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.
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!
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 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
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?
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.
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.
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?
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.
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.
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.
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.
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"
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
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?
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.
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
+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.
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
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
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 🙏🙏🙏
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 bet those dislikes are from grumpy physics teachers who dont want student's to pass LOLOLOLOL !
100%
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
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 !!
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.
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 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
Great your videos, always perfect explanations. Thank you!
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.
sir..this was the exact same question i was looking for.. couldn't have got any better explanation...thank u very much sir..
You have perfected the art of teaching.
The most simple and awsome explanation of this topin that i can get just cleared my doubts
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).
i love your videos !!! Thank you!
Your videos help so much
so helpful, thank you for your videos!
Best one out of all!! Thanks for the video Mr! You're truly a lifesaver.
Glad it was helpful for you.
Thank you so much I have a mid term on this and this helped me so much
You are sent from the sky to save me! GREAT VIDEOS ! MUCH APPRECIATED
BIG UP!!
Sir Your Lesson is just Amazing
Thank you so much, you are the reason I am passing physics!
Absolutely brilliant. Straight to the point!
thank you so much! such a helpful clear explanation
Thanks for the lesson. It actually helped alot!!
Thank you so so much ... I was enlightened .. U answered my confusions ... Best teacher ever
I'm thankful that I can help you through add revenue! thanks sir
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!
You, sir, are a gentleman and a scholar
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
great teaching! I finally understand how to do this problem.
Great to hear!
Thanks alot teacher u did helped me alot ☺
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?
You are amazingly great!! thank you so much
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?
This video is lifesaver.. thank you so much
Glad it helped!
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.
Beautiful.
Great video, very well explained :)
This guy sounds like Christophe Waltz and I love it/him.
Also--a clear, succinct explanation.
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!
What is the value of electric field strength at the centre of dipole
Ishfaq,
Take the final equation and set "a" equal to zero.
Thanks once again..this what i requested last time 🙏🏼🙏🏼👍
Most welcome 😊
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
Best ever,,, thanks Dr.
Thank you! 🙂
so nice i understood very well thankzzzzzz
Thank you ✊you're very excellent
Thank you my man
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.
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.
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!
Very nice video.
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?
thanks, it was helpful.
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 ?
une trés bonne explication,merci
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.
thanks for this
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!!
love you sir
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.
thank you so much
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"
thanks sir, you helped me a lot.
Glad to hear that
@@MichelvanBiezen love from Pakistan sir.
THANK YOU!
Thank you, sir
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).
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
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 the charges r different
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?
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.
Thank you!
You're welcome!
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!!
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
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
Professor Van Biezen is the OG!!!
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
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.
Thank you
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
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.
Gotta love that bow tho!
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.
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.
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)
What if both charges are negative??
Then you don't have a dipole.
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
lifesaver
Thank you sir
Welcome
Thanks.
You're welcome
Sir u r from which country?
I grew up in Belgium, but now I live in the US
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 ?
Cool man
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!
ÇOK İYİ Bİ HESAP TEŞEKKÜRLER
How to find the value of a?
a is a "given" quantity. a can have any value as it doesn't change how you solve the problem.
@@MichelvanBiezen sir. The only given in our task is the q of the dipole and and the d. How can I find the electric field halfway? Thankyou
@@MichelvanBiezen I hope you could help sir. Thank you. I'll wait for your reply
Odd flashing of black marks on the whiteboard from 4:30 to end of video....
I like the new bow tie :)
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