Physics 36 Electric Field (6 of 18) Infinite Line Charge

In this video I will find the electric field of an infinite line charge.

[riteshbhartiya]

Who else watched video in hope of explanation of solving the integral.

[thelastcoolbrandon]

pro status! Note to self  - always trust physics advice from a man in a bowtie.

[HaiderKhanPakistan]

Loved how casually the guy said infinity over infinity is one xD xD. If you want to succeed in life, you need this much of confidence

[anasghaffar]

if you'd solved the integral it would've been perfect.

[winenoob]

THANK YOU. you are seriously a life saver to so, so many students out here. Most of the time, physics doesn't make sense to me until I watch your videos

[yusufakyldrm]

this integral is wrong(in my opininon ) where is from a^2

[dalenassar]

comment to the infinity / infinity:
On your calculator, let a number like 1, 000, 000 represent infinity and a much smaller number for a: like a=2. Plug them in at the integral evaluations and you will find you get 1- (-1)=2...this PARTICULAR one is defined as stated!

[kulsoomhussaini]

literally youre the only reason I am not failing physics, thank you!!!

[umiturgutaswwsa]

Thanks again it looks like i will be with you until i graduate from Mechatronics Engineering.Greetings from Türkiye.

[ajathreya]

I guess it will be easy if we take
x= a Tan (theta) for integration.

[josueesquivel]

Hi, Thank you so much for your instructional videos! I just had a particular question in your process. At around 7:27, when you defined dEx, why is it that you didn't multiply the equation by 2 given that while the vertical (y) components of the charges above and below the horizontal lines cancel out, their horizontal components match and double?

Again thanks for your videos! I always come to watch them after having read the book to reinforce my understanding!

[valeriereid]

Watching this a second time and it is more beautiful on the second viewing. Thanks so much for making these lectures available. Certainly is helping with E&M.

[emirhanylmazguney]

infinity over infinity is not 1 at all. You didn't even tell about limit... Completely wrong.

[vansf]

It is so obvious that when you assume r becomes equal to x and remove a, then r will no longe intersect the rod at tge differential mass element delta x nor any other point on the rod, and therefore r will no longer be the distance between the rod and the point of interest P becomes parallel to the rod or overlapping the rod when the straight segment" a " which is connecting the rod and the point p and perpendicular to the rod and representing tge distance between the rod and tgat point P becomes zero which is totally absolutely nonsensical

[deniz-gunay]

yesss.. you should be careful when a physicist begins to integrate..

[NA-udqm]

While inf/inf I don't casually agree with, what I find amazing about this is what was NOT calculated.

My thought experiments are making me question the following:

(1) if the line charge is symmetrically spherical/circular, the result, I assume, would be the same.

(2) (more importantly) if the line charge is NOT perfectly symmetrically (i.e., contains jagged edges), then the Electric field would contain Electric Field components in the Y-Direction, which, effectively, takes away x-component electric field values, REDUCING the overall result.

I find the (2) thought experiment more important because, while we want an overall "Electric Field Measurement" that is easy to calculate, we should account for how any impurity would affect our initial measurements. A "kink" on the top of the line charge would cause an E-Y component to exist and that Y-Component would take an electric field value from E_X.

[sandlertossone]

Doing L'opitals will not work or at least I just kept going in a big circle. I divided through by x before taking the limit using infinity and negative infinity. For the life of me, I get 1 for the positive infinity and 1 for negative infinity which subtracting equals zero. Maybe it is me but treating infinity like a real or any kind of number and doing operations with it probably saves time if you know the outcome but confuses the hell out of everybody. Limits in indeterminate forms do take time but I would really like to know how you determined the second fraction as negative. Thanks again

[rodri]

Thank you very much, i was searching exactly for this, everywere i looked it was explained with Gauss Law and the gaussian surface, i had to search in english to find this video whith the Coulomb's law solution. Greetings from Argentina.

[vansf]

You cannot just ignore a and use infinity so fallaciously conveniently, because no matter how small the value of a can be it is the reason why r can never be the same as x, as the value of x approaches infinity . If r became equal to x it would mean that r would be parallel to x, and thus r would no longer the distance between the rod of charge and the point of interest p at which we want to find the net magnitude of the electric field set up by the infinite rod of charge

Your argument is nonsensical without any concrete proof, because you can never prove why you can just delete the value of a to suit your own subjective beliefs.You have to prove it
Now let see how fallacious your argument is by dividing both the numerator and the denominator by x, you will have
Lim (x/x)/ (squrt of x^2+a^2)/x) as x approaches + infinity
Now moveping x into the squrt will give :

Lim 1/ (squrt of x^2/x^2 + a^2/x^2) as x approaches + infinity
Lim 1/ (sqrt of 1 + a^2/x^2) as x approaches + infinity
So result will be 1/ ( squrt of 1 +0) =1
Similarly you have
Lim x/ sqrt of x^2 + a^2 as x approaches negative infinity
Now dividing both the numerator and the denominator by x
Lim x/x /(squrt of x^2/ x^2 +a^2/x^2) as x approaches - infinity
= Lim 1/ (squrt of 1 + a^2/x^2) as x approaches - infinity
= 1/ sqrt of 1 + 0 = 1
Now you subtract the result of limit evaluation for the expression as x approaches - infinity from the result for the expression as x approaches + infinity
1 - 1 =0, but not 2
How you can say anything about such a contradiction when obviously my limit evaluation is accurately correct ?

[MrSarimjavaid]

Thanks for making it super easy and fun :)