0
$\begingroup$

The first configuration

Given a spherical shell of a conducting material with radius $a$, surrounded by a poorly conducting material between it and another conducting shell with radius $b$, you can calculate the resistance between the two shells as

$$ R = \frac{1}{4\pi\sigma}\left(\frac{1}{a}-\frac{1}{b}\right) $$ Plot of resistance

Where $\sigma$ is the conductivity of the poorly conducting material. As $b$ gets larger the value of $R$ converges to $\frac{1}{4\pi\sigma a}$, as shown above. Now suppose that we immerse two of these shells with radius $a$ in the ocean, making $b$ very large, with a potential difference of $V_0$, my textbook claims that the resistance between the two shells is twice $\frac{1}{4\pi\sigma a}$.

Immersed in water

I fail to see how this could be case. Doesn't the other shell in the water act like the outer layer in the first case? What is different in the two configurations?

$\endgroup$
  • $\begingroup$ What is that purple line? $\endgroup$ – DanielSank Aug 28 '15 at 8:12
  • $\begingroup$ It is a plot of R(b) $\endgroup$ – V.Vocor Aug 28 '15 at 8:30
1
$\begingroup$

As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$.

We can think of the situation in which there are two shells in the infinite sea of poorly conducting material as a circuit connected in series: from one shell to infinity, from infinity to other shell. Since the resistance between one shell and the infinity is $1/4 \pi \sigma a$, the resistance between the two shells will be twice that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.