Self Study - Foundations of Electromagnetic Theory Problem 2-1 (two charged particles hanging on strings)

[Usiia]

Homework Statement

Two Particles, each of mass m and having charges q and 2q respectively, are suspended from strings of length l from a common point. Find the angle θ that each string makes with the vertical.

Relevant Equations

Coulomb force definition with permittivity variable present.

I'd like someone to check that this solution is right, and if not, give me the point of failure and not the answer.

Thanks!

 

[AntSC]

Error in your first equation. Otherwise, all the components are there and looks good

 

[Usiia]

Thanks! Will take another look and report back!

 

[Usiia]

Ok, second try:

 

[BvU]

Attachment disappeared !

 

[Usiia]

needed a correction, making that now, having camera/file issues.

EDIT: No, I the right one I meant to upload was wrong. One sec.

ok, try number 4:

 

[BvU]

Picture OK now (well, a bit of $LaTeX$ would be a lot better ) . What about $r_{12}$ ?

 

[Usiia]

Tried using latex, mathjax wouldn't work

 

[Usiia]

Also, I'm incredibly good at LaTeX as well as a CS person, but I don't like it for HW type problems. I just get caught up making it prefect looking instead of right.

If it's a space thing, I'll try harder :)

 

[Usiia]

So, I left out r_1,2 normalized in this because it wasn't necessary to figure out magnitudes of the force, the direction laterally didn't seem important.

 

[AntSC]

needed a correction, making that now, having camera/file issues.

EDIT: No, I the right one I meant to upload was wrong. One sec.

ok, try number 4:

View attachment 294532

Looks good to me. Yes, it was your rearrangement of $F=mg\tan \theta$ that was the issue. Everything else seems good

 

[Usiia]

Thank you so much! I'll work on getting the LaTeX working for presenting my answers better and preserving the server storage!

 

[AntSC]

Thank you so much! I'll work on getting the LaTeX working for presenting my answers better and preserving the server storage!

There is a thread on the use of Latex for PF if needed.

 

[Usiia]

I'm going to work on this a little more because I feel some identities could simplify this result.

 

[Usiia]

So I am using an arctan identity

arctan(2mg*pi*permit*r^2/q^2)= -pi/2 - arctan(2mg*pi*permit*r^2/q^2)

and so

-pi/2 - arctan(2mg*pi*permit*r^2/q^2)

EDITED: So, if we took the limit m->0, g->0 or we'd get that this system wants to be at no angle with no force (which is right) and otherwise, q->inf, it's going to be some amount less than 180 degrees.That interpretation is correct right?

 

[mjc123]

So arctan(x) = -π/2 - arctan(x)?
Did you mean arctan(x) = π/2 - arctan(1/x)?
You are being very sloppy, as is shown by missing terms out in some of the equations you have posted. This is not the way to get right answers.

EDITED: So, if we took the limit m->0, g->0 or we'd get that this system wants to be at no angle with no force (which is right)

Which is wrong. If m or g = 0, there is no vertical force, so θ = 90°, where the particles are as far apart as possible.

What about r? You are not given r, but you are given l, the length of the string. What is r in terms of l and θ? (This makes things more complicated!)

 

[Usiia]

Thank you for your reply.

The identity I found had - π/2 and I had inverted the internal expression.

I believed that this was correct with my choice of origin. Also, -π/2 is 0 degrees in the orientation I choose where g points down.

I apologize if I had been sloppy. I am just getting back into this work. I will work on my presentation and draft better.

 

[Usiia]

What about r? You are not given r, but you are given l, the length of the string. What is r in terms of l and θ? (This makes things more complicated!)

Figured out the block quote, so yes, I agree and will work on that as well. Thank you!

 

[mjc123]

Good. Kudos to you for liking my post though it criticised you; that's the way to learn! Good luck.

 

[Usiia]

Good. Kudos to you for liking my post though it criticised you; that's the way to learn! Good luck.

I'm sadly aware of how far I've slipped.

 

[kuruman]

The problem gives you the length of the string $l$ not the separation between the masses. Therefore your answer must be in terms of $l$, i.e. given $l$ in meters and $q$ in Coulombs, find $\theta$ in radians. @BvU pointed that out in post #7.

 

[Usiia]

The problem gives you the length of the string $l$ not the separation between the masses. Therefore your answer must be in terms of $l$, i.e. given $l$ in meters and $q$ in Coulombs, find $\theta$ in radians. @BvU pointed that out in post #7.

Yes, I see and realize that now. Thank you.

 

[Usiia]

So, to start I have two related triangles. One of the forces and one of the lengths.

$$ F_{1,2} \cos \theta - mg \sin \theta = 0 $$
$$ r_{1,2} = l \cos \theta$$
$$ F_{1,2} = \dfrac {1}{4 \pi \epsilon_0} \dfrac {2q^2}{r_{1,2}^2} $$

And then
$$ \dfrac{q^2 \cos \theta}{2 \pi \epsilon_0 ( l \cos \theta )^2 } = mgl \sin \theta$$

$$ \dfrac{q^2 }{2 \pi \epsilon_0 (\cos \theta l) ^2 } = mgl \sin \theta$$

Is this right so far or should I be treating this like a related rates problem?

 

[PeroK]

I must admit I don't understand your diagram or your equations. Why not use $T$ for the tension in the string; $F$ for the electrostatic force; and, $r$ for the distance between the masses?

I suggest you show both masses on the diagram, so you can see how $r$ relates to $\cos \theta$ (or, should that be $\sin \theta$?)

 

[Usiia]

I must admit I don't understand your diagram or your equations. Why not use $T$ for the tension in the string; $F$ for the electrostatic force; and, $r$ for the distance between the masses?

I suggest you show both masses on the diagram, so you can see how $r$ relates to $\cos \theta$ (or, should that be $\sin \theta$?)

I will draw out the masses and take another look at the trig functions, thanks!

 

[Usiia]

I will draw out the masses and take another look at the trig functions, thanks!

Ok, here's a more complete diagram.

ignore the wrong note on tan to the right.

 

[PeroK]

That doesn't look right. The gravitational force on each mass is simply $mg$. The same force $F$ acts on each particle (hence the symmetry).

 

[Usiia]

You're right. I got caught up in finding out where to place that sin.

 

[Usiia]

Is this more accurate?

 

[PeroK]

Just to say that I'm getting an "unsolvable" cubic equation in $\tan \theta$. Does the question expect an exact solution?

 

[Usiia]

Just to say that I'm getting an "unsolvable" cubic equation in $\tan \theta$. Does the question expect an exact solution?

It is a cubic relation.

 

[Usiia]

Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

 

[Usiia]

Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

Of course this means nothing to me. I am supposing that this is a related rates problem and that I need more than what I think I have so far in terms of the tangent.

 

[PeroK]

Of course this means nothing to me. I am supposing that this is a related rates problem and that I need more than what I think I have so far in terms of the tangent.

First, generate the force balancing equations for $T, F$ and $mg$.

 

[Usiia]

First, generate the force balancing equations for $T, F$ and $mg$.

Need to go to work now, but will come back to this with this approach.

Thank you so much!