CALC The region between two concentric conducting spheres with radii and is filled with a conducting material with resistivity $\mathit{\rho }$. (a) Show that the resistance between the spheres is given by

$\mathit{R}\mathbf{=}\frac{\mathbf{\rho }}{\mathbf{4}\mathbf{\pi }}\mathbf{(}\frac{\mathbf{1}}{\mathbf{a}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{b}}\mathbf{)}$

(b) Derive an expression for the current density as a function of radius, in terms of the potential difference${\mathit{V}}_{\mathbf{a}\mathbf{b}}$ between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation $\mathbf{L}\mathbf{=}\mathbf{b}\mathbf{-}\mathbf{a}$between the spheres is small.

Consider the expression for the ohm’s law.

$V=IR$

Here,$I$ is current in ampere$\left(A\right)$ ,$R$ is resistance in ohms $\left(\mathrm{\Omega }\right)$and $V$is the potential difference volt $\left(V\right)$.

Consider the ratio of role="math" localid="1655790599013" $\mathit{V}$to$\mathit{I}$ for a particular conductor is called its resistance $\mathit{R}$given as follow:

$$\begin{gathered} R=\frac{V}{I} \\ =\frac{\rho L}{A} \end{gathered}$$

Here, $\rho$is resistivity $\mathrm{\Omega }·\mathrm{m},L$, is length in m and $A$is area in ${\mathrm{m}}^2$.

The equation of current density is as follow:

$J=\frac{1}{4\pi r^2}$

(a)

Consider very thin shell of radius $r$and thickness $dr$.

Consider the resistance of the thin shell as follows:

$dR=\frac{\rho dr}{4\pi r^2}$

Integrate both sides when $r$is from $a$to $b$:

$$\begin{gathered} \int dR=\frac{\rho }{4\pi }{\int }_a^b\frac{dr}{r^2} \\ R=\frac{\rho }{4\pi }{\left[\frac{-1}{r}\right]}_a^b \\ R=\frac{\rho }{4\pi }\left[-\frac{1}{b}-\left(-\frac{1}{a}\right)\right] \\ R=\frac{\rho }{4\pi }\left[\frac{1}{a}-\frac{1}{b}\right] \end{gathered}$$

Hence, the resistance between the spheres is $R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$.

(b)

Consider the potential difference between the two sphere is $V_{ab}$and current flowing through spherical shell is $I$.

The area of spherical shell is $4\pi r^2$and the resistance is $R$.

Solve for the potential difference as follows:

$V_{ab}=IR$

Substitute the expression for the resistance and solve as:

$$\begin{gathered} V_{ab}=I\frac{\rho }{4\mathrm{\pi }}\left(\frac{1}{a}-\frac{1}{b}\right) \\ I=\frac{\mathit{4}\pi V_{\mathit{a}\mathit{b}}}{\rho \left({\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}\right)} \end{gathered}$$

So, the current density of spherical shell:

$J=\frac{I}{4\pi r^2}$

Substitute the expression for the current and solve:

$$\begin{gathered} J\mathit{=}\frac{\mathit{4}\mathit{\pi }{\mathit{V}}_{\mathit{a}\mathit{b}}}{\mathit{4}\mathit{\pi }{\mathit{r}}^{\mathit{2}}\mathit{\rho }\left[{\displaystyle \frac{\mathrm{1}}{\mathrm{a}}}\mathrm{-}{\displaystyle \frac{\mathrm{1}}{\mathrm{b}}}\right]} \\ \mathit{}\mathit{}\mathit{=}\frac{{\mathit{V}}_{\mathit{a}\mathit{b}}}{{\mathit{r}}^{\mathit{2}}\mathit{\rho }\left[{\displaystyle \frac{\mathrm{1}}{\mathrm{a}}}\mathrm{-}{\displaystyle \frac{\mathrm{1}}{\mathrm{b}}}\right]} \end{gathered}$$

Hence, the expression for the current density as a function of radius in terms of the potential difference $V_{ab}$between to spheres is $J=\frac{V_{ab}}{\rho \left({\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}\right)}\frac{1}{r^2}$.

(c)

Substitute$b-a$ for $L$in the equation of resistance.

$$\begin{gathered} R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right) \\ =\frac{\rho }{4\pi }\left(\frac{b-a}{ab}\right) \\ =\frac{\rho L}{4\pi \left(ab\right)} \end{gathered}$$

Consider the separation between the sphere is very small such that $a=b=r$. Then, the equation for the resistance is as follows:

$$\begin{gathered} R=\frac{\rho L}{4{\mathrm{\pi r}}^2} \\ =\frac{\rho L}{A}\left\{A=4\pi r^2\right\} \end{gathered}$$

Here, cross-sectional are is $A$.

Hence, for the separation$L=b-a$ between the spheres is small then the equation$R=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$ is reduced to the form $R=\frac{\rho L}{A}$.