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Author Topic: Coulombs law and motion.  (Read 6628 times)
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Mardoxx
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on: April 22, 2010, 12:48:56 am » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

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Imagine two charged particles are placed a small distance, say 1 mm, apart and held there. The charge on each of the particles  is 0.02C.

One of the particles is released. Describe the motion of the free particle and calculate the total work done.

so;
q_1 = q_2 = 0.02 \mathrm{C}
r = 1\times10^{-3} \mathrm{m}
F(r) = \frac{k \cdot q_1 q_2}{r^2}\mathrm{N} \text{ where } k = \frac{1}{4\pi\epsilon_0} = 8.9875517873681764\times10^9 \mathrm{N \cdot m^2\cdot C^{-2}}

We know that as r increases F decreases so,
F(r) > F(r+\delta r)

We shall assume it is q_2 that is released and q_1 is held by an imaginary force.
When q_2 is released a force of F(r) = \frac{k \cdot 0.02 \times 0.02}{(1\times10^{-3})^2}\mathrm{N} is pushing it in the positive direction.

After a tiny bit of time, \delta t, the particle will have moved \delta r and so the force will have decreased meaning that the acceleration of the particle will have decreased...

This is where I get stuck.
how can we express the particles motion if it has no set mass...?

I was thinking change in momentum, but that still depends on mass...

This is an interesting one!
Please can you give me a pointer or two?

Is it actually possible to work it out without mass???
« Last Edit: April 22, 2010, 01:20:29 am by Mardoxx » Logged
Fu-Kwun Hwang
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Reply #1 on: April 22, 2010, 02:47:49 pm » posted from:Taipei,T'ai-pei,Taiwan

The force \vec{F}=\frac{kq_1 q_2}{r^2}\hat{r}
So the work done from r_0 to r is

W=\int \vec{F}\cdot d\vec{s}=\int_{r_0}^{r} \frac{kq_1 q_2}{r'^2}dr'=-\frac{kq_1 q_2}{r}|_{r_0}^{r}=kq_1 q_2 (\frac{1}{r_0}-\frac{1}{r})
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Mardoxx
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Reply #2 on: April 22, 2010, 06:30:20 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

Thanks Smiley
but how can I describe its motion without a mass?

Also, 0.02C ... that's pretty huge to be held at 1mm!!! The initial force comes out as... 355920000N!!!!
« Last Edit: April 22, 2010, 08:06:56 pm by Mardoxx » Logged
Fu-Kwun Hwang
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Reply #3 on: April 22, 2010, 09:06:04 pm » posted from:Taipei,T\'ai-pei,Taiwan

Yes. the mass m is needed.

Initial velocity is zero, so W=\tfrac{1}{2}mv^2=kq_1 q_2 (\frac{1}{r_0}-\frac{1}{r})

so v(r)= \sqrt{\frac{2kq_1 q_2}{m}(\frac{1}{r_0}-\frac{1}{r})}
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Mardoxx
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Reply #4 on: April 22, 2010, 09:22:53 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

but again, I need this so I can do
position_new = position_old + velocity*dtime

this is confusing me lots Sad
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Fu-Kwun Hwang
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Reply #5 on: April 22, 2010, 09:26:41 pm » posted from:Taipei,T\'ai-pei,Taiwan

You need to ask the one who gave you the question.
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Mardoxx
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Reply #6 on: April 22, 2010, 10:24:50 pm » posted from:Kingston Upon Hull,Kingston upon Hull, City of,United Kingdom

I think i've got it

if I use F=ma(r) I can get
a(r) = (k*q1*q2)/(m*r**2)

then if i use euler/RK I can get velocity and then distance at time t Cheesy


actually, no I don't know
because still, doing this... itsays it would get to the moon in a few seconds - and I'm not sure that's right....
« Last Edit: April 22, 2010, 11:55:41 pm by Mardoxx » Logged
siamon
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Reply #7 on: April 10, 2015, 02:22:42 pm » posted from:,,Satellite Provider

 Your physics simulations are so impressive . I downloaded EJS and tried to play with it. I have already seen you flash tutorial.  However, could you give me some suggestions  - where and how  can I get some books or web tutorials  to learn EJS? I just feel that it would take me longer time to play with this software and understand it without reading a book or a help manual.
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lookang
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http://weelookang.blogspot.com


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Reply #8 on: April 10, 2015, 04:44:49 pm » posted from:,,Satellite Provider

http://www.um.es/fem/EjsWiki/Main/Webcasts
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DieterLowe
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Reply #9 on: April 17, 2015, 06:18:38 pm » posted from:Kannur,Kerala,India

The quantitative expression for the effect of these three variables on electric force is known as Coulomb's law. Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

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Louise88
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Reply #10 on: October 08, 2015, 04:27:09 pm » posted from:Bangalore,Karnataka,India

Thank you it is a relevant information.
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