Question

2 SPHERICAL CONDUCTORS Two concentric spherical conductors or radii a, b (with b> a) have an Ohmic material with conductivity ơ placed between them. 2.1 20 POINTS Find the resistance between the shells. 2.2 15 POINTS Now consider two spherical conductors of radius a whose centres are a distance R apart as in the figure with R » a. They are submerged together in the same bath of (liquid) Ohmic material with conductivity o. Find the resistance be the same bath of between the

Answer 1

Let Q be the charge on the inner shell. Then

$\overrightarrow{E}=\frac{1}{4\pi \epsilon_{o}}\frac{Q}{r^2}\hat{r}$

in the space between them, and

$V_{a}-V_{b}=-\int_{b}^{a}\overrightarrow{E}.\overrightarrow{dr}$

or,

$V_{a}-V_{b}=-\int_{b}^{a}\frac{1}{4\pi \epsilon_{o}}\frac{dr}{r^2}$

or,

$V_{a}-V_{b}=\frac{Q}{4\pi \epsilon_{o}}(\frac{1}{a}-\frac{1}{b})$

now

$I=\sigma \frac{Q}{\epsilon_{o}}=\frac{\sigma}{\epsilon_{o}}\frac{4\pi \epsilon_{o}(V_{a}-V_{b})}{\frac{1}{a}-\frac{1}{b}}$

so, the resistance between the shell is given by

$R=\frac{V_{a}-V_{b}}{I}=\frac{1}{4\pi \sigma}(\frac{1}{a}-\frac{1}{b})$

Now if b is very large the

$R=\frac{1}{4\pi \sigma a}$

Essentially all of the resistance is in the region right around the inner sphere. Successive shells contribute less and less because the cross-section area   $4\pi r^2$ gets large and large. For the two submerged spheres

$R=\frac{2}{4\pi \sigma a}=\frac{1}{2\pi \sigma a}$