Plastic Balls in a Test Tube with Mass

Click For Summary

Homework Help Overview

The problem involves two charged plastic balls in a test tube, where one ball is positioned above the other. The scenario describes the equilibrium condition between gravitational and electrostatic forces, particularly focusing on how changing the mass of the upper ball affects the distance between the two balls.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between charge and mass, questioning how these factors influence the forces acting on the balls. There is an exploration of gravitational and electrostatic forces, with attempts to relate them mathematically.

Discussion Status

The discussion is active, with participants offering insights into the forces at play and attempting to derive relationships between the variables involved. Some participants have suggested comparing initial and final conditions to understand the changes in distance due to the mass reduction.

Contextual Notes

Participants are considering the implications of changing the mass of one ball and how it affects the equilibrium distance, while also noting the need for a formula for the force of repulsion. There is an acknowledgment of the lack of explicit consensus on the final distance outcome.

PeachBanana
Messages
189
Reaction score
0

Homework Statement



One charged plastic ball is vertically above another in a test tube as shown. The balls are in equilibrium a distance "d" apart. If the mass of the upper ball is reduced by a factor of four, the distance between the balls in the test tube would become :

1) 4d
2) 8d
3) 2d
4) square root of 2d

Homework Equations



?

The Attempt at a Solution



How are charge and mass related? That's what I can't understand.
 

Attachments

  • physics.png
    physics.png
    4.3 KB · Views: 592
Physics news on Phys.org
Charge decides the repulsive force and mass decides the gravitational attractive force. In equilibrium position they are equal amid opposite.
 
Ok. I'm thinking along the lines of

F initial = G m1m2 / r^2

F final = G 1/4m1m2/ r^2

But I need to put those in terms of "r" and compare, right?
 
By changing mass of tthe plastic ball, the weight mg changes. Now compare the two weights with the force of repulsion.
 
Last edited:
Ok. What I just did was

Gm1m2
--------
1/4 Gm1m2

to get a 4:1 ratio. Yet I didn't use anything with "R."
 
When mass decreases, the weight decreases. To balance this the charges must move farther. What is the formula for the force of repulsion?
 
F = kq1q2/r^2 ?
 
Curret. Now if the initial distance for equilibrium is d for mg , what is new distance to balance mg/4?
 
I'm thinking:

d initial = mg
--------------
d final 1/4 mg

solving for d initial I get:

d initial = 4d final

But there's a square involved so shouldn't the answer be "2d?"
 
  • #10
Yes.
 

Similar threads

Replies
3
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
9K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
10
Views
2K