Metal sphere torn apart in electric field

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Discussion Overview

The discussion revolves around the problem of determining the electric field strength required to break apart a hollow metal sphere placed in a uniform electric field, particularly when the radius of the sphere is doubled while keeping the wall thickness constant. Participants explore the implications of this change on the forces acting on the sphere and the nature of its failure.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the charge density on the sphere is proportional to the electric field strength (E) and the cosine of the angle (theta), leading to equal but opposite charges on each hemisphere, suggesting that the force is proportional to E squared and radius squared.
  • Another participant suggests that the force per unit length on each hemisphere should be considered, indicating a relationship of force to E squared and radius (R).
  • A different viewpoint questions why the tensile force required to break the sphere should remain constant, given that the circumference and cross-sectional area of the sphere are proportional to its radius.
  • One participant expresses an intuition that the electric field required might be four times stronger, based on the phrasing of the question, though they do not provide mathematical backing.
  • A question is raised about whether the failure mechanism is due to tensile failure or buckling, suggesting that this distinction could affect the analysis.
  • Another participant proposes that the failure is likely tensile, as the forces act in opposite directions on the positive and negative parts of the sphere.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the radius of the sphere and the required electric field strength, with no consensus reached on the correct approach or conclusion regarding the failure mechanism.

Contextual Notes

Participants have not resolved the mathematical relationships involved, and there are assumptions regarding the nature of the forces and failure modes that remain unexamined.

krishna mohan
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I read a question from a question paper..which went something like this...

There is a hollow metal sphere, initially uncharged, put in a uniform field. A field strength E0 is required to break the sphere. If the radius of the sphere is doubled, but the thickness of its walls is not changed, then what is the value of the electric field needed to break it up?

In Griffiths 3rd ed, eg 3.8, I saw that a charge density proportional to E and to cos(theta) is built up in the sphere..on integrating this over each hemisphere, I found that one hemisphere is positive and the other negative..but with equal magnitude of charge proportional to E and radius(r) squared. Thus, force will be proportional to E squared and r squared.

Since, in the second case, the thickness of the walls remains the same, I reasoned that the force to break it apart should be the same as before.
Now, force depends on (E x r)^2...
So..if r doubles,,we need to use only E0/2...


But my answer does not match with the given solution..where am I going wrong?
 
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You need to find the force per unit length on each hemisphere.
This is F/(2 pi R)~E^2 R.
 
krishna mohan said:
Since, in the second case, the thickness of the walls remains the same, I reasoned that the force to break it apart should be the same as before.
The circumference of the sphere at the waist, and the cross section (circumference times thickness) of metal are proportional to the radius of the sphere. So why should the tensile force to break it apart remain the same?
 
i can't back it up with math yet, but I am getting the strange feeling that the field must be four times stronger just from the phrasing of the question.
 
Thanks clem and bob..
 
Is the failure do to tensile failure or buckling? It seems to make a difference.
 
Guess it should be tensile failure...the positive part of the sphere is pulled along the electric field and the negative part in the opposite direction...both forces being away from centre of sphere..
 

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