2. Advance Illustration | Electric Force and Electric Field | Expansion of Ring due to Charge

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[deepikaverma]

This is how sir got T=Fl2pi
Basically, we know that
2Tsin(x/2)=F
And for some small x, sin(x/2) can be written as x/2
So we get 2T(x/2)=F
Now, rearranging it, T=F/x( this is for our small element
Applying it to the ring, we get x(theta) as 2pi radians,
Hence, T=F/2pi
Hope this helped you

[prakritprasoon]

While i am not in complete denial of the comments below me, i think one method to understand the division of F by 2pi is to understand that the repulsion between the point charge and the ring is due to the electric field generated by the point charge(q). Now, the electric field lines are uniform across the entire circumference and thus dF force is acting on every small element in a perpendicular direction. So the F/2pi represents the force as a scalar divided across the circumference. Now, if we apply the method to find tension in the ring it is matching well with what sir has written.

P.S. I don't know if this was what the comments below were trying to explain but it took me a while to find this perspective.

[varshinilolla]

Is it possible to do the same with field energy density?

[kakalichakraborty]

Can anyone help me that how Tension = Radial force /2π ????

Where 2π came from ???

[theamithota]

Precise explanation: force on a small part can be written 2Tsin(©)/2 = dF
Theta small so 2Tש/2=T©=dF
Integrate T.2π=F

[nikitachakraborty]

Sir didn't understand why u divided tension by 2 pi please tell sir

[TargetIITBatch]

For those who didt get the feel of the question should ... first find the tension ...then just imagine to cut the ring after cutting it how much force do you need to maintain its shape it will be T so then apply the formula of youngs modulus and solve the question...

[iota]

Tension will also be evenly distributed in the ring .

[dots]

I think i can help some folks out there

Let consider a small dl element which subtends 2θ at center
So total tension in that string is 2Tθ which must be equalize by electrostatic repulsion

Let assume that small element has dq charge ( say dq =λdl because ring is uniform) so electrostatic repulsion is Q (dq) /4πeR^2 = Q(λdl) /4πeR^2

And dl = 2Rθ ( we can check it by drawings the diagram) so

2Tθ = Qλ 2Rθ / 4πeR^2 => T = Qλ /4πe R
Stress = T/A ( A is cross sectional area) = Qλ/ 4πeRA

∆R/R = Qλ/Y4πeRA now λ = total charge on ring / circumference


Here e = epsilon
Just substitute it and get the answer 2.25 mm

[OmPrakash-pvum]

sir you are like god who want to crack advance with sellf study🙏🙏🙏🙏

[makeki]

The tension is not acting outwards but inwards which in return causing expansion of the ring. So why have you taken delta R, with my reasoning it should have been delta L that is the length of wire expanding.

[satvikdeep]

What Sir has done is some kind of short cut or direct trick, to do it yourself, you need to the elemental analysis by talking a small element of the rope.

[shreyanshrajput]

Sir, but E field at centre of uniformly charged ring is 0. So, how force is exerted on ring if ring is not exerting any force on charge at centre?

[h.vats_]

Sir the formula for tension you have written is used at the equilibrium stage i.e. after the ring has been stretched, so the radius must be 0.1+∆r, but you have taken only 0.1 why??

[siddhantrajoriya]

@physicsgalaxy sir why we have used increment in tension instead of total tension in hooke's law? Please reply as soon as possible.

[grandson_f_phixis]

Sir can you please reply why increment in tension is considered and not actual tension??

[rachitparwanda]

Sir, but i am getting T=k.q1.q2/π.r^2.. plz clarify...
I have reached here by solving from the general method to find temsion in circles/rings...(by taking small element of angular width d(theta) and balancing invard tension component and outer electrostatic repulsion )

[raviupadhyay]

sir i do not understand that why does the area remains constant while radius is increased.

[SomewhereLost]

Logic :
Agr kisi m mass ko circular motion mei rotate krnge toh T= mrw^2 aayega .
And agr kisi ring ko ghumaye toh fir T= mrw^2/2pie .


And agr kisi dow charge ke beech tension nikale toh electrostatic force ke brabar hoga. Tension = kqq`/r^2

And agr kisi dow charge ke beech ring mei increment dekhe toh

By observation upar wali 2 equation ko
2pie se divide horha hai
Toh isme bhi F electrostatic se 2pie se divide hoga ..
🥂

[naveenkhola]

Sir i didn't get the idea why u put r and delta r instead of 2πr and 2π delta r, i think if we are taking force as tention and are of cross section then change in length should be of wire