Electric Field Due to a Ring of Charge, Linear Charge Density, Physics Practice Problems

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7 жыл бұрын

This physics video tutorial explains how to calculate the electric field of a ring of charge. It explains why the y components of the electric field cancels and how to calculate the linear charge density given the total charge of the ring, the radius, and the distance between the center of the ring and a point P that lies on the x axis perpendicular to the surface of the ring. This video contains plenty of examples and practice problems.

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@TheOrganicChemistryTutor 6 ай бұрын

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@dwwwwwwwwwwwwwwwwwwwwwww 4 жыл бұрын

The book was complicated and demotivating, why is this so much simpler? Great video as always, my friend you are a saint.

@vernardoortega6822 3 жыл бұрын

I been struggling with this section. My teacher makes it, so complicated and sometimes frustrating the way he explains it. Thank you, very much.

@yvansugira8120 2 жыл бұрын

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@user-to3do8ee9z 5 жыл бұрын

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@user-fs7kx6gl2m 4 жыл бұрын

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@der_ABV 4 жыл бұрын

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@nothysaint27 5 жыл бұрын

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 👍👍👍👍

@arcadiapeek6856 2 жыл бұрын

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@nickcarriero8274 2 жыл бұрын

Thank you so much, this is so much easier to understand than the way my professor explains it. You're a real life saver

@musicatation2254 4 жыл бұрын

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@MasterSpencer2483 5 жыл бұрын

Really helpful, thanks so much! You helped my avoid a frustrating night

@rotemdagan3722 4 жыл бұрын

I read this material from my book for like 3 hrs and yet didnt understand until this video, so thanks dude

@user-gi4ei5ln3j 2 ай бұрын

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@tudong8649 Жыл бұрын

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@user-gw2vz5gh2n 10 ай бұрын

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@whitehole4460 3 жыл бұрын

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@saneleqwabe8424 3 жыл бұрын

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@theroadnottaken_ 5 жыл бұрын

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@ahmedalblooshi8206 3 жыл бұрын

Thank you mister ... it was way simpler than the book

@analauravestibulanda1576 7 жыл бұрын

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@MARTINUS_DAVIE 3 жыл бұрын

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@KarausMiller 4 жыл бұрын

Awesome! Helpful tutorial

@cekulamtina6663 5 жыл бұрын

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@abktv1054 2 жыл бұрын

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@yapzhilin9857 3 жыл бұрын

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@jaykhosia9353 7 жыл бұрын

Thank you so much, sir, for making it easy for us. (Y)

@daniellaughoc4003 2 жыл бұрын

How is this different from the one with the formula of the denominator raised to 3/2……?

@jesseb5378 Жыл бұрын

A savior to students for sure!

@theanonimo26able 3 жыл бұрын

Thank you so much man you really help a lot

@fur7i236 5 жыл бұрын

bro u r a life saver.....thanks

@lefterisspanoudakis3152 Жыл бұрын

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?

@NeoTechVision101 2 ай бұрын

Clearest explanation

@oscarkholopha2959 6 жыл бұрын

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@bhavitilwani1545 4 жыл бұрын

Thanks bro you the real MVP

@Pages_Perfected 4 жыл бұрын

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@Pages_Perfected 4 жыл бұрын

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@Walid-xx1xr 2 жыл бұрын

Thanks for the save, Like Always♥️

@jasmine9581 3 жыл бұрын

omg first one is pretty much one of my homework questions due soon

@saligia0743 Жыл бұрын

That's help me soo much but I have question in the Last equation we got if x=0,x>>0, x

@ronny_ranks 5 жыл бұрын

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@The053199 2 жыл бұрын

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

@themoon8671 6 жыл бұрын

Thanks alot U R GREAT

@jesusmrosario-claudio4104 3 жыл бұрын

Thank you once again.

@denizsoydan8235 3 жыл бұрын

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?

@sunildhaul2397 3 жыл бұрын

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.

@emxoco 7 ай бұрын

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@hassanabdulmezeed6965 2 жыл бұрын

thank you this really helped me

@beyzanursahin2159 4 жыл бұрын

helped a lot thanks :)

@wahidhamidy6886 7 жыл бұрын

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@philangezwitsabedze9483 4 жыл бұрын

wow, this is very helpful

@KinGxWolF 7 жыл бұрын

Thank you!!!

@sofianetakouk540 3 жыл бұрын

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@verrawar7906 6 жыл бұрын

Thank you so much .

@AHMEDALFourjani1998 6 жыл бұрын

Thank you so much

@sphystius9031 3 жыл бұрын

Wow ... this is so much simpler

@elvisvasquez7553 9 ай бұрын

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?

@ashadulislamrony7692 2 ай бұрын

Nice Class

@AmanKumar-pj4bb 5 жыл бұрын

Thanks a lot

@tobiaschapinda6771 2 жыл бұрын

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.

@user-ui4tr5sv7m 10 ай бұрын

thank you

@SamSam-tn9uh Жыл бұрын

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@petermoon9473 Жыл бұрын

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@ahmadshayain 4 жыл бұрын

Thanks man

@Ali_Albahrani 22 күн бұрын

Thanks

@alexmccann101 4 жыл бұрын

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@relaxationvideoszzmedozz341 4 жыл бұрын

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@regnas7868 5 жыл бұрын

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.

@tagekumarpaul7474 5 жыл бұрын

No matter where you draw the line from the ring to point p R will be the same value.

@regnas7868 5 жыл бұрын

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

@tagekumarpaul7474 5 жыл бұрын

Haha I was struggling with the calculus portion of this too. Cheers mate

@regnas7868 5 жыл бұрын

Good luck man :)

@tagekumarpaul7474 5 жыл бұрын

Thanks u too

@hasanshirvanov5290 3 жыл бұрын

Təşəkkürlər

@abd1428 3 жыл бұрын

thank u

@malakfalkoun1280 Жыл бұрын

Sir I love you 😔

@arinzeanthony7447 2 жыл бұрын

Please when is E=p/£and when is it p/£. Where E= Electric field P= charge density £= permittivity of free space.

@hossiluc 4 жыл бұрын

I love it

@orangeguy9166 Жыл бұрын

love you

@v.m.5850 4 жыл бұрын

Why isn’t the linear charge density inserted before integrating?

@sunildhaul2397 3 жыл бұрын

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 2 жыл бұрын

@@sunildhaul2397 How is this different from the one with the formula of the denominator raised to 3/2……?

@sunildhaul2397 2 жыл бұрын

@@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.

@loucimouadi3983 2 жыл бұрын

I don't understand why he didn't integrate directly dE?

@gnyszbr4187 2 жыл бұрын

@@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?

@metalks3716 3 жыл бұрын

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@ahmadahmadi5468 3 жыл бұрын

❤

@user-wg9iy2nk3t 4 ай бұрын

anyone who knows his name?

@troymills2614 6 жыл бұрын

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 !!

@NikkaPleeease 6 жыл бұрын

The denominator should be (x^2+a^2)^3/2, not R^3.... I think you're right!

@v.m.5850 4 жыл бұрын

I have the same question, I know this is late but have you figured out why?

@daniellaughoc4003 2 жыл бұрын

@@NikkaPleeease How is this different from the one with the formula of the denominator raised to 3/2……?

@Sarah-jr8qv 2 жыл бұрын

@@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

@EggPuffsEdge 5 жыл бұрын

Please do behind the scene video sir

@hasny9688 Ай бұрын

adam ya adam

@alexmichaelsafayan1814 Жыл бұрын

How is R a constant? 7:00 - the logic here does not make sense

@mobilegameplay2090 Жыл бұрын

Yep I did not get it as well

@clioosesli4646 Жыл бұрын

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

@alexsafayan7684 Жыл бұрын

@@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

@clioosesli4646 Жыл бұрын

@@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

@alexmichaelsafayan1814 Жыл бұрын

@@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?