Author Topic: Coulombs law and motion.  (Read 9119 times)

Mardoxx

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Coulombs law and motion.
« on: April 22, 2010, 01:48:56 am »
Quote
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;
$q_1 = q_2 = 0.02 mathrm{C}$
$r = 1   imes10^{-3} mathrm{m}$
$F(r) = frac{k cdot q_1 q_2}{r^2}mathrm{N}   ext{ where } k = frac{1}{4piepsilon_0} $ $= 8.9875517873681764   imes10^9 mathrm{N cdot m^2cdot C^{-2}}$

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 $F(r) = frac{k cdot 0.02   imes 0.02}{(1   imes10^{-3})^2}mathrm{N}$ 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, 02:20:29 am by Mardoxx »

Fu-Kwun Hwang

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Re: Coulombs law and motion.
« Reply #1 on: April 22, 2010, 03:47:49 pm »
The force $vec{F}=frac{kq_1 q_2}{r^2}hat{r}$
So the work done from $r_0$ to $r$ is

$W=int vec{F}cdot dvec{s}=int_{r_0}^{r} frac{kq_1 q_2}{r'^2}dr'=-frac{kq_1 q_2}{r}|_{r_0}^{r}=kq_1 q_2 (frac{1}{r_0}-frac{1}{r})$

Mardoxx

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Re: Coulombs law and motion.
« Reply #2 on: April 22, 2010, 07:30:20 pm »
Thanks :)
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, 09:06:56 pm by Mardoxx »

Fu-Kwun Hwang

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Re: Coulombs law and motion.
« Reply #3 on: April 22, 2010, 10:06:04 pm »
Yes. the mass m is needed.

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

so $v(r)= sqrt{frac{2kq_1 q_2}{m}(frac{1}{r_0}-frac{1}{r})}$

Mardoxx

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Re: Coulombs law and motion.
« Reply #4 on: April 22, 2010, 10:22:53 pm »
but again, I need this so I can do
position_new = position_old + velocity*dtime

this is confusing me lots :(

Fu-Kwun Hwang

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Re: Coulombs law and motion.
« Reply #5 on: April 22, 2010, 10:26:41 pm »
You need to ask the one who gave you the question.

Mardoxx

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Re: Coulombs law and motion.
« Reply #6 on: April 22, 2010, 11:24:50 pm »
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 :D


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 23, 2010, 12:55:41 am by Mardoxx »

siamon

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Re: Coulombs law and motion.
« Reply #7 on: April 10, 2015, 03:22:42 pm »
 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.

lookang

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Re: Coulombs law and motion.
« Reply #8 on: April 10, 2015, 05:44:49 pm »

DieterLowe

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Re: Coulombs law and motion.
« Reply #9 on: April 17, 2015, 07:18:38 pm »
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.


Louise88

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Re: Coulombs law and motion.
« Reply #10 on: October 08, 2015, 05:27:09 pm »
Thank you it is a relevant information.