Estimate Electric Potential of Soap Bubble Drop

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Homework Help Overview

The problem involves estimating the electric potential of a soap bubble drop after it bursts. The soap bubble has a specified radius and wall thickness, and it is initially charged to a certain potential. The transition from a bubble to a drop raises questions about how to calculate the potential in this new form.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the formula V=kq/r for both the bubble and the drop, questioning whether the change in geometry affects the calculations. There are considerations about estimating the density of the soap solution to determine the new radius of the drop. Some participants express uncertainty about the need to calculate charge density and how to approach the integration of the electric field for the potential.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants suggest using the same potential formula for the drop, while others are exploring the implications of the change from a shell to a solid sphere. There is no explicit consensus yet, but participants are engaging with the problem and raising relevant questions.

Contextual Notes

Participants note the importance of estimating the density of the soap/water solution and the implications of the spherical symmetry of the drop in their calculations. There is also a mention of the volume calculation of the soap bubble, which may influence the subsequent steps.

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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?
 
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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.
 
Delphi51 said:
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
 
No, doesn't matter that it is solid. As long as it is spherically symmetrical.
 

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