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- Thread starter ermia
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In summary: The whole thing is easy but why force per unit length is equal in both cases while the thicknesses are equal?The force per unit length is the same because the area is the same.

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hutchphd

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Knowing this please show us your solution to the problem

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ermia

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Why?hutchphd said:The strength of any glue joint is proportional to the contact area of the glue

And why this happens to electric field

Why? t is not a differential to say $$A=2\pi r t$$hutchphd said:So if the hemisphere shells of thickness t and radius r are glued together the rending force required will be proportional to Frend α rt

b.t.w: I don't have doubt about whole question. Just the part I asked.

I thought the answer of my question was sth obvious cause all solutions in the internet or the book refused explaining it but I couldn't understand why this happend?

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I would have thought that obvious. If a joint area A can take force F, then a second joint area A in parallel can take F. Between them they can withstand a force 2F.ermia said:Why?

It has nothing to do with what the source of the force is.ermia said:And why this happens to electric field

t is the a thickness, which we are told is small, so the area of the annulus is approximately ##2\pi r t##.ermia said:Why? t is not a differential to say $$A=2\pi r t$$

b.t.w: I don't have doubt about whole question. Just the part I asked.

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ermia

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Why the pressures for two different spheres at the area middle of themselves should be equal to each other??haruspex said:I would have thought that obvious. If a joint area A can take force F, then a second joint area A in parallel can take F. Between them they can withstand a force 2F.

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The LaTeX you are looking for is \propto:hutchphd said:will be proportional to $$ F_{rend}~ \alpha~rt $$

$$

F_{\rm rend} \propto rt

$$

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Since you have not posted the solution I struggle to understand what you are asking.ermia said:Why the pressures for two different spheres at the area middle of themselves should be equal to each other??

Please post the solution, indicating exactly which step is the problem.

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ermia

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I uploaded it here.haruspex said:Since you have not posted the solution I struggle to understand what you are asking.

Please post the solution, indicating exactly which step is the problem.

It has nothing special for me except the part I asked.

but why?

Last edited:

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vela

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It seems obvious why it wouldn't change. Why do you think it would change?ermia said:It has nothing special for me except the part I asked.

Since the thickness of the sphere walls remains unchanged, the force tearing the sphere per unit length must remain unchange

but why?

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Lnewqban

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Because that is the maximum stress that the material of that sphere allows before failing.ermia said:I uploaded it here.

It has nothing special for me except the part I asked.

Since the thickness of the sphere walls remains unchanged, the force tearing the sphere per unit length must remain unchange

but why?

More perimetral area requires stronger pulling force to separate the sphere.

Please, see:

https://en.wikipedia.org/wiki/Stress_(mechanics)#Simple_stress

https://en.wikipedia.org/wiki/Ultimate_tensile_strength

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TSny

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As others have stated, it will take twice as much force to break apart twice as much joined area. That is, the forceermia said:Why the pressures for two different spheres at the area middle of themselves should be equal to each other??

So, P = (cA)/A = c =

The value of the constant

In your original post, you asked

ermia said:The whole thing is easy but why force per unit length is equal in both cases while the thicknesses are equal?

For a patch length ##l## and thickness ##t## as shown, the area of the patch is ##A = lt##.

So, the force required to disjoin this area is ##F = cA = clt##. Thus, the force per unit length is ##F/l = clt/l = ct##. Since ##c## and ##t## are the same for the two spheres, the force per unit length required to split the sphere is the same for the two spheres.

The electric field needed to tear a conducting sphere is known as the critical electric field. It is the minimum electric field required to break the bonds holding the sphere together and cause it to tear apart.

The critical electric field can be calculated using the formula E = 2σ/r, where E is the electric field, σ is the surface charge density of the sphere, and r is the radius of the sphere.

The critical electric field is affected by the surface charge density of the sphere, the radius of the sphere, and the material properties of the sphere such as its conductivity and strength.

Yes, it is possible for the electric field to exceed the critical value. In this case, the sphere will experience electrical breakdown and may even disintegrate.

Understanding the critical electric field is important in various industries, such as electronics and materials science. It can also be used in the design and testing of high-voltage equipment and for predicting the behavior of materials under extreme electric fields.

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