91Ó°ÊÓ

a) The graph of the electric potential versus position.

b) Conductivity of section 3,${\mathrm{σ}}_3=3.00×{10}^7{\left(\mathrm{Î\copyright }.\mathrm{m}\right)}^{-1}$

The current density is the rate of flow of current per unit cross-section area. It is also defined as the rate of flow of charges per unit time per unit area. We can find the conductivities of sections by using the equations of current density in terms of current and conductivity.

Formulae:

The current density of a material passing through the area,$J=i=A$ …(¾±)

Here,J is current density, i is current and A is area of cross-section of the conductor.

The current density in terms of conductivity, $J=\mathrm{σ}E$ …(¾±¾±)

Since, the absolute values of the slopes are equal to the respective electric field magnitudes; thus, from the given graph we can find the electric fields of each section are given as:

For section 1,

$$\begin{gathered} {\mathrm{E}}_1=\frac{\left(9-7\right)\mathrm{V}}{\left(4-0\right)\mathrm{mm}} \\ =\frac{2\mathrm{V}}{4\mathrm{mm}} \\ =0.5\frac{\mathrm{V}}{\mathrm{mm}} \\ =0.5×{10}^3\frac{\mathrm{V}}{\mathrm{m}} \end{gathered}$$

For section 2,

$$\begin{gathered} {\mathrm{E}}_2=\frac{\left(7-3\right)\mathrm{V}}{\left(5-4\right)\mathrm{mm}} \\ =\frac{4\mathrm{V}}{1\mathrm{mm}} \\ =\frac{4\mathrm{V}}{\mathrm{mm}} \\ =4×{10}^3\frac{\mathrm{V}}{\mathrm{m}} \end{gathered}$$

For section 3,

$$\begin{gathered} E_3=\frac{\left(3-0\right)\mathrm{V}}{\left(8-5\right)\mathrm{mm}} \\ =\frac{3\mathrm{V}}{3\mathrm{mm}} \\ =1\frac{\mathrm{V}}{\mathrm{mm}} \\ =1×{10}^3\frac{\mathrm{V}}{\mathrm{m}} \end{gathered}$$

As values ofcurrent and area are same, current densities must also be the same considering equation (i), i.e.

$J_1=J_2=?J_3$ …(¾±¾±¾±)

Thus, the value of the conductivity of section 1 can be given using equation (iii) and equation (ii) as follows:

$$\begin{gathered} {\mathrm{σ}}_1\left(0.50×{10}^3\frac{\mathrm{V}}{\mathrm{m}}\right)={\mathrm{σ}}_3\left(1.0×{10}^3\frac{\mathrm{V}}{\mathrm{m}}\right) \\ \mathrm{σ}=\frac{3.00×{10}^7{\left(\mathrm{Î\copyright }.\mathrm{m}\right)}^1\left(1.0×{10}^3\mathrm{V}/\mathrm{m}\right)}{\left(0.50×{10}^3\mathrm{V}/\mathrm{m}\right)} \\ =6.00×{10}^7{\left(\mathrm{Î\copyright }.\mathrm{m}\right)}^{-1} \end{gathered}$$

Therefore, the conductivity in section 1 is $6.00×{10}^7{\left(\mathrm{Î\copyright }.\mathrm{m}\right)}^{-1}$.

Similarly, the value of the conductivity of section 2 can be given using equation (iii) and equation (ii) as follows:

$$\begin{gathered} {\mathrm{σ}}_2\left(4.0×{10}^3\frac{\mathrm{V}}{\mathrm{m}}\right)={\mathrm{σ}}_3\left(1.0×{10}^3\frac{\mathrm{V}}{\mathrm{m}}\right) \\ \mathrm{σ}=\frac{3.00×{10}^7{\left(\mathrm{Î\copyright }.\mathrm{m}\right)}^{-1}\left(1.0×{10}^3\mathrm{V}/\mathrm{m}\right)}{\left(4.0×{10}^3\mathrm{V}/\mathrm{m}\right)} \\ =7.50×{10}^6\left(\mathrm{Î\copyright }.\mathrm{m}\right) \end{gathered}$$

Therefore, the conductivity in section 2 is $7.50×{10}^6\left(\mathrm{Î\copyright }.\mathrm{m}\right)$.