| 6 comments ]

The ball is a conductor, so all of the charge is on its surface. The electrical field on its surface is



Because it's placed inside a material with conductivity σ, the current density on its surface is




Hence, the total current flow out from the ball is





The current equals to the decreasing rate of charge.



Hence,




Integrate both sides of the equation.






That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

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6 comments

Anonymous said... @ March 14, 2009 at 1:24 AM

ε is permittivity or emissivity ?? plz explain

Copycat91 said... @ March 14, 2009 at 9:59 PM

Sorry, it's permittivity.
I was very tired when I wrote it...

Anonymous said... @ March 15, 2009 at 1:17 AM

no probs i got it [:)]

Anonymous said... @ April 28, 2009 at 11:59 PM

There's a little problem with this:
"The ball is a conductor, so all of the charge is on its surface"
So if a current goes through a wire (or a conductor in general), the charges must be on the surface??? This only happens with AC current with high frequency. In my opinion, the reason is not only that the ball is a conductor.

There's also another problem of the statement "The electrical field on its surface is E=...". This formula is valid with a condition: the charge density is equal in every part of the surface. The reason is about symmetry.

Actually I think we have a better way to get the formula of E - use the vector:
_ Say {E} is electrical field strength vector on the surface, and {J} is the current density vector, we have: {J} = sigma x {E}
_ The current going out of the surface is: I = integral of {J}x{dS}, where {dS} is the "surface area" vector.
Hence: I = integral of (sigma x {E}{dS}) = sigma x (integral of {E}{dS})
_ Notice that integral of {E}{dS} = electrostatic flux through the surface = Q/epsilon.
We now have the same result, but with a more general method :)

Anonymous said... @ December 16, 2009 at 1:40 PM

the solution is looking appropriate......

mks said... @ December 23, 2010 at 7:55 PM

will sigma not change with time?

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