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Like the charged rubber balloon described at the end of section 1.14, a charged soap bubble experiences an outward electrical force on every bit of its surface. given the total charge q on a bubble of radius r, what is the magnitude of the resultant force tending to pull any hemispherical half of the bubble away from the other half? (should this force divided by 2πr exceed the surface tension of the soap film, interesting behavior might be expected!)

Respuesta :

Force on hemispherical half is the force due to surface tension

This force is tangential to the surface and it is always towards the other half

This surface tension force is balance by the force due to excess pressure in the drop

Force due to surface tension property is given by

[tex]F = T*L[/tex]

here L is the length of the circumference

[tex]F = T*2\pi R[/tex]

now for the excess pressure force we can use

[tex]F = \Delta P * \pi R^2[/tex]

now since these forces are counter balanced so

[tex]\Delta P * \pi R^2 = T * 2\pi R[/tex]

[tex]\Delta P = \frac{2T}{R}[/tex]

so this way we can find excess pressure

So force due to surface tension is just product of surface tension and length and given by [tex]T*2\pi R{/tex]

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