If you could look closely enough, then, you're approximating the electric field due to an infinite plane of charge, hence the different answer. This would be a great question to post on the APlusPhysics site and see what others say!
Hi Jiang. For the square with a uniformly distributed positive charge in the yz plane, at the origin, the net electric field will be 0 in the x-direction because the force can only push or pull in the direction of the line of the charge to the point. There's no way to get an x-component of force. For the finite planar charge distribution, imagine as you get very close to the plane... that planar charge starts to look infinitely big compared to the distance (think imaginary microscope).
When you say the symmetry cancels out in the X-direction, are you talking about the symmetry due to the fact that point C is in the midpoint of the circular arc?
Hi! No, you can't cancel out the dQ in the differential. The d in the dQ means differential of Q, the d's in the denominator refer to distances. An unfortunate choice of symbology, perhaps, but one that you'll run across quite regularly.
Thanks so much for an awesome video!! I was only confused by one thing.. Where did you extract (Q/L)dy to be equal to delta Q?? I understand all math and reasoning before and after, but I'm failing to understand this one point. it's probably super easy and I'm just missing it..
You're welcome Annie. (Q/L)dy is the amount of charge contained in our tiny little bit of length (Q/L is the linear charge density, dy is the length, multiple them to get the charge contained in that length). So then Delta Q=(Q/L)dy is the small amount of charge contained in that length. And when I go to make that length infinitesimally small, I call it dQ instead of Delta Q. Hope that helps!
This may have already been mentioned but you can use trigonometric substitution to solve the integral from straight wire problem, I think that is much better than looking up and memorizing some integrals from a table, even for AP students.
True, and I actually had a video where I did it that way previously, but it turned into a 40-some minute video, and got us doing much more calculus than physics, so eventually I decided on this version. But you're absolutely right, there are other methods of solving the problem, and it's definitely worth walking through the trigonometric substitution solution at least once to see how it's done.
But it seems that in the steps after that you did cancel the d on the numerator with the d^2 in the denominator, thus leaving you with a d in the denominator. Right?
Hi, Mr. Fullerton, i just have a quick question. I'm so confused that why when a square of side length a with uniformly distributed positive charge lies on the yz plane with its center at the origin, the electric field along the x-axis is zero when x =0 but the electric field due to a finite planar circular charge distribution lying on the yz plane will have a constant value (λ/2ε0) when x approaches zero?
I have a question to the instructor, Dan Fullerton. For the integration on a thin straight insulating wave question could we have not argued since the the wire is symmetric with respect to the y distances the value achieved from the integration could simply be done on the bound 0 to L/2 multiplied by 2. Therefore, 2 * the integration of dy/(y^2 + d^2)^(3/2) on the limits of integration 0 to L/2? (Of course with all the constants in there?) This way when you plug in L/2 after integrating you only have to do it once. If there is something preventing this please explain to me.
Will you be asked to do calculations like these on the actual exam or will they be less complex? Sorry if it may seem like a stupid question but I just want to know how much I should focus on actually knowing how to do the full calculations or if I should just know the general concepts behind it.
Well then thank you sir because you did a great job explaining it. Now I just need to watch it again and practice it a few times on scratch paper to get it down😂
Not exactly. You have to be careful of your symbols here. dQ is the differential of Q. The d's on the right refer to the distance d. Once you've done your integration, there are no more differentials, so all the d's are distance, and you can treat them with your standard algebraic rules.
Hi! At 3:25, when you’re taking the differential of the y component of the electric field, wouldn’t it require multiplication with Rsin(theta) rather than sin(theta)?
I'll pull the physics teacher card and answer that the same way I would with my students in class... you tell me what would happen if you integrated from 0 to L instead of -L/2 to L/2. Would you expect a different result just by changing your references? Would your answer look different, even if it had the same physical interpretation?
Dan Fullerton I know this is late, but you can do this problem using radians as the limits, and converting what's in the integrand to polar coordinates.
After 14:05: You obviously can not cancel a d from d^2 (in the denominator) with the d from dQ (= a differential in the numerator)! Physicists can do a lot, but not this ;-)
Hi, So I used Gauss's law to solve the question at 6:12 and I got the 2nd solution (if l goes to infinity), I was wondering, why did Gauss's law not work for me. Note: I used a cylinder as the Gaussian Surface.
I think the problem you're going to have with Gauss's Law and the finite wire is going to involve symmetry. I haven't tried it myself yet, but you're going to have some net flux through the caps of the cylinder, making Gauss's Law still true, but not overly helpful in calculating the E field...
In my book, I have a similar problem like this but instead of an insulating wire, its called a nonconducting wire and I'm guessing nonconducting is another word for semiconducting.
It can have charge distributed across it, charge just doesn't flow through it very well. As an example, you can charge up an insulator such as a rubber rod fairly easily by rubbing it with fur.
They're close. Delta Q is a small bit of charge. dQ is a differential of charge (i.e. take delta q, and make it smaller and smaller and smaller heading toward infinitesimally small).
Thank you. I really appreciate that you could devote your time to help us. I'm taking the AP Physics C this May. This is really helpful
Ah, I love your videos, but I have trouble differentiating between Q and Theta
If you could look closely enough, then, you're approximating the electric field due to an infinite plane of charge, hence the different answer. This would be a great question to post on the APlusPhysics site and see what others say!
Hi Jiang. For the square with a uniformly distributed positive charge in the yz plane, at the origin, the net electric field will be 0 in the x-direction because the force can only push or pull in the direction of the line of the charge to the point. There's no way to get an x-component of force. For the finite planar charge distribution, imagine as you get very close to the plane... that planar charge starts to look infinitely big compared to the distance (think imaginary microscope).
When you say the symmetry cancels out in the X-direction, are you talking about the symmetry due to the fact that point C is in the midpoint of the circular arc?
Hi Xi Chen. No, it won't be zero when d approaches infinity if it's an infinite line charge.
Hi! No, you can't cancel out the dQ in the differential. The d in the dQ means differential of Q, the d's in the denominator refer to distances. An unfortunate choice of symbology, perhaps, but one that you'll run across quite regularly.
Thanks so much for an awesome video!! I was only confused by one thing.. Where did you extract (Q/L)dy to be equal to delta Q?? I understand all math and reasoning before and after, but I'm failing to understand this one point. it's probably super easy and I'm just missing it..
You're welcome Annie. (Q/L)dy is the amount of charge contained in our tiny little bit of length (Q/L is the linear charge density, dy is the length, multiple them to get the charge contained in that length). So then Delta Q=(Q/L)dy is the small amount of charge contained in that length. And when I go to make that length infinitesimally small, I call it dQ instead of Delta Q. Hope that helps!
Thanks so much for responding!! Your videos have been a massive help to me for my college E&M class :)
Annie Greer My pleasure!
This may have already been mentioned but you can use trigonometric substitution to solve the integral from straight wire problem, I think that is much better than looking up and memorizing some integrals from a table, even for AP students.
True, and I actually had a video where I did it that way previously, but it turned into a 40-some minute video, and got us doing much more calculus than physics, so eventually I decided on this version. But you're absolutely right, there are other methods of solving the problem, and it's definitely worth walking through the trigonometric substitution solution at least once to see how it's done.
You're welcome!
@8:45, why did you write ∆Q and not dQ? What's the difference?
But it seems that in the steps after that you did cancel the d on the numerator with the d^2 in the denominator, thus leaving you with a d in the denominator. Right?
wish i discovered this series in the beginning of the semester. at least i will do well on my final. official john cena alert
Hi, Mr. Fullerton, i just have a quick question. I'm so confused that why when a square of side length a with uniformly distributed positive charge lies on the yz plane with its center at the origin, the electric field along the x-axis is zero when x =0 but the electric field due to a finite planar circular charge distribution lying on the yz plane will have a constant value (λ/2ε0) when x approaches zero?
@7:43, how do you know the electric field is directed downward? Is it assumed that the total charge of the wire is positive?
I have a question to the instructor, Dan Fullerton. For the integration on a thin straight insulating wave question could we have not argued since the the wire is symmetric with respect to the y distances the value achieved from the integration could simply be done on the bound 0 to L/2 multiplied by 2. Therefore, 2 * the integration of dy/(y^2 + d^2)^(3/2) on the limits of integration 0 to L/2? (Of course with all the constants in there?) This way when you plug in L/2 after integrating you only have to do it once. If there is something preventing this please explain to me.
Sounds reasonable to me!
Sure!
Will you be asked to do calculations like these on the actual exam or will they be less complex? Sorry if it may seem like a stupid question but I just want to know how much I should focus on actually knowing how to do the full calculations or if I should just know the general concepts behind it.
Yes, you can be asked to do calculations like this on the exam.
Well then thank you sir because you did a great job explaining it. Now I just need to watch it again and practice it a few times on scratch paper to get it down😂
@@umarpatel3517 It does take some practice... good luck!
at 14:05 the d in the differential dQ, can that be cancelled out with a variable distance squared d^2? you forgot the equal sign at the end btw
Not exactly. You have to be careful of your symbols here. dQ is the differential of Q. The d's on the right refer to the distance d. Once you've done your integration, there are no more differentials, so all the d's are distance, and you can treat them with your standard algebraic rules.
you dont use d as a variable when you are using differentials
come on man
I do apparently -- perhaps not my all-time wisest decision. ;-)
Dan Fullerton Haha I actually was complaining about that when trying to follow. Otherwise I love the video and appreciate you! :D
Hi! At 3:25, when you’re taking the differential of the y component of the electric field, wouldn’t it require multiplication with Rsin(theta) rather than sin(theta)?
That's already taken into account from our electric field formula.
Could I also use Gauss's law to determine the electric field of the insulating semi circle?
I'll pull the physics teacher card and answer that the same way I would with my students in class... you tell me what would happen if you integrated from 0 to L instead of -L/2 to L/2. Would you expect a different result just by changing your references? Would your answer look different, even if it had the same physical interpretation?
Dan Fullerton I know this is late, but you can do this problem using radians as the limits, and converting what's in the integrand to polar coordinates.
After 14:05: You obviously can not cancel a d from d^2 (in the denominator) with the d from dQ (= a differential in the numerator)! Physicists can do a lot, but not this ;-)
Obviously. Which is why it's a good thing that dQ in the numerator isn't a differential, it's d*Q.
@@DanFullerton Yep, you're right! The notation fooled me :-(
For the first question, it asks for the electric field at point C. Why did you only find the y component and not the x component?
Because I can observe that by symmetry any net x-component will cancel out.
oh I see thank you
Why does delta Q equal (Q/L) * dy
Hi,
So I used Gauss's law to solve the question at 6:12 and I got the 2nd solution (if l goes to infinity), I was wondering, why did Gauss's law not work for me. Note: I used a cylinder as the Gaussian Surface.
I think the problem you're going to have with Gauss's Law and the finite wire is going to involve symmetry. I haven't tried it myself yet, but you're going to have some net flux through the caps of the cylinder, making Gauss's Law still true, but not overly helpful in calculating the E field...
I understand. Alright thanks a lot!
At 19:00 isn't E supposed to be zero when d approaches infinity?
How can an insulating wire carry charge?
In my book, I have a similar problem like this but instead of an insulating wire, its called a nonconducting wire and I'm guessing nonconducting is another word for semiconducting.
It can have charge distributed across it, charge just doesn't flow through it very well. As an example, you can charge up an insulator such as a rubber rod fairly easily by rubbing it with fur.
Are deltaQ and dQ the same thing?
They're close. Delta Q is a small bit of charge. dQ is a differential of charge (i.e. take delta q, and make it smaller and smaller and smaller heading toward infinitesimally small).
Dan Fullerton So why is the DeltaQ in the second example equal to (Q/L)*(dy)? This means that deltaQ = lamda(dy) since lamda=Q/L?
I am giving the physics c AP. Will this course cover all the topics mentioned in the syllabus ??
+shloak patil Yes
why is ∆Q = (Q/L)dy? Shouldn't it be ramda times (2/L) according to the linear charge density?
I try doing it didn't get the right answer. i
Cheers :)