3 Charges and unsuccessful attempt

Here is a problem , which I was solving today
Here is my attempt:
I think the toral electrostatic energy of system will be conversed.
It will interchange between kinetic energy and electrostatic potential energy.
This is starting energy of systemETotal=1912Kq2dE_{Total }=\frac{19}{12} \frac{Kq^{2}}{d}
After that the net external force in the system is zero.
Si the centre of mass will be in constant, and will lie in x axis only [x_{com}=\frac{4d}{3}]
When the system is released , force will start acting simultaneously on all 3 charge, therefore all will acquire different velolicty in different direction.
Let the velocity of 1st, 2nd, 3rd charge be v1,v2,v3\vec{v_{1}},\vec{ v_{2}}, \vec{v_{3}} .
No forces is acting on the system , so at any time the velocity of center of mass will be zerovcom=mv1+mv2+mv3m+m+m=0v_{com}=\frac{m \vec{v_{1}}+m \vec{v_{2}}+m \vec{v_{3}}}{m+m+m}=0
v1+v2+v3=0\vec{v_{1}}+\vec{v_{2}}+ \vec{v_{3}}=0 Now, how to proceed?

Share your views.
Thanks in advance.

Note by Talulah Riley
1 month ago

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@Steven Chase @Karan Chatrath @Uros Stojkovic Your inputs would be helpful.

Talulah Riley - 1 month ago

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@Lil Doug

I'll try the problem soon, despite the fact that you already have a solution.

Krishna Karthik - 1 month ago

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@Krishna Karthik in case you are interested.

Talulah Riley - 1 month ago

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Is the answer :

q29πϵomd\sqrt{\frac{q^2}{9 \pi \epsilon_o md}}

Karan Chatrath - 1 month ago

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@Karan Chatrath Yes you are CORRECT\Huge CORRECT

Talulah Riley - 1 month ago

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@Karan Chatrath please share your solution

Talulah Riley - 1 month ago

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The above note is edited.

Talulah Riley - 1 month ago

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Let the X coordinate of the leftmost charge be x1=xx_1=x, that of the middle charge be x2=x+sx_2=x+s and that of the rightmost charge be x3=x+4dx_3=x +4d. Applying Newton's second law to each of the masses gives the following:

mx¨1=kq2s2kq216d2m\ddot{x}_1 = -\frac{kq^2}{s^2} -\frac{kq^2}{16d^2} mx¨2=kq2s2kq2(4ds)2m\ddot{x}_2 = \frac{kq^2}{s^2} -\frac{kq^2}{(4d-s)^2} mx¨3=kq216d2+kq2(4ds)2m\ddot{x}_3 = \frac{kq^2}{16d^2} +\frac{kq^2}{(4d-s)^2}

Adding all of the above equations gives:

x¨1+x¨2+x¨3=0\ddot{x}_1 + \ddot{x}_2 + \ddot{x}_3 =0     3x¨+s¨=0\implies 3\ddot{x} + \ddot{s} = 0

    x˙=s˙3\implies \dot{x} = -\frac{\dot{s}}{3}

Plugging this result in the 2nd equation of motion gives:

s¨=3kq22(1s21(4ds)2)\ddot{s} = \frac{3kq^2}{2}\left(\frac{1}{s^2} - \frac{1}{(4d-s)^2} \right)

s(0)=3ds(0) = 3d s˙(0)=0\dot{s}(0) = 0

The speed of the second particle is:

x˙2=x˙+s˙=2s˙3\dot{x}_2 = \dot{x}+\dot{s} = \frac{2\dot{s}}{3}

Can you proceed from here?

Karan Chatrath - 1 month ago

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Edited comment above to remove typos. Hope this helps.

Karan Chatrath - 1 month ago

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@Karan Chatrath thanks, let me try to proceed.

Talulah Riley - 1 month ago

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@Karan Chatrath I just found an alternative approch, it was a very question also.
Basically I thought that author wants to say that at t=0t=0 all balls are released freely.
But after t=0t=0 , extreme balls are still rigidly attached.
Here is an alternative method:
Electrostatic energy E=Kq2s+Kq2(4ds)+Kq24dE=\frac{Kq^{2}}{s}+\frac{Kq^{2}}{(4d-s)}+\frac{Kq^{2}}{4d}

Whenever that middle ball will acquire maximum velocity, at that instant will have minimum energy.
dEds=0\frac{dE}{ds}=0
After solving above equation gives s=2d\boxed{s=2d}

I am posting a new note within a hour showing a another attempt of a problem.
Thanks in advance.

Talulah Riley - 1 month ago

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Nice alternative method. I will look at your other note later.

Karan Chatrath - 1 month ago

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