Title: Coulombs law and motion. Post by: Mardoxx on April 22, 2010, 12:48:56 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; $q_1 = q_2 = 0.02 \mathrm{C}$ $r = 1\times10^{3} \mathrm{m}$ $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??? Title: Re: Coulombs law and motion. Post by: FuKwun Hwang on April 22, 2010, 02:47:49 pm The force $\vec{F}=\frac{kq_1 q_2}{r^2}\hat{r}$
So the work done from $r_0$ to $r$ is $W=\int \vec{F}\cdot d\vec{s}=\int_{r_0}^{r} \frac{kq_1 q_2}{r'^2}dr'=\frac{kq_1 q_2}{r}_{r_0}^{r}=kq_1 q_2 (\frac{1}{r_0}\frac{1}{r})$ Title: Re: Coulombs law and motion. Post by: Mardoxx on April 22, 2010, 06:30:20 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!!!! Title: Re: Coulombs law and motion. Post by: FuKwun Hwang on April 22, 2010, 09:06:04 pm 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 $v(r)= \sqrt{\frac{2kq_1 q_2}{m}(\frac{1}{r_0}\frac{1}{r})}$ Title: Re: Coulombs law and motion. Post by: Mardoxx on April 22, 2010, 09:22:53 pm but again, I need this so I can do
position_new = position_old + velocity*dtime this is confusing me lots :( Title: Re: Coulombs law and motion. Post by: FuKwun Hwang on April 22, 2010, 09:26:41 pm You need to ask the one who gave you the question.
Title: Re: Coulombs law and motion. Post by: Mardoxx on April 22, 2010, 10:24:50 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.... Title: Re: Coulombs law and motion. Post by: siamon on April 10, 2015, 02:22:42 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.
Title: Re: Coulombs law and motion. Post by: lookang on April 10, 2015, 04:44:49 pm http://www.um.es/fem/EjsWiki/Main/Webcasts
Title: Re: Coulombs law and motion. Post by: DieterLowe on April 17, 2015, 06:18:38 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.
Title: Re: Coulombs law and motion. Post by: Louise88 on October 08, 2015, 04:27:09 pm Thank you it is a relevant information.
