91Ó°ÊÓ

Consider the expression for the ohm’s law.

$V=IR$

Here,$I$ is current in ampere$\left(A\right)$ ,$R$ is resistance in ohms $\left(\mathrm{Î\copyright }\right)$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:

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

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

The equation of current density is as follow:

$J=\frac{1}{4\mathrm{Ï€}{r}^2}$

(a)

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

Consider the resistance of the thin shell as follows:

$dR=\frac{\mathrm{ÒÏ}dr}{4\mathrm{Ï€}{r}^2}$

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

$$\begin{gathered} ∫dR=\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}{∫}_a^b\frac{dr}{r^2} \\ R=\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}{\left[\frac{-1}{r}\right]}_a^b \\ R=\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}\left[-\frac{1}{b}-\left(-\frac{1}{a}\right)\right] \\ R=\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}\left[\frac{1}{a}-\frac{1}{b}\right] \end{gathered}$$

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

(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 $4\mathrm{Ï€}{r}^2$and the resistance is $R$.

Solve for the potential difference as follows:

$V_{ab}=IR$

Substitute the expression for the resistance and solve as:

$$\begin{gathered} V_{ab}=I\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}\left(\frac{1}{a}-\frac{1}{b}\right) \\ I=\frac{\mathit{4}\mathrm{Ï€}{V}_{\mathit{a}\mathit{b}}}{\mathrm{ÒÏ}\left({\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}\right)} \end{gathered}$$

So, the current density of spherical shell:

$J=\frac{I}{4\mathrm{Ï€}{r}^2}$

Substitute the expression for the current and solve:

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

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

(c)

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

$$\begin{gathered} R=\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}\left(\frac{1}{a}-\frac{1}{b}\right) \\ =\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}}\left(\frac{b-a}{ab}\right) \\ =\frac{\mathrm{ÒÏ}L}{4\mathrm{Ï€}\left(ab\right)} \end{gathered}$$

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

$$\begin{gathered} R=\frac{\mathrm{ÒÏ}L}{4{\mathrm{Ï€°ù}}^2} \\ =\frac{\mathrm{ÒÏ}L}{A}\left\{A=4\mathrm{Ï€}{r}^2\right\} \end{gathered}$$

Here, cross-sectional are is $A$.

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