Сurrent through spherical capacitor

[sergiokapone]

Homework Statement

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 $\Lambda = \frac{4\pi\lambda}{\epsilon} \frac{R_1R_2}{R_1-R_2}$

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 $\frac{1}{\epsilon}$?

 

[mfb]

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

 

[sergiokapone]

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

It must be "specific conductivity of the dielectric".