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.

## 6 comments

ε is permittivity or emissivity ?? plz explain

Sorry, it's permittivity.

I was very tired when I wrote it...

no probs i got it [:)]

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 :)

the solution is looking appropriate......

will sigma not change with time?

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