Consider the configuration consisting a +q charge....

Click For Summary

Homework Help Overview

The discussion revolves around a configuration of charges consisting of a +q charge and two -q charges aligned along the x-axis. Participants are exploring the forces acting on the +q charge when it is displaced both perpendicularly and along the x-axis. The problem involves Taylor expansion and vector components of forces.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the Taylor expansion of the force acting on the +q charge when displaced vertically and horizontally. There are attempts to clarify the vector components of the forces and the correct application of the Taylor expansion.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about vector addition and the setup of the forces. Some guidance has been offered regarding the proper treatment of vector components and the implications of symmetry in the system.

Contextual Notes

There are indications of confusion regarding the use of variables and the setup of the problem, particularly in how forces are represented and calculated. Participants are also navigating the complexities of vector notation and the geometry involved in the problem.

Decadohedron
Messages
17
Reaction score
0
Member advised to use the homework template for posts in the homework sections of PF.
and two -q charges aligned along the x-axis as follows...

-q__________q__________-q <--- the space between them being d

a) Suppose the +q charge is displaced perpendicularly by a small vertical distance dy. What is the total force F acting on it? Taylor expand your answer to leading order in dy
b) Next suppose instead the +q charge is displaced along the x-axis by a small dx distance. What is the force acting on it?

--------------

So for a) I'm kind of stuck at the Taylor expansion on the force.

F = (q)(-q)/4piε0d2 yhat + (q)(-q)/4piε0d2

where d= sqrt(x^2+dy^2)

I'm at a point where I'm trying to simplify something but feel that I"m doing something wrong in the math

Same for b)

Thanks!
 
Physics news on Phys.org
Decadohedron said:
and two -q charges aligned along the x-axis as follows...

-q__________q__________-q <--- the space between them being d

a) Suppose the +q charge is displaced perpendicularly by a small vertical distance dy. What is the total force F acting on it? Taylor expand your answer to leading order in dy
b) Next suppose instead the +q charge is displaced along the x-axis by a small dx distance. What is the force acting on it?

--------------

So for a) I'm kind of stuck at the Taylor expansion on the force.

F = (q)(-q)/4piε0d2 yhat + (q)(-q)/4piε0d2

where d= sqrt(x^2+dy^2)
One obvious problem here is you're trying to add a vector to a scalar. A less obvious one is you're using the variable ##d## in two different ways. You essentially have the magnitude of the force between the pairs of charges right:
$$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{x^2+y^2}$$ What should ##x## equal here?

What are the x-component and y-component of the net force on the middle charge going to be?

Once you have those expressions, then use the Taylor expansion about ##y=0##.

I'm at a point where I'm trying to simplify something but feel that I"m doing something wrong in the math

Same for b)

Thanks!
 
  • Like
Likes   Reactions: Decadohedron and mfb
vela said:
One obvious problem here is you're trying to add a vector to a scalar. A less obvious one is you're using the variable ##d## in two different ways. You essentially have the magnitude of the force between the pairs of charges right:
$$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{x^2+y^2}$$ What should ##x## equal here?

What are the x-component and y-component of the net force on the middle charge going to be?

Once you have those expressions, then use the Taylor expansion about ##y=0##.
The second expression should also have a yhat - I was posting in a bit of a hurry my bad.

Net force in the x direction is going to be 0 due to symmetry. The restoring force would be pointing back down to the origin.
I guess I'm stuck at getting that expression.

I should have
F = -2(q^2/4piε0(1/dy^2)yhat)EDIT: Pardon the bad formatting and all - trying to get used to how the forum works. Also, pardon the mad noob question - I guess I understand the logic of where the problem is going I just can't put it into math.

Thanks again.
 
You were closer before. When the middle charge is collinear with the others, there should be no vertical force. Is your latest expression for the force consistent with that?
 
  • Like
Likes   Reactions: Decadohedron
I"m supposed to calculate the force once it's shifted in delta y.

https://drive.google.com/open?id=0B76TBRMyBuffbkVPY3hjZU0xV3M

The picture above is where I always get to before not knowing what to do anymore.
 
Last edited by a moderator:
From your image it doesn't look to me like you've added the vectors of the forces properly.
upload_2016-9-17_8-29-36.png

By symmetry only the y-components of the forces survive the vector summation, but you haven't shown how those y-components are determined from the force magnitudes. Your work seems to be taking the full force magnitudes and just calling them y-components. Unless there's something special about your ##\hat{y}## vectors?

What do you need to multiply the force magnitudes by in order to "extract" the magnitude of their y-components?
 
  • Like
Likes   Reactions: Decadohedron
gneill said:
From your image it doesn't look to me like you've added the vectors of the forces properly.
View attachment 106122
By symmetry only the y-components of the forces survive the vector summation, but you haven't shown how those y-components are determined from the force magnitudes. Your work seems to be taking the full force magnitudes and just calling them y-components. Unless there's something special about your ##\hat{y}## vectors?

What do you need to multiply the force magnitudes by in order to "extract" the magnitude of their y-components?

I should have called it something else but

##\hat{r}## = rvector/|r| = (0, y)/sqrt(x^2+y^2)
 
Decadohedron said:
##\hat{r}## = rvector/|r| = (0, y)/sqrt(x^2+y^2)
The numerator of what you wrote suggests ##\vec{r} = (0,y)## while the denominator suggests ##\vec{r} = (x,y)##.

The vector ##\hat r = \frac{\vec{r}}{\|\vec{r}\|}## has to have a magnitude of 1. The vector ##\frac{(0,y)}{\sqrt{x^2+y^2}}## clearly doesn't, so the two can't be equal.

Forget about vector notation for right now. By what do you need to multiply the magnitude of the force by to get just the y-component? Note that ##x## won't appear in the final expression because ##x=0## for the middle charge.
 
  • Like
Likes   Reactions: Decadohedron
vela said:
The numerator of what you wrote suggests ##\vec{r} = (0,y)## while the denominator suggests ##\vec{r} = (x,y)##.

The vector ##\hat r = \frac{\vec{r}}{\|\vec{r}\|}## has to have a magnitude of 1. The vector ##\frac{(0,y)}{\sqrt{x^2+y^2}}## clearly doesn't, so the two can't be equal.

Forget about vector notation for right now. By what do you need to multiply the magnitude of the force by to get just the y-component? Note that ##x## won't appear in the final expression because ##x=0## for the middle charge.

r sin theta would give me that magnitude of y?

I really appreciate the help guys - sorry that I"m being a bit dense at the moment.
 
  • #10
Decadohedron said:
r sin theta would give me that magnitude of y?
If r is the magnitude of the vector of interest and θ is the angle that r makes with the positive x-axis, then yes, r sin(θ) would give you the magnitude of its y-component.

Now, see if you can "construct" sin(θ) from the given geometry using the triangle sides.
 
  • Like
Likes   Reactions: Decadohedron
  • #11
I"m not quite sure what you mean by construct here, so I assumed this

rsin theta = y

sin theta = y/sqrt(d^2+y^2)
So then I taylor expanded 1/sqrt(d^2+y^2) and since it's to the highest order of y, I ended up with

sin theta approximately equals y/dWhich when I put that back into the original force equation I come out with -2q^2y/4piepsilon0d^3
 
  • #12
I don't think it matters in this case, but I'd wait to Taylor expand until after you get the expression for the force because the magnitude of the force also depends on ##y##.
 
  • Like
Likes   Reactions: Decadohedron
  • #13
There was no other y in the expression so I figured it'd be safe to Taylor expand then.

Glad I finally got this - thanks to you guys! My head was getting sore from beating it on the desk so much.

Now onto part b
 

Similar threads

Replies
1
Views
904
Replies
1
Views
2K
Replies
21
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K