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This reduces my stress by 99%. I am doing this topic right now but in chinese. So complicated that I can't understand what my professor is teaching us. Thanks man, this is great video for me 👍👍👍👍
My professor spent two weeks on this and I still didnt get it, one night after sleeping on this video and I can do the super hard problems my professor gives us. My professor teaches a super basic level understanding of the topics and then expects us to solve super complex problems. This video literally is saving me on my midterm cause this is the only topic I was struggling with in physics 2
@@clioosesli4646 but that is simply not true… take, for example, the left most point on the ring vs the right most point on the ring. the distance from each of those points to the point of interest is different
@@alexsafayan7684 how ? Okay it's a ring right? Okay just take a ring shaped bangle or anything and a wooden stick or a pen then pass it through the centre of the ring then you will see that no matter it's from the right left or top bottom or any point the distance is same
A thin circular ring, of radius 20cm, is charged, with a uniform charge density(rho). If a small section, of 1cm length, is removed from the ring. Calculate the electric field intensity at the center of the ring.
You're correct. No definite integral was taken therefore some constant c should come into the picture. We can then determine c by substituting E=0 when Q=0, which gives us c=0. Note that after indefinite integration, we get E as a function of Q i.e E = f(Q) + c. Another way of doing this is to look at an indefinite integral as always being a sum of all infinitesimal parts when applied to a given situation. In this case, we do not need to use definite integral since we know the integral of small E i.e. dE in this situation is the sum of all dEs over the cirumference and the integral of dQ is the sum of all dQs over the cirumference, which is clearly E (net electric field) and Q (total charge on ring) respectively. So, either use definite integral or indefinite integral sum approach.
I don't know why R doesn't change?? I understand that R is in terms of a and x which they are constant; However, if dq will be in another position on the ring, u will not have a right triangle to use the Pythagorean theorem to find R in terms of X and a; therefore, R is not constant. Please correct me. Thank you.
I figured it out after looking at it again, the problem was that i was looking at it in 2 dimension. However, in a 3 dimension space, the point P is above the circle which will give a right triangle in all cases. It was just a issue of imagination lol
How can we prove using math, and not symmetry, that the y-components are add up to zero? Can we take the integral of dE_y from 0 to 2πα or we must calculate two or more integrals because the sign of the y-component is changing?
this is the problem with our society, the video that has a lot of informative and useful content has only a few thousand views, but a video of bear climbing mountain has millions of views.what do you think.??
You could do that also and still you will get the same answer. In place of dQ you would use charge per unit length X dl where dl is an infinitesimal length along the circumference. You would then get E net in terms of linear charge density.
@@daniellaughoc4003 , it will ultimately give the same end result except that now your final formula obtained after integration would be having linear charge density in it.
@@sunildhaul2397 will that be the same for all uniformly distributed charges? Is it ok if I do not use density and just integrate dq for those type of questions but with different shapes with uniformly distributed charges?
Why wasn't the linear charge density inserted into the point charge equation before integration, after pulling out the constants you should have been integrating d(theta) = theta. Which would have cancelled out the 2pi leaving only KQ/(x^2+a^2)^3/2 = 2.05*10^5 N/C !!
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I been struggling with this section. My teacher makes it, so complicated and sometimes frustrating the way he explains it. Thank you, very much.
Good one.
It is just so surprising how this teacher of ours is called the Organic Chemistry Tutor and he is teaching a wide range of stuff ; Math,Physics....
Thank teach you are a blessing
you are the best!!!! noone could make physics interesting to me, but with you İ can study without getting bored. and you have the most precious voice İ have ever heard))))
bro your voice make us calm and this is the reason why we learn !!! god save you for us
I'm in my second year in physics and i like your videos A LOT!! also we share it between us to understand our topics and have examples on it. thank you very much teacher!!!
You are such a positive influence on STEM education in the 21st century that I genuinely believe you should have your own Wikipedia page.
Interesting!!! This makes the one in textbook a piece of cake. Thank you!!!
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I read this material from my book for like 3 hrs and yet didnt understand until this video, so thanks dude
This reduces my stress by 99%. I am doing this topic right now but in chinese. So complicated that I can't understand what my professor is teaching us. Thanks man, this is great video for me 👍👍👍👍
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Thank you mister ... it was way simpler than the book
My professor spent two weeks on this and I still didnt get it, one night after sleeping on this video and I can do the super hard problems my professor gives us. My professor teaches a super basic level understanding of the topics and then expects us to solve super complex problems. This video literally is saving me on my midterm cause this is the only topic I was struggling with in physics 2
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How is R a constant? 7:00 - the logic here does not make sense
Yep I did not get it as well
Because the distance from any point of the ring to the point where we are finding the electric field won't change that's why R is a constant
@@clioosesli4646 but that is simply not true… take, for example, the left most point on the ring vs the right most point on the ring. the distance from each of those points to the point of interest is different
@@alexsafayan7684 how ?
Okay it's a ring right?
Okay just take a ring shaped bangle or anything and a wooden stick or a pen then pass it through the centre of the ring then you will see that no matter it's from the right left or top bottom or any point the distance is same
@@clioosesli4646 Oh wow, I think I've been interpreting this as a 2D problem, when in reality, the ring is "facing" the point of interest, correct?
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A thin circular ring, of radius 20cm, is charged, with a uniform charge density(rho). If a small section, of 1cm length, is removed from the ring. Calculate the electric field intensity at the center of the ring.
But.. didn't we ignore the "+c" for the indefinite integral? Was it somehow definite, did I miss it? Did we say from 0 to Q or something like that?
You're correct. No definite integral was taken therefore some constant c should come into the picture. We can then determine c by substituting E=0 when Q=0, which gives us c=0. Note that after indefinite integration, we get E as a function of Q i.e E = f(Q) + c.
Another way of doing this is to look at an indefinite integral as always being a sum of all infinitesimal parts when applied to a given situation. In this case, we do not need to use definite integral since we know the integral of small E i.e. dE in this situation is the sum of all dEs over the cirumference and the integral of dQ is the sum of all dQs over the cirumference, which is clearly E (net electric field) and Q (total charge on ring) respectively. So, either use definite integral or indefinite integral sum approach.
How is this different from the one with the formula of the denominator raised to 3/2……?
I don't know why R doesn't change?? I understand that R is in terms of a and x which they are constant; However, if dq will be in another position on the ring, u will not have a right triangle to use the Pythagorean theorem to find R in terms of X and a; therefore, R is not constant. Please correct me. Thank you.
No matter where you draw the line from the ring to point p R will be the same value.
I figured it out after looking at it again, the problem was that i was looking at it in 2 dimension. However, in a 3 dimension space, the point P is above the circle which will give a right triangle in all cases. It was just a issue of imagination lol
Haha I was struggling with the calculus portion of this too. Cheers mate
Good luck man :)
Thanks u too
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Thank you once again.
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That's help me soo much but I have question in the Last equation we got if x=0,x>>0, x
helped a lot thanks :)
One thing I never understood is why is λ ds = dQ. If λ is already the amount of charge per unit of length, why do we need to multiply it by ds?
How can we prove using math, and not symmetry, that the y-components are add up to zero? Can we take the integral of dE_y from 0 to 2πα or we must calculate two or more integrals because the sign of the y-component is changing?
Thanks alot
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this is the problem with our society, the video that has a lot of informative and useful content has only a few thousand views, but a video of bear climbing mountain has millions of views.what do you think.??
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Please when is E=p/£and when is it p/£.
Where E= Electric field
P= charge density
£= permittivity of free space.
Why isn’t the linear charge density inserted before integrating?
You could do that also and still you will get the same answer. In place of dQ you would use charge per unit length X dl where dl is an infinitesimal length along the circumference. You would then get E net in terms of linear charge density.
@@sunildhaul2397 How is this different from the one with the formula of the denominator raised to 3/2……?
@@daniellaughoc4003 , it will ultimately give the same end result except that now your final formula obtained after integration would be having linear charge density in it.
I don't understand why he didn't integrate directly dE?
@@sunildhaul2397 will that be the same for all uniformly distributed charges? Is it ok if I do not use density and just integrate dq for those type of questions but with different shapes with uniformly distributed charges?
Thanks a lot
Thanks man
Thanks buddy
Təşəkkürlər
Why wasn't the linear charge density inserted into the point charge equation before integration, after pulling out the constants you should have been integrating d(theta) = theta. Which would have cancelled out the 2pi leaving only KQ/(x^2+a^2)^3/2 = 2.05*10^5 N/C !!
The denominator should be (x^2+a^2)^3/2, not R^3.... I think you're right!
I have the same question, I know this is late but have you figured out why?
@@NikkaPleeease How is this different from the one with the formula of the denominator raised to 3/2……?
@@NikkaPleeease it's the same because R= sqrt of x^2+a^2 which equals to (x^2+a^2)^1/2 so R^3 = (x^2+a^2)^3/2
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