example, I wanted to reference the distance x from the origin in terms of a and b. You can try it in a different way.
I multiplied the integral by (-1) twice in order not to change the value. It was done to produce the correct differential in order to integrate it correctly.
We multiplied by "-1" twice, so we didn't change anything. But we do need a proper differential in order to integrate. The differential of (a - x)^-2 is - dx. Therefore we needed the negative there.
Thanks a lot for those very concise and didactally awesome lessons.
Although I studied electrical engineering those lessons made contents even more clearer to me.
I am retired but a lifelong learner and even sat down trying to solve the challenging tests.
Due to your godd lessons I was able to see things more clearly.
Keep on the good work,
Uwe Janssen
Uwe,Great to see you are staying active with this intellectual stimulation. I find that as I am getting older, I am still making new discoveries on material I never fully comprehended. Enjoy.Michel
multiplies by -1 because he just skipped the u-substitution since it was simple enough.
Great video
Very cool video. I'm excited that I am at my point in my engineering journey such that all of this made perfect sense and I could probably replicate a similar problem pretty easily. Thank you for the video sir, enjoy your week!
Glad you liked it. 🙂
God Bless you Michel van Biezen, God Bless you.
3:54 why we didnt mulitply the force with the total distance traveled while the force was applied??
The force is not constant and that is why we must integrate.
@@MichelvanBiezen thank you sir
Sir what if both electron and alpha move towards each other??plz make a video on this..
before 5th april, can anyone explain why is it that we want to find the force in terms of c like used in 1:55
This problem can be more easily solved using the energy conservation equation.
First Thx for all your videos. I really did enjoy watching it. I wanna know that are there any textbooks or sources that have the challenging problems like this, because I love doing difficult problems
I like the way you solve questions. I appreciate you!
Sir Michel van Beizen, I really don't understand why you put a -1 (5:13 - 5:15). Is it part of the rule? What I know is Work done = negative Kinetic Energy. Thank you sir. BTW, a good example tho.
I'm not sure if it is going to help you right now, but that -1 comes from the U substitution. He didn't show the U substitution. u= (a-x) du= -1dx . As a result, dx= (-1) du.
Hi charlie wyeth falcon,
He put -1 to get the integration of form ((x)^n)(x').
Why is force not constant that we had to use calclus? When we calculate attraction force between two particles we never do use calclus so I am curious.
If the particles are at a fixed distance and you want to calculate the force between them you don't need calculus. But if you want to calculate the work or energy, or something like that which depends on the force as the particles move, the force will be constantly changing.
hey, would the work done by the electron be negative.. in that the +ve to -ve is an attractive force, and therefore for the electron to go there to those alpha particles would be negative work. conversely if it were the same like charges it would take work to go to each other and therefore the work is positive
Since the electron will be in a lower energy state, the work done moving the electron closer to the positive charge is indeed negative. W = change in energy
Is there any way to calculate this for a water molecule.
I do not understand why you chose a random position with a distance from initial point is x and then let Fc with R = a - x ? Maybe to solve dW?
the particle is gonna move to b he called that distance x and since x starts at 0 means that (a-0) = a and should end at a-b means a-(a-b)=b and thats the whole work done from (a) distance down to (b) distance. he just described a function
It's a method that allows anyone to express a variable distance. In this case, X becomes the variable distance. If the distance between the 2 charges was 1 meter, we can describe the distance of one particle moving across that space as (1 - x). He substituted that for r, which he doesn't need. (1 - x) allows us to then apply differentiation and integration techniques.
Can i directly find the acceleration of the electron by using a=F/m and than use v^2=u^2+2as for finding final velocity as u=0 ?
Not in this case since the acceleration is not constant.
Thanks Sir...!! You are really awesome !!
Is there a way to do this without using intergral ? I haven’t learnt it yet
Not this example. But we have plenty of other examples where you don't need calculus in these playlists on Coulomb's law.
why cant we call the whole distance x and then integrate? I tried to solve like that and couldnt get a valid answer
No, it must be solved as shown in the video.
Hi sir,
I did not understand the first step of the exercise or rather see it as illogical
Because the change in work is impossible to equal the change in the distance times the force
For example between two distances of 5 meters(f=5n) and 6meters(f=6n)
the Change in work will be equal 11
But with dw = f * dx the change in work will be 5.5 between the two points
so i think that the change in the work is dw = f(x)*dx + df(x)*x =(5.5n)*1m + 1n * 5.5m
please sir could you correct my way of thinking if i am wong
Whenever you use differentials, they will give you the instantaneous work. Since the work is not linear, you cannot compare it to a linear approximation. That is the key to differentiation and differentials. The concept of that can be found here: CALCULUS 1 CH 1 LIMITS & DERIVATIVES and here: CALCULUS 1 CH 6 DIFFERENTIALS
Thank you sir
Hey man what does the Big E with the subzero stand for and what's it's value(Is it a constant)?
It represents a constant value.
sir,why do we use this method..?we can take the same answer easily by using electric potential energy change = kinetic energy gain..
That should work.
How do we apply the electric potential energy charge=kinetic energy formula may u explain it to m
@@jworldtrading6490 they are talking about finding the strength of the electric field or voltage and calculate how much energy it takes to move the charge and then relate it to 1/2 mv^2
Thank u so much from algeria ^^
Welcome to the channel!
why we did integration for this question?
I thoght , we just need to think W=F.dx in this question we know dx=(a-b) F=k.q.Q/R^2 What is R for force equation?
R is changing continuosly.So we need to integrate.
(That comment for viewers who do not understand integration)
I didn't understand why you choose a_x as a distance why you didn't work with a directly please answer me !!
There are always multiple ways in which you can set up a problem like this. It this example, I wanted to reference the distance x from the origin in terms of a and b. You can try it in a different way.
sir why did we multiply -1 for integration is that rule to integrate?
he did a u sub in his head. he said u = a- x therefor du = -dx
Hello Sir, I didn`t understand why you multiply by -1 in the integral part. Thank you
I multiplied the integral by (-1) twice in order not to change the value. It was done to produce the correct differential in order to integrate it correctly.
Is this calculus based physics or algebra based physics?
It is both, but emphasizes calculus based.
Hi, Is there a way to calculate the work easier and faster?
fenash1,
This is a more advanced problem.
Take a look at the physics 35 coulomb's law playlist for more straightforward problems.
I really didnt get the point that you multiplied by -1 while integrating.
We multiplied by "-1" twice, so we didn't change anything. But we do need a proper differential in order to integrate. The differential of (a - x)^-2 is - dx. Therefore we needed the negative there.
Why you omitted the mines
Can this also be solved using Kinematics?
Robb G. Yes ,but it would become more complex still on one stage you would have to apply this method.
The audio is not very clear oh man. Interesting topic
Yes, our early videos had some audio problems. We fixed that for the later videos.
X, dx and b confuse me 😢🙎♀️
wakapenga chibaba
your integration is wrong
You are blocking your own explanation
Yes, it took me a while to learn how not to do that. 🙂
👍