Q26P - Fundamentals Of Physics

Question

In Figure-a, a $\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{V}$battery is connected to a resistive strip that consists of three sections with the same cross-sectional areas but different conductivities. Figure-b gives the electric potential V(x) versus position x along the strip. The horizontal scale is set by ${\mathbf{x}}_{\mathbf{s}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{mm}$. Section 3 has conductivity $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{ }{\left(\mathrm{\Omega m}\right)}^{\mathbf{-}\mathbf{1}}$. (a) What is the conductivity of section 1 and (b) What is the conductivity of section 2?

Step-by-Step Solution

Answer

a)   The conductivity of section $1 \mathrm{is} 6.00\times {10}^7{\left(\mathrm{\Omega }.\mathrm{m}\right)}^{-1}$

b)   The conductivity of section 2 is $7.50\times {10}^6{\left(\mathrm{\Omega }.\mathrm{m}\right)}^{-1}$

Step 1: Identification of given data

a)  The graph of the electric potential versus position.

b)  Conductivity of section 3, ${\mathrm{\sigma }}_3=3.00\times {10}^7{\left(\mathrm{\Omega }.\mathrm{m}\right)}^{-1}$

Step 2: Significance of conductivity

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, $J=i=A$              …(i)

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, $J=\sigma E$                                       …(ii)

Step 3: (a) Determining the conductivity of section 1

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,

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

For section 2, 

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

For section 3, 

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

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

$J_1=J_2=?J_3$                                                                                                       …(iii)

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

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

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

Step 4: (b) Determining the conductivity of section 2

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

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

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