Charged Metallic Balloon ( understanding a concept)

AI Thread Summary
The discussion revolves around calculating the electric potential on the surface of a metallic balloon with a radius of 34.5 cm and a charge of 1.05 × 10^-8 C. The application of Gauss's law is emphasized to determine the electric field, leading to the formula Voltage = Ed, where d is the distance from the center to the surface. Clarification is provided that the distance d is indeed 34.5 cm, as it represents the radius of the balloon, and the potential is derived from integrating the electric field from infinity to the surface. The concept of electric potential is explained as the work done to bring a unit positive charge from infinity to the balloon's surface.
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 guass 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 guass 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: E = kq/r^2.
-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:

\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)

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