Interesting Charge Separation Problem

Here is a problem sent by Neeraj.

Let $x = 0$ be the vertical line right in between the particles. Both particles have the same mass and experience the same force magnitude, so they will be equidistant from $x = 0$ at all times. Let $x$ be the distance of each particle away from the axis. The differential equation for this system is:

$\frac{k q^2}{(2x)^2} = \frac{k q^2}{4x^2} = m \ddot{x}$

This differential equation seems pretty nasty. I looked around online for analytical solutions and didn't find much. Nevertheless, the simulation is very easy to run numerically. The strategy is to run the base case first and call the final time $t_0$ . Then run again with parameters $a, b , \eta$ changed from their original values, and see how the final time changes.

Multiplying the charge magnitudes by $a$ and $b$ causes the final time to be multiplied by $\frac{1}{\sqrt{a b}}$ . Multiplying the initial separation by $\eta$ causes the final time to be multiplied by $\sqrt{\eta^3}$ . So the final result is:

$t_1 = t_0 \sqrt{\frac{\eta^3}{a b}}$

Simulation code is attached:

import math

dt = 10.0**(-5.0)

m = 1.0
k = 1.0
q = 1.0
x0 = 1.0

###########################

# Base case

t = 0.0

x = x0
xd = 0.0
xdd = 0.0

while x <= 2.0*x0:

    x = x + xd*dt
    xd = xd + xdd*dt

    F = k*q*q/((2.0*x)**2.0)

    xdd = F/m

    t = t + dt

t0 = t

###########################

# Modified case

a = 1.0
b = 1.0
n = 1.0

t = 0.0

x = n*x0
xd = 0.0
xdd = 0.0

while x <= 2.0*n*x0:

    x = x + xd*dt
    xd = xd + xdd*dt

    F = k*(a*q)*(b*q)/((2.0*x)**2.0)

    xdd = F/m

    t = t + dt

t1 = t

###########################

print dt
print (t1/t0)

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
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Comments

@Steven Chase Thanks I am very grateful to you.

Talulah Riley - 9 months, 3 weeks ago

I was solving this with anayltical method.
Will you take a look in my rough work/attempt???

Sure, I'm curious to see what you did

Steven Chase - 9 months, 3 weeks ago

@Steven Chase Nowadays I am studying mathematics only, therefore I am little bit not connected with physics, so I am not able to how to do this problem

Nice method. I was about to use the hand differential equation solving method posted by Flammable Maths:

https://www.youtube.com/watch?v=OtUdGdcfUcE

And thereby create an equation for the time taken. After all, the nature of gravitational and electrostatic force is very similar; essentially we are solving the same differential equation, so we can use the same method:

$\displaystyle \ddot{r} = \pm \frac{C}{r^2}$

It is a hard differential equation to solve analytically, since it is a nonlinear second order ODE.

Krishna Karthik - 9 months, 3 weeks ago

@Steven Chase the above comment is edited.

Take a look at this video @Lil Doug

The video above shows how much time it takes for two gravitating bodies to collide. Essentially, the differential equation that is to be solved is the same; an inverse square non-linear ODE.

You can apply this to electrostatics as well, despite the fact that the video is about gravity.

@Steven Chase I think I know how to solve the non-linear ODE above; I sent a video to Neeraj. Here's the video:

It's by Flammable Maths (Jens Genau) who solves a similar ODE above (gravitation). I think the same idea can be applied to electrostatics. So there is a way to do it analytically, I think.

@Krishna Karthik I am damn sure I can explain better than that guy.

Flammable maths is an absolutely epic mathematician. He's covered some really interesting videos.

Thanks for sending that. Looks like a very involved process

@Steven Chase Hey sir see my method above.
Around 1 year ago i have solved this types of problems that how much time it will take to collide two particle but I don't no why I am not able to do it now .
See my solution above
In the first case, I think my process correct.

@Steven Chase I think you are the only solver my problem?

Maybe I could have solved it using a numerical method, but yes, he still his the only solver. Nicely done, too.

Of course, we both know that Steven Chase taught me how to do a time-domain simulation. In retrospect, he, and no one else, taught me how to do a physics simulation. And I'm obviously still not as experienced as him in that (or for that matter, anything in physics).

So, there's a gem of knowledge I think I learnt from him. I mostly looked at his code and learnt how to do it by looking at his code.

In terms of pure programming or coding, I learnt from my father. So, thanks, Steven Chase, for teaching me how to do a time-domain simulation. Cheers.

@Krishna Karthik I am not saying for this problem bro.
I have posted a another problem in E and M section, I am talking about that.

@Talulah Riley – Oh, ok. Lol.

@Talulah Riley – Yeah; that looks mortifyingly hard.

@Krishna Karthik – @Krishna Karthik And what about your test. You are not asking doubts.
It seems you have solved all problems.
Do you want more problem?

@Talulah Riley – Yeah; gimme a Newtonian mechanics problem.

@Krishna Karthik The problems are taken from my physics book.
There are total 36 problems , Part A contains 33 and Part B contains 3 problems.

Yo I need some help😫

I'm not able to do no 6.

Pls help.

Krishna Karthik - 9 months, 1 week ago

@Krishna Karthik I have posted a note.

Talulah Riley - 9 months, 1 week ago

@Steven Chase I have a good attempt for the above problem now.
Would you like to see that?

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}`	$\boxed{123}$