1. Sep 9, 2014

### ex81

Charged Particle thread with linear charge force

1. The problem statement, all variables and given/known data
A uniformly charged thread with linear charge density 5 C/m lies along the curve y = (3/m) x^2 for 0≤x7m. The y component of the electric force it exerts on a 6 C point charge located at (3m, 0m, 2m) is given by:

2. Relevant equations

line integral equation

∫ f(x, f(x)) √([x(t)]^2 +[y(t)]^2)
fyi, the x(t), and y(t) are both derivatives, I just didn't see the symbol

F = k Qq /r^2

3. The attempt at a solution

As written I'm assuming that both charges are supposed to be positive. So they will be pushing away from each other. Since it is a point vs a thread, the force of the thread against the point will not create a secondary equation.
I'm not exactly sure where to go.

My point of integration will be very likely be from 0 to 7 as that is the value of x.

Last edited: Sep 9, 2014
2. Sep 9, 2014

### ex81

Wow I was tired when I posted that.

Starting from the basic force equation.

F = k Qq / r^2

F = k(6) (the tiny cumulative charge as we go across the integral) /(change in distance) ^2

F= 6(5)k(3x^2(1+36x^2)^(1/2))÷(r^2)

F= 90k(x^2(1+36x^2)^(1/2))÷(r^2)

Last edited: Sep 9, 2014
3. Sep 9, 2014

### rude man

I would try to compute the potential ψ, then E = - ψ.

4. Sep 9, 2014

### ex81

Not sure how I would calculate that atm. My brain is a bit mush. But that does explain a bit of the sign issue I was having with this. I figured it should be negative, but I was just not seeing where it was coming from.

The r value should look something like

([x-3]^2+[3x^2]^2+[2^2])
So
X^2-6x+9 +9x^4+4
9x^4+x^2-6x+13

Therefore the answer should be :

F= -90k integral (x^2(1+36x^2)^(1/2))/(9x^4+x^2-6x+13)^(3/2) with x:0 to 7

Last edited: Sep 9, 2014
5. Sep 9, 2014

### rude man

Not sure what you mean by 'negative'. I just gave the formula for the E field, given the potential distribution ψ.

The differential potential dψ would be kλds/r2
where r is the distance between (3,0,2) and an element of charge along the thread dq = λ ds, ds2 = dx2 + dy2 along the parabola. The parabola relates dy to dx. Then ψ = ∫ψ dx. The result is ψ(x0, y0, z0).

You can then drop the subscripts on x, y and z and compute E(x,y,z) = - ψ(x,y,z).

Actually, looking at it some, I suspect you might well be better off by evaluating the three components of E separately, bypassing ψ altogether. If you're really gung-ho you could do both and check the answers against each other.

6. Sep 9, 2014

### rude man

The distance r between a point (x,y,0) on the parabola y = 3x2 (the thread) and the observation point P(x0, y0, z0) is given by

r2 = (x0 - x)2 + (y0 - y)2 + z02.

If you go with potential you can't substitute x0 = 3, y0 = 0, z0 = 2 until after you have computed the potential. See my previous post.

As I said, I woud on second thoughts probably try to compute Ex, Ey and Ez directly, forgetting about potential altogether, at least at first.

7. Sep 9, 2014

### ex81

I'm assuming you are using
using U = kqq/r for the potential electro static equation. It isn't in my text, and I just saw that in another related post.

8. Sep 9, 2014

### rude man

Yes. If it isn't in your book then maybe you're not supposed to know about it ... in which case again you should maybe stick with E = Ex i + Ey j + Ez k. Then the force on the charge Q is of course just QE.

I've looked at Ex and I don't like the looks of that integral! I may be missing the boat here ... let's see if others also offer suggestiuons.

9. Sep 9, 2014

### Staff: Mentor

The question statement as presented looks like the prelude to a multiple choice selection. Is that the case?

As noted by rude man, it looks like you'll be running into a rather nasty integral if you want to solve symbolically. A numerical integration would be doable though, if you just need the numerical value.

10. Sep 10, 2014

### ex81

That is correct, there were 5 possible answers but they all had 90k integral 0to7

How my professor proceeded.

Given:

Lambda =5c/m
Aofx =0, Bofx =, q = 6c, xp =3, yp=0, zp=2, y=3x^2,
y=y(x), dy(x)/dx = 6c

Fqy = 5kq integral 0to7 [(yp-y(x))(1+yprime^2)^(1/2)]/[(3-x)^2 -y(x)^2 +(2)^2]

Then he jumped a bit, and got my answer.
Going back a bit, it looks like he started with the following equation :

Vector of Fq = kq integral over Q for the vector Fdq/r^3