91Ó°ÊÓ

Consider the expression for the ohm’s law.

$\$V=IR\$$

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

Consider the ratio of to for a particular conductor is called its resistance given as follow:

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

Here, $\mathrm{ÒÏ}$is resistivity , L is length in m and A is area in m².

The equation of current density is as follow:

$J=\frac{I}{4\mathrm{Ï€}{r}^2}$

(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 is from to :

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

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 .

The area of spherical shell is $4\mathrm{Ï€}{r}^2$and the resistance is .

Solve for the potential difference as follows:

$V_{ab}=IR$

Substitute the expression for the resistance and solve as:

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

So, the current density of spherical shell:

$J=\frac{I}{4\mathrm{Ï€}{r}^2}$

Substitute the expression for the current and solve:

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

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

(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:

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

Here, cross-sectional are is A.

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