Collection of Physics Problems: Solution

Thursday, March 13, 2003

Solution - Leak Charge

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

$E=\frac{1}{4\pi\epsilon}\frac{Q}{R^2}$

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

$J=\sigma E$

$J=\sigma\frac{1}{4\pi\epsilon}\frac{Q}{R^2}$

Hence, the total current flow out from the ball is

$I = \frac{\sigma}{4\pi\epsilon}\frac{Q}{R^2}(4\pi R^2)$

$I = \frac{\sigma}{\epsilon}Q$

The current equals to the decreasing rate of charge.

$I=-\frac{dQ}{dt}$

Hence,

$\frac{dQ}{dt} = -\frac{\sigma}{\epsilon}Q$

$\frac{dQ}{Q} = -\frac{\sigma}{\epsilon}dt$

Integrate both sides of the equation.

$\int_{Q_0}^Q\frac{dQ}{Q} = -\int_0^t\frac{\sigma}{\epsilon}dt$

$\ln\left(\frac{Q}{Q_0}\right)=-\frac{\sigma}{\epsilon}t$

$Q=Q_0e^{-\frac{\sigma}{\epsilon}t}$

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

6 comments:

AnonymousMarch 14, 2009 at 1:24 AM
ε is permittivity or emissivity ?? plz explain
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mfkasimMarch 14, 2009 at 9:59 PM
Sorry, it's permittivity.
I was very tired when I wrote it...
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AnonymousMarch 15, 2009 at 1:17 AM
no probs i got it [:)]
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AnonymousApril 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 :)
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AnonymousDecember 16, 2009 at 1:40 PM
the solution is looking appropriate......
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mksDecember 23, 2010 at 7:55 PM
will sigma not change with time?
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