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 system $E_{Total }=\frac{19}{12} \frac{Kq^{2}}{d}$
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 $\vec{v_{1}},\vec{ v_{2}}, \vec{v_{3}}$ .
No forces is acting on the system , so at any time the velocity of center of mass will be zero $v_{com}=\frac{m \vec{v_{1}}+m \vec{v_{2}}+m \vec{v_{3}}}{m+m+m}=0$
$\vec{v_{1}}+\vec{v_{2}}+ \vec{v_{3}}=0$ Now, how to proceed?

Share your views.
Thanks in advance.

#ElectricityAndMagnetism

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Comments

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

Talulah Riley - 8 months, 4 weeks ago

@Lil Doug

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

Krishna Karthik - 8 months, 4 weeks ago

@Krishna Karthik in case you are interested.

Talulah Riley - 8 months, 4 weeks ago

Is the answer :

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

Karan Chatrath - 8 months, 4 weeks ago

@Karan Chatrath Yes you are $\Huge CORRECT$

Talulah Riley - 8 months, 4 weeks ago

@Karan Chatrath please share your solution

Talulah Riley - 8 months, 4 weeks ago

The above note is edited.

Talulah Riley - 8 months, 4 weeks ago

Let the X coordinate of the leftmost charge be $x_1=x$ , that of the middle charge be $x_2=x+s$ and that of the rightmost charge be $x_3=x +4d$ . Applying Newton's second law to each of the masses gives the following:

$m\ddot{x}_1 = -\frac{kq^2}{s^2} -\frac{kq^2}{16d^2}$ $m\ddot{x}_2 = \frac{kq^2}{s^2} -\frac{kq^2}{(4d-s)^2}$ $m\ddot{x}_3 = \frac{kq^2}{16d^2} +\frac{kq^2}{(4d-s)^2}$

Adding all of the above equations gives:

$\ddot{x}_1 + \ddot{x}_2 + \ddot{x}_3 =0$ $\implies 3\ddot{x} + \ddot{s} = 0$

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

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

$\ddot{s} = \frac{3kq^2}{2}\left(\frac{1}{s^2} - \frac{1}{(4d-s)^2} \right)$

$s(0) = 3d$ $\dot{s}(0) = 0$

The speed of the second particle is:

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

Can you proceed from here?

Karan Chatrath - 8 months, 4 weeks ago

Edited comment above to remove typos. Hope this helps.

Karan Chatrath - 8 months, 4 weeks ago

@Karan Chatrath thanks, let me try to proceed.

Talulah Riley - 8 months, 4 weeks ago

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

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

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

Talulah Riley - 8 months, 4 weeks ago

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

Karan Chatrath - 8 months, 4 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$