Calculating Electric Force of a Point Charge on a Charged Particle Thread

[ex81]

Charged Particle thread with linear charge force

Homework Statement

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:

Homework 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

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.

 

[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)

 

[rude man]

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

 

[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

 

[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.

 

[rude man]

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

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 E_x, E_y and E_z directly, forgetting about potential altogether, at least at first.

 

[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.

 

[rude man]

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.

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 = E_x i + E_y j + E_z k. Then the force on the charge Q is of course just QE.

I've looked at E_x 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.

 

[gneill]

Homework Statement

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:

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.

 

[ex81]

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

My answer was correct. :-)

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