Estimate Electric Potential of Soap Bubble Drop

[Liquidxlax]

Homework Statement

A soap bubble 10cm in radius with a wall thickness of 3.3x10^-6cm is charged to a 100V potential. The bubble bursts and falls as a spherical drop. Estimate the potential of the drop.

Homework Equations

V=kq/r

U=qV

-dV= E

r=10cm
a=3.3x10^-6

The Attempt at a Solution

solved for the volume of the soap

(4pi(r+a)³)/3 - (4pi(r)³)/3 = 4.1469x10^-9

Now I'm not sure what to do. Do i need to calculate Q from the potential given and then divide the Q with the volume to find charge density and solve for potential?

If so do i need to integrate the electric field thick shell equation to get the potential to solve for q, then use the equation for charged sphere?

 

[Delphi51]

I think you just use V=kq/r again for the collapsed sphere.
You'll have to estimate the density of the soap/water solution to get the radius of the drop.

 

[Liquidxlax]

I think you just use V=kq/r again for the collapsed sphere.
You'll have to estimate the density of the soap/water solution to get the radius of the drop.

it doesn't matter that ones a shell and ones a sphere? I did calculate the volume which i can get a new radius from

 

[Delphi51]

No, doesn't matter that it is solid. As long as it is spherically symmetrical.