# Metal sphere torn apart in electric field!

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?

Meir Achuz
Homework Helper
Gold Member
You need to find the force per unit length on each hemisphere.
This is F/(2 pi R)~E^2 R.

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 im 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..