Calculating Electric Forces with Dangling Objects

Click For Summary

Homework Help Overview

The discussion revolves around calculating electric forces involving two identical metal spheres, one neutral and one charged, that are suspended from an insulating wire. The problem involves understanding the forces acting on the spheres and how to relate them to the charges involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss drawing free body diagrams (FBD) to analyze the forces acting on the spheres, including electric force, tension, and gravitational force. There are attempts to relate these forces using trigonometric functions.

Discussion Status

Some participants are exploring the relationships between the forces and questioning the correct application of trigonometric functions to solve for the electric force. There is a recognition of the need to find the distance between the spheres and how it relates to the forces involved.

Contextual Notes

There is mention of missing information regarding the radius or distance between the two spheres, which is crucial for the calculations. Participants are also navigating the implications of using the length of the rope in their reasoning.

Cory
Messages
10
Reaction score
0
How do I calculate Electric forces when objects are dangling from insulated wire. For example
Neutral metal sphere A, of mass 0.10kg hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge -q, is brough into contact with the sphere A. The spheres repel and stelle down as shown in the following figure

.\ <) = 12
...\
...\
...\
(B)...(A)
* <) AB(top) = 90 degrees.
Calculate the initial charge on B.

The formula's are Fe =
k*q1*q2
r^2
 
Physics news on Phys.org
Ok Cory,a are you able to draw a FBD of the forces acting on A?
 
I did with electric force outward, Force of tension angled at 12degrees, and force of gravity. Then I rearranged them to make a triangle.
ie
^
.\ =Ft
...\
...\
...0---->Fe
...|
...| Fg
...\/
 
That's perfectly fine, what's your problem with solving the question then?
 
I can't seem to understand how to solve the FBD for fe. I tried using
tan(12) = Fe
.....m*g
but that doesn't seem to work, and I am really just not sure how to go about this.
 
Last edited:
That doesn't seem to be right, why would you use sin(12)=Fe?

:)

sin(12)=Fe/Ft isn't it? Think of a triangle. Thus Fe=Ft*sin(12) You can find Ft using Fg
 
Sorry, I edited my last post, I typed the wrong thing. Should I still try Fe = ft*sin12? I don't really understand
 
Ft*cos12=Fg help you?
 
I don't think so.. I'm trying to find the initial charge on A
 
  • #10
Nm I see your problem, you do know Fe.

The spheres are equal so q1 = q2

Thus:
Fe = Fg*tan12
and
Fe =k*q^2*r^2
 
  • #11
so find Fe with mg*tan12 (which is: 0.20830543) then use that to solve for q in Fe =k*q^2*r^2? why does it change from divided by r^2 to multiplied?
 
  • #12
Cory said:
so find Fe with mg*tan12 (which is: 0.20830543) then use that to solve for q in Fe =k*q^2*r^2? why does it change from divided by r^2 to multiplied?

Oh it doesn't, misread your equation sorry ^^
 
  • #13
I seem to still be missing radius
 
  • #14
Heh sorry, it does not mean radius, r means the distance between the two particles. Are you able to find it using trigonometrics?
 
  • #15
well yea, but then it's the same as Fe = Fg*tan12 which doesn't work. Maybe I should use the length of the rope?
 
  • #16
What do you mean?

k*q^2*r^2=Fg*tan12

q=sqrt(Fg*tan12/r^2/k)
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
9K