Charged Metallic Balloon ( understanding a concept)

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Homework Help Overview

The discussion revolves around a physics problem involving a charged metallic balloon, specifically focusing on calculating the electric potential at the surface of the balloon using Gauss's law and related concepts.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss's law to determine the electric field and the subsequent calculation of voltage. Questions arise regarding the distance used in the voltage calculation, particularly why it is set to the radius of the balloon.

Discussion Status

Some participants have provided insights into the relationship between electric field and potential, suggesting the need for integration from a reference point. Others express confusion about the assumptions regarding charge movement and the implications for the distance in the voltage formula.

Contextual Notes

There is an ongoing exploration of the definitions and implications of electric potential and field in the context of a conducting sphere, with participants questioning the setup and assumptions made in the problem.

Bailey
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the questioon is this:
A balloon of radius 34.5 cm is sprayed with a metallic coating so that the surface is conducting. A charge of 1.05 × 10-08 C is placed on the surface. What is the potential on the balloon's surface?

how i solved it:
-apply the Gauss law, which gives me the Electrical Field.

-apply the formula : Voltage = Ed ,

where E = electrical field , & d = distance traveled by charges (parrallel to field line)

n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?

its it b/c the positive charge move to the center, which travel a distance of 34.5cm? since if the surface is positively charged, it would repel the " + " charge to the centre & attact the " - " charge to surface...right?
 
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Bailey said:
how i solved it:
-apply the Gauss law, which gives me the Electrical Field.
No problem. You should get the field as a function of distance from the center, for points outside of the balloon: [itex]E = kq/r^2[/itex].
-apply the formula : Voltage = Ed ,
You must integrate from a reference point (usually infinity = 0 potential) to the balloon's surface.

where E = electrical field , & d = distance traveled by charges (parrallel to field line)

n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?
It's not. It turns out that the potential at a point a distance r from the center of the balloon (where r >= the balloon's radius) will equal kq/r.
 
Bailey said:
n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?

its it b/c the positive charge move to the center, which travel a distance of 34.5cm? since if the surface is positively charged, it would repel the " + " charge to the centre & attact the " - " charge to surface...right?
Electrical potential is the work done to bring a unit +charge from infinity to the surface of the sphere, which is:

[tex]\int_{\infty}^{34.5} E\cdot ds = kQ(\frac{1}{R}-\frac{1}{\infty}) = \frac{1}{4\pi \epsilon_0}(\frac{Q}{34.5}-0)[/tex]

AM
 
ah, ic, thanks guy. it seem to clear thing up quite a bit.
 

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