Coulomb Force Between A Charge Rod & A Particle

Adam Teasdale Hartshorne

Hi All,

I have a simple physics question, which I wonder if somebody could help
me with as I am a computer science person trying to use Coulomb force as
an error metric.

In 3D space, if I have a charged rod, (length l with start and end
points, r1 and r2) and an oppositely charged particle positined at p1,
what is the coulomb force between them?

I am assuming you have to integrate the standard coulomb force between
two particles along the rod (Excuse the poor english),

Any help is much appreciated,

Adam

Cl.Massé

"Adam Teasdale Hartshorne" <adam@dcs.warwick.ac.uk> a écrit dans le message
de news: ed4ndk$stj$1@mail.dcs.warwick.ac.uk

> I have a simple physics question, which I wonder if somebody could help
> me with as I am a computer science person trying to use Coulomb force as
> an error metric.
>
> In 3D space, if I have a charged rod, (length l with start and end
> points, r1 and r2) and an oppositely charged particle positined at p1,
> what is the coulomb force between them?
>
> I am assuming you have to integrate the standard coulomb force between
> two particles along the rod (Excuse the poor english),

The result depends on only two variables: the radial distance between the
point and the rod, and the axial distance. Let's take a reference frame X,
Y, Z such that the rod is along the Z axis and centred at the origin, thus
starting at z = -l / 2 and ending at z = l / 2. If x, y, z are the
coordinates of the point, the radial and axial distance are respectively
r = sqrt(x^2 + y^2) and h = z.

Each infinitesimal element of the rod gives a force which radial and axial
components are:

dF_r = k r dz / sqrt((h-z)^2 + r^2) / ((h-z)^2 + r^2)

dF_a = k (h-z) dz / sqrt((h-z)^2 + r^2) / ((h-z)^2 + r^2)

Finally the total force is their integral over z between -l / 2 and l / 2

F_r = k r Int dz / ((z-h)^2 + r^2)^3/2

F_a = k Int (h-z) dz / ((z-h)^2 + r^2)^3/2

Combining them, the force is:

F = sqrt(F_r^2 + F_a^2)

Good luck.

Your may also consider an infinite rod. It is then much simpler since the
axial component is zero and the force is proportional to 1 / r.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.