You are a life saver. I have been following you since last september and man, your explanation are spot on. Better than some of our teachers here in Montreal's polytechnique. Merci beaucoup!
have a physics exam for this tomorrow and you sir are an excellent teacher. great explanations and give pointers in areas where most educators would not bother to. thanks for your videos you are changing the world a student at time.
THANK YOU SO MUCH for all your videos you've made for physics. Explaining the concepts behind everything has made University Physics so much easier for me.
Your videos are amazing! Please keep it going!! I would love to see more conceptual based videos as well (in addition to the numerical/computational based problems that you do).
Nice lecture. It is relevant to stress, however, that a long distance (a>>d) the dipole behaves as a point charge with effective charge of q.d; and the decay of the electric field magnitude is to the power of minus three (instead of minus two), i.e, decay faster
Thank you so much sir! It was so easy an explanation for so hard a topic (for me for the last 2 years)! You are an excellent teacher. Thank you again sir
i literally pay hundreds to learn concepts that are explained in such a shit way. then there's this channel where i learn concepts just like that in a matter of 10 minutes. thanks bruh
please help, A water molecule has a permanent dipole moment of magnitude 6.2*10^-30 Cm. A water molecule located 10 nm from a Na+ ion in a saltwater solution. what force does the ion exert on the molecule?
will the Force be in the same direction as the electric field, and will we only have to multiply the formula derived for the electric field by q to find the force?
So is this true for ALL dipole? The y components will always cancel? PS Thanks for all the great videos. They got me through rotational kinematics and now E&M.
+Rahul Tiwari An electric dipole is a particular arrangement of two opposite charges. When a positive charge is placed closed to a negative charge, the two charges close together like that is called an electric dipole. The electric field perpendicular to the line connecting the two charges has a particular equation as shown in the video.
Hey there. Awesome video. Your teaching style/method matches up with my learning style/method. Maybe you're just a good teacher, I don't know. But hey, I'm having trouble understanding around 5:30 why x^2+a^2 square root x^2+a^2 became (x^2+a^2)^(3/2). Maybe that's an algebra rule I didn't understand. Would you or someone care to clarify? Please and thanks yo
Ok, thank you. Though, I'm still not seeing the addition equaling (3/2) 🤔. You have an exponent of (1/1) and you're adding it to the exponent of (1/2). I don't see what I'm missing
you can do these equations very fast! thats impressive. for future reference i think it would be more effective for teaching first time learners if you went a bit slower
We did by choosing the correct direction of the electric field caused by the second charge. Then we found the magnitude of the resulting electric field.
How do i do this problem if the Qs are not equal in magnitude but different numbers and do not have the same sign? That would mean their corresponding vertical components would not cancel and would I have to calculate those vertical components as well?
in the previous video,when we calculate the total magnitude of electric field ET, we say ET = square root of ETx^2 + ETy^2 ,but in this video we calculate the magnitude of ET as the sum of two x components .why we do that and what is the difference between two problems?
That is what is so interesting about dipoles. The y-components of the electric field due to the 2 charges are equal in magnitude and opposite in direction and thus cancel out. The x-components however are equal in magnitude, BUT in the same direction, thus they are added.
Since the two x-components point in the same direction we can simply add them. Note that you only have to use the square root of the sum of the squares method if the components are perpendicular.
Back in the day, master mathematicians called The "perpendicular bisector" of a line the "MEDIATOR". I wish I saw it more often...-hint- ps I think it originated in the field of geometry, then went general. Does anyone else know anything about this?
soulseeker Either charge. (They both have the same magnitude, one is -q and the other is +q) Two charges with the same magnitude (but opposite sign) spaced closely together forms a "dipole"
really good explanation. But what surprises me is that why do we ignore the sign of q. I am used to working in vectors so when i add them as vectors, not as magnitudes, I am supposed to get a 0 due to opposite charges. Could you explain this..
We don't actually ignore the direction component of the vector. In the beginning we draw the electric field component contributed by each charge. We labeled them E1 and E2. We then added them as vectors, and realized that only the vertical components cancel and that the horizontal components are additive. All we have to do then is add the magnitudes of the horizontal components to find the magnitude of the total electric field at that point. And the direction is in the positive x-axis.
Usually axis is perpendicular to whatever we are considering. eg. in case of the earth (almost perpendicular to the equator). But in case of the axis of the electric dipole, why do we consider it along the dipole ?
+ANNA KAVALAM In more advanced courses we calculate the effect of the dipole from every direction. The simple case we only look at the dipole from a point perpendicular to the line connecting the two charges or along the line connecting the two charges.
This explanation has some serious flaws: 1.The angle for the negative charge isn't the same as that for the positive one as is clear from the diagram 2.The electric field of the negative charge is considered negative but he chose it to be positive 3.The distance towards the left of a should be negative
Point charges 𝑞1 = +12 nC and 𝑞2 = -12 nC are 0.100 m apart. (Such pairs of point charges with equal magnitude and opposite sign are called electric dipoles.) Compute the electric field caused by 𝑞1, the field caused by 𝑞2 , and the total field (a) at point a; (b) at point b; and (c) at point c. Kidly solve sir
I wonder if I can simply find the angle using the inverse tangent. I've been doing that and my answer are always a bit off. I'm not sure if that's what I"m doing wrong.
Without seeing what you are doing it is not possible to respond to this question, but if you know the opposite side and you know the adjacent side and it is a right triangle then it should work.
Ah, you are asking about the sign of the charge? The sign of the charge only determines the DIRECTION of the electric field. The magnitude of the electric field only depends on the magnitude of the charge.
Sir I don’t find the lectures on dipole dipole interaction will you be uploading later or not please let me know sir your lectures are very helpful to us so kindly upload this topic too sir 🙏🙏🙏
With a dipole you typically pick the point along the perpendicular bisector. But if you want to fine the electric field at any other point, the vertical components would not cancel out and you would have perform an integration twice, once for each direction.
sir where can i find the video for electric field expression for a electrical dipole for axial position and any random position other than axial or vertical . please help . i need those urgently .
That said, it is worked out exactly the same except that the y-component doesn't cancel out and you'll have to calculate both the x and y component of the electric field.
@@MichelvanBiezen sir its easier said than done for students like me ... it would have been of great help if u could have put another video explaining that ... anyways thanks a lot for your existing videos ... they help a lot
6 years later and still helping students all over the world.thanks a bunch
7 years** , thank you!
I swear MVB is the best teacher I've ever had
9 years
I bet those dislikes are from grumpy physics teachers who dont want student's to pass LOLOLOLOL !
100%
You are a life saver. I have been following you since last september and man, your explanation are spot on. Better than some of our teachers here in Montreal's polytechnique. Merci beaucoup!
+Cdk Kamungaz C'est mon plaisir.
have a physics exam for this tomorrow and you sir are an excellent teacher. great explanations and give pointers in areas where most educators would not bother to. thanks for your videos you are changing the world a student at time.
So true !!
MY SYMMETRICALLY-HEARING EARS ARE HAPPY, FINALLY.
indeed, thank you for this comment :(
y r u sad?
YESSS
I have my physics exam tomorrow and this equation was never derived for me, thank you SO MUCH. I'm gonna look at the next ones too :D
THANK YOU SO MUCH for all your videos you've made for physics. Explaining the concepts behind everything has made University Physics so much easier for me.
You have perfected the art of teaching.
I appreciate another excellent lecture. Taking E&M this summer, I must rely this channel to get me through it.
You are welcome. You will find all the topics covered in E&M explained in detail on this channel. 🙂
@@MichelvanBiezen Thank you!
sir..this was the exact same question i was looking for.. couldn't have got any better explanation...thank u very much sir..
Absolutely brilliant. Straight to the point!
You are sent from the sky to save me! GREAT VIDEOS ! MUCH APPRECIATED
BIG UP!!
Your videos are amazing! Please keep it going!! I would love to see more conceptual based videos as well (in addition to the numerical/computational based problems that you do).
You, sir, are a gentleman and a scholar
great teaching! I finally understand how to do this problem.
Great to hear!
Nice lecture. It is relevant to stress, however, that a long distance (a>>d) the dipole behaves as a point charge with effective charge of q.d; and the decay of the electric field magnitude is to the power of minus three (instead of minus two), i.e, decay faster
The most simple and awsome explanation of this topin that i can get just cleared my doubts
Thanks for the lesson. It actually helped alot!!
Best one out of all!! Thanks for the video Mr! You're truly a lifesaver.
Glad it was helpful for you.
This guy sounds like Christophe Waltz and I love it/him.
Also--a clear, succinct explanation.
Thank you so much sir! It was so easy an explanation for so hard a topic (for me for the last 2 years)! You are an excellent teacher. Thank you again sir
Thank you so much, you are the reason I am passing physics!
i love your videos !!! Thank you!
Your videos help so much
Sir Your Lesson is just Amazing
Thank you so so much ... I was enlightened .. U answered my confusions ... Best teacher ever
the video of when i was 9 yrs still at primary grade 4 😂😂now using it at varsity😂...no need for a tutor 😂❤❤,,Merci beaucoup 👨🎓❤🙏
You are welcome.
Thank you so much I have a mid term on this and this helped me so much
Great your videos, always perfect explanations. Thank you!
i literally pay hundreds to learn concepts that are explained in such a shit way. then there's this channel where i learn concepts just like that in a matter of 10 minutes. thanks bruh
Thank's doctor
You're the best
please help,
A water molecule has a permanent dipole moment of magnitude 6.2*10^-30 Cm. A water molecule located 10 nm from a Na+ ion in a saltwater solution. what force does the ion exert on the molecule?
This video is lifesaver.. thank you so much
Glad it helped!
Thanks once again..this what i requested last time 🙏🏼🙏🏼👍
Most welcome 😊
oh my god,this the most very nice video how to explain physics on you tube..
i'll tell to my friendsss
love,
Indonesia
Welcome to the channel!
so nice i understood very well thankzzzzzz
so helpful, thank you for your videos!
I'm thankful that I can help you through add revenue! thanks sir
Best ever,,, thanks Dr.
Thank you! 🙂
will the Force be in the same direction as the electric field, and will we only have to multiply the formula derived for the electric field by q to find the force?
The direction of the force on a positive test charge placed in an electric field is the same as the direction of the electric field.
@@MichelvanBiezen thank you for your reply, Sir!
Professor Van Biezen is the OG!!!
Thanks alot teacher u did helped me alot ☺
So is this true for ALL dipole? The y components will always cancel? PS Thanks for all the great videos. They got me through rotational kinematics and now E&M.
Sir what exactly is this electric dipole moment.Is it a force ???
+Rahul Tiwari
An electric dipole is a particular arrangement of two opposite charges. When a positive charge is placed closed to a negative charge, the two charges close together like that is called an electric dipole. The electric field perpendicular to the line connecting the two charges has a particular equation as shown in the video.
+Michel van Biezen Sir but what is the significance of that word moment in electric dipole moment
Hey there. Awesome video. Your teaching style/method matches up with my learning style/method. Maybe you're just a good teacher, I don't know. But hey, I'm having trouble understanding around 5:30 why x^2+a^2 square root x^2+a^2 became (x^2+a^2)^(3/2). Maybe that's an algebra rule I didn't understand. Would you or someone care to clarify? Please and thanks yo
The rule is: When the base is the same and you multiply you add the exponents. Example a^1 * a^(1/2) = a^(3/2)
Ok, thank you. Though, I'm still not seeing the addition equaling (3/2) 🤔. You have an exponent of (1/1) and you're adding it to the exponent of (1/2). I don't see what I'm missing
1 + (1/2) = 3/2
Michel van Biezen haha, oh yes. Thank you.
Isn't it strange how our brains sometimes just don't see something so obvious? (Happens to me as well).
you can do these equations very fast! thats impressive. for future reference i think it would be more effective for teaching first time learners if you went a bit slower
Thank you for the feedback.
thanks sir, you helped me a lot.
Glad to hear that
@@MichelvanBiezen love from Pakistan sir.
What is the value of electric field strength at the centre of dipole
Ishfaq,
Take the final equation and set "a" equal to zero.
mr. van beizen why didnt you consider the negative sign for E2 ?
We did by choosing the correct direction of the electric field caused by the second charge. Then we found the magnitude of the resulting electric field.
Do you have any examples for electric dipole questions?
Just the one you are looking at. What specific examples are you thinking of, or is there a homework problem you are looking for?
How do i do this problem if the Qs are not equal in magnitude but different numbers and do not have the same sign? That would mean their corresponding
vertical components would not cancel and would I have to calculate those vertical components as well?
Yes, you would have to calculate both components.
it wouldn't be a dipole
in the previous video,when we calculate the total magnitude of electric field ET, we say ET = square root of ETx^2 + ETy^2 ,but in this video we calculate the magnitude of ET as the sum of two x components .why we do that and what is the difference between two problems?
That is what is so interesting about dipoles. The y-components of the electric field due to the 2 charges are equal in magnitude and opposite in direction and thus cancel out. The x-components however are equal in magnitude, BUT in the same direction, thus they are added.
okay but why we don't use the square root when we calculate the magnitude of ET ? ... when we can say that magnitude of ET = E1 + E2 ?
Since the two x-components point in the same direction we can simply add them. Note that you only have to use the square root of the sum of the squares method if the components are perpendicular.
thank you so much! such a helpful clear explanation
Back in the day, master mathematicians called The "perpendicular bisector" of a line the "MEDIATOR". I wish I saw it more often...-hint-
ps
I think it originated in the field of geometry, then went general.
Does anyone else know anything about this?
first of all thank you for videos. but i wonder something why dont you use 1/4piE0
my teacher obsessed with it
It is just easier to use k instead.
what if one if the electric field on y-axis and the other one we should devaided it into y and x component, how to get the resultant of them?
Thank you!
You're welcome!
What will the result be if the two charges are positive????
Then the components of the electric field in the horizontal direction will cancel and there would only be a component in the vertical direction.
love you sir
Thanks.
You're welcome
I'm curious sir, but at your final answer, the q, you say its the magnitude of the charge? Which charge, is it the magnitude of the two given charges?
soulseeker
Either charge. (They both have the same magnitude, one is -q and the other is +q)
Two charges with the same magnitude (but opposite sign) spaced closely together forms a "dipole"
why didnt you take -q as negative in when adding their cosines
Daniel Joseph he's already considered the negative sign while determining the direction it would act in
Very nice video.
Great video, very well explained :)
really good explanation. But what surprises me is that why do we ignore the sign of q. I am used to working in vectors so when i add them as vectors, not as magnitudes, I am supposed to get a 0 due to opposite charges. Could you explain this..
We don't actually ignore the direction component of the vector. In the beginning we draw the electric field component contributed by each charge. We labeled them E1 and E2. We then added them as vectors, and realized that only the vertical components cancel and that the horizontal components are additive. All we have to do then is add the magnitudes of the horizontal components to find the magnitude of the total electric field at that point. And the direction is in the positive x-axis.
You are amazingly great!! thank you so much
Usually axis is perpendicular to whatever we are considering. eg. in case of the earth (almost perpendicular to the equator). But in case of the axis of the electric dipole, why do we consider it along the dipole ?
+ANNA KAVALAM
In more advanced courses we calculate the effect of the dipole from every direction. The simple case we only look at the dipole from a point perpendicular to the line connecting the two charges or along the line connecting the two charges.
+Michel van Biezen But Sir, my doubt is just about why the line connecting the two charges are named as axial ?
Gotta love that bow tho!
Beautiful.
maybe i will not fail physics 102 after all.... thanks sir you are great
If you watch the videos and learn how to do this problems (without watching) you will succeed!
I know cos θ = adj/hyp.But why is that hyp is x, not d?
+WeiFeng Ma
The adjacent side is x and the hypotenuse is R in the sketch.
This explanation has some serious flaws:
1.The angle for the negative charge isn't the same as that for the positive one as is clear from the diagram
2.The electric field of the negative charge is considered negative but he chose it to be positive
3.The distance towards the left of a should be negative
Nothing wrong with the diagram, the vectors, or the explanation. Not sure why you seem to think otherwise.
can someone reply to me i need it ASAP, why did we disregard the direction and focus on the magnitude?
Sorry, I didn't understand your question. ( I don't think we disregarded anything)
@@MichelvanBiezen sorry for the confusion prof
What if the charges r different
Sir u r from which country?
I grew up in Belgium, but now I live in the US
Point charges 𝑞1 = +12 nC and 𝑞2 = -12 nC are
0.100 m apart. (Such pairs of point charges
with equal magnitude and opposite sign are
called electric dipoles.) Compute the electric
field caused by 𝑞1, the field caused by 𝑞2 , and
the total field (a) at point a; (b) at point b; and
(c) at point c.
Kidly solve sir
Where are the positions a, b, and c?
Positions a b c are not given
But in diagram
A is negative
B is positive
Sir can u give me your
Email address
Or
WhatsApp number
I will share problem screenshot
I wonder if I can simply find the angle using the inverse tangent. I've been doing that and my answer are always a bit off. I'm not sure if that's what I"m doing wrong.
Without seeing what you are doing it is not possible to respond to this question, but if you know the opposite side and you know the adjacent side and it is a right triangle then it should work.
@@MichelvanBiezen Thanks!!
Why do you express the equation in vector just put x head?
Since the vector only points in the x-direction we only need to show the x-component (using the x-unit vector)
thank you so much
Thank you sir
Welcome
Thank you ✊you're very excellent
excuse me but why we dont have to consider the negative charge in the expression of E2?
We did, so I am not sure about your question.
so in the expression of E we have to take the absolute value of q that's it?
Ah, you are asking about the sign of the charge? The sign of the charge only determines the DIRECTION of the electric field. The magnitude of the electric field only depends on the magnitude of the charge.
thanks a lot
thanks for this
Sir I don’t find the lectures on dipole dipole interaction will you be uploading later or not please let me know sir your lectures are very helpful to us so kindly upload this topic too sir 🙏🙏🙏
I don't think we have dipole dipole interaction on video.
Michel van Biezen could you please upload it sir ?
What would be the answer if both were positive charges
The the x-components of the vectors would cancel out the the y-components would add. Then it would no longer be a "dipole".
@@MichelvanBiezen thank you.
une trés bonne explication,merci
What if both charges are negative??
Then you don't have a dipole.
Thank you
Is it possible to obtain the zero electric field for this system? Do you have any idea? Please, help me
Since the 2 charges have the same magnitude, there is no point anywhere where the electric field = zero
@@MichelvanBiezen thank you so much
thanks, it was helpful.
How would we determine the field if it was at some point P, instead of on a bisector. How do we solve this if we cannot exploit symmetry?
With a dipole you typically pick the point along the perpendicular bisector. But if you want to fine the electric field at any other point, the vertical components would not cancel out and you would have perform an integration twice, once for each direction.
Ok... i tried it earlier and ended up with a trig sub! And a two part equation!
Yes, that is a difficult problem. I should do an example or two like that. (I'll put it on the list of videos to do).
Awesome! Thank you so muuch!!
THANK YOU!
Thank you, sir
I like the new bow tie :)
Thank you my man
ÇOK İYİ Bİ HESAP TEŞEKKÜRLER
sir where can i find the video for electric field expression for a electrical dipole for axial position and any random position other than axial or vertical . please help . i need those urgently .
I don't believe we have such an example.
That said, it is worked out exactly the same except that the y-component doesn't cancel out and you'll have to calculate both the x and y component of the electric field.
@@MichelvanBiezen sir its easier said than done for students like me ... it would have been of great help if u could have put another video explaining that ... anyways thanks a lot for your existing videos ... they help a lot
Yes, we realize that. We are still working on adding more videos as time permits.
Odd flashing of black marks on the whiteboard from 4:30 to end of video....
Cool man
lifesaver