Suppose the conductivity of the material separating the cylinders in Ex. 7.2 is not uniform; specifically, $\mathbf{\sigma }\left(s\right)\mathbf{=}\mathbf{k}\mathbf{/}\mathbf{s}$, for some constant . Find the resistance between the cylinders. [Hint: Because a is a function of position, Eq. 7.5 does not hold, the charge density is not zero in the resistive medium, and E does not go like 1/s. But we do know that for steady currents is the same across each cylindrical surface. Take it from there.]

The conductivity of the material,$\sigma \left(s\right)=\frac{k}{s}$

Here k is the constant.

The current is l.

The equation to calculate the surface current density is given as follows.

$\mathbf{J}\left(s\right)\mathbf{=}\frac{\mathbf{l}}{\mathbf{A}}$ …… (1)

Here, A is the area of surface perpendicular to the current.

The equation to calculate the area of the surface perpendicular to the current is given as follows.

$\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{\tau \tau aL}$ (2)

Here, is the radius of the cylinder and is the length of cylinder.

The surface current density also given as follows.

$\mathbf{J}\left(s\right)\mathbf{=}\mathbf{E\sigma }$ …… (3)

Here, E is the electric field intensity.

The equation to calculate the potential difference between the cylinder is given as follows.

$\mathbf{V}\mathbf{=}\mathbf{-}{\mathbf{\int }}_{\mathbf{b}}^{\mathbf{a}}\mathbf{E}\mathbf{.}\mathbf{dl}$ …… (4)

The equation to calculate the resistance between the cylinder is given as follows.

$\mathbf{R}\mathbf{=}\frac{\mathbf{V}}{\mathbf{l}}$ …… (5)

Consider the gaussian cylinder having the radius and length .

Equate the equation (1), equation (2) and (3),

$E\sigma =\frac{l}{A}$

Substitute for A and $\frac{k}{s}$for $\sigma$into above equation.

$$\begin{gathered} E\times \frac{k}{s}=\frac{l}{2\mathrm{\pi sL}} \\ E\times \frac{k}{s}=\frac{l}{2\mathrm{\pi L}} \\ E=\frac{l}{2\mathrm{\pi kL}} \end{gathered}$$

Calculate the potential difference between the cylinder.

Substitute $\frac{l}{2\mathrm{\pi kL}}$for E into equation (4).

$$\begin{gathered} V=-{\int }_b^a\frac{l}{2\mathrm{\pi kL}}.dl \\ V=-\frac{l}{2\mathrm{\pi kL}}{\int }_b^{a}dl \\ V=-\frac{l}{2\mathrm{\pi kL}}\left(a-b\right) \\ V=\frac{l}{2\mathrm{\pi kL}}\left(b-a\right) \end{gathered}$$

Calculate the expression for the resistance of the cylinder.

Substitute$\frac{l}{2\mathrm{\pi kL}}\left(b-a\right)$ for V into equation (5).

$R=\frac{{\displaystyle \frac{l}{2\mathrm{\pi kL}}}\left(b-a\right)}{l}$

role="math" localid="1657699971864" $R=\frac{l}{2\mathrm{\pi kL}}\left(b-a\right)$

Hence the resistance between the cylinder is$\frac{l}{2\mathrm{\pi kL}}\left(b-a\right)$ .