2 spheres connected by wire, find tension

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Homework Help Overview

The problem involves two conducting spheres connected by a wire, with a charge applied to one sphere. Participants are tasked with finding the tension in the wire, considering the distribution of charge and the forces acting on the spheres.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply concepts from Gauss's Law and considers the forces acting on the spheres to determine the tension in the wire. Some participants question the understanding of how tension is generated in a wire and discuss the implications of unequal forces on each end.

Discussion Status

Participants are actively clarifying the concept of tension and its relationship to the forces acting on the spheres. There is a recognition that the tension in the wire is equal to the force exerted by the spheres, and some guidance has been provided regarding the nature of tension in a massless rope.

Contextual Notes

Participants are exploring the implications of different forces acting on the spheres and how that affects the tension in the wire. The discussion includes assumptions about the distribution of charge and the setup of the problem.

vladimir69
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Homework Statement


2 conducting spheres of radius r=0.5cm are connected by a long thin piece of conducting wire of length L=2metres. A charge of 60 micro Coulombs is put onto one of the spheres, find the tension T in the wire. Assume the charge is evenly distributed along the surface of the two spheres.


Homework Equations


[tex]F=k_{e}\frac{Qq}{r^2}[/tex]
[tex]Q=30\mu C[/tex]
Q is 30 micro Coulombs because I assumed that the charge would distribute evenly onto the surface of both spheres.

The Attempt at a Solution


This question was in the Gauss Law chapter of my book however I didn't use Gauss Law for this one.
I said
[tex]F_{12}=k_{e}\frac{Q^2}{L^2}[/tex]
[tex]F_{21}=-F_{12}[/tex]
So according to me each sphere is "pulling" the rope in opposite directions to produce a combined force of
[tex]F=2F_{12}[/tex]
The tension in the wire must balance this force otherwise it will break
So
[tex]F_{wire} = F=4N[/tex]
However the answer gives 2 Newtons which I can't convince myself is right because each sphere is contributing its own force on the string.
 
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You are misunderstanding how tension is created. To create a tension F in a rope, for example, both ends must be pulled with the force F. The tension in the rope--or wire--is the force with which it pulls. The rope pulls each sphere with a force F, thus the tension is F, not 2F.
 
OK thanks for that. I suppose one way of looking at it would be to consider the forces on one of the spheres, there is only the electric force and the tension, for the sphere to remain where it is these forces would have to be equal.

How about the case when the force is different on each end? What would be the tension then?

I always thought of tension as a mass suspended by a string with the string pulling up with the same force as the mass is pulling down.
 
vladimir69 said:
I suppose one way of looking at it would be to consider the forces on one of the spheres, there is only the electric force and the tension, for the sphere to remain where it is these forces would have to be equal.
Sounds good.

How about the case when the force is different on each end? What would be the tension then?
If the external forces on the spheres were different, you'd have to calculate the tension based on the net force and acceleration. But there would be a single tension throughout the rope, assuming it's massless.

I always thought of tension as a mass suspended by a string with the string pulling up with the same force as the mass is pulling down.
Good, that works. Realize that whatever's connected to the other end of the rope--the ceiling, perhaps--is also pulling up with the same force.
 
Thanks for clearing that up, the bit at the end was what I was missing - the ceiling is pulling up with the same force.
 

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