Homework Help: Сurrent through spherical capacitor

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1. Nov 8, 2015

sergiokapone

1. The problem statement, all variables and given/known data
Determine the conductivity of the insulator in a spherical
capacitor filled with weakly conductive dielectric. Specific conductivity of the dielectric is λ, the dielectric permittivity ε.

Ansver in book is $\Lambda = \frac{4\pi\lambda}{\epsilon} \frac{R_1R_2}{R_1-R_2}$

3. The attempt at a solution
My solution is to use following law's

$div \vec j = 0$ (1)

but

$\vec j = \lambda \vec E$(2)

then

$div \vec j = \lambda div \vec E =0$ (3)

in spherical coordinates $div \vec E =0$ leads to
$Er^2 = const$

then current $I=j4\pi r^2 = 4\pi \lambda \frac{const}{r^2} r^2 = 4\pi \lambda \cdot const$.
Now find the voltage:
$V = \int\limits_{R_1}^{R_2} E dr = \int\limits_{R_1}^{R_2} \frac{const}{r^2} dr = const \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Then, from the Ohm's law
$\Lambda = \frac{I}{V} =4\pi\lambda \frac{R_1R_2}{R_1-R_2}$

My answer is differ from book, where have I missed $\frac{1}{\epsilon}$?

Last edited: Nov 8, 2015
2. Nov 8, 2015

Staff: Mentor

Apparently "specific dielectric conductivity" (never heard of that concept) differs from the specific conductivity by this factor of ε.

3. Nov 8, 2015

sergiokapone

It must be "specific conductivity of the dielectric".