91Ó°ÊÓ

Figure which gives the electrical potential versus position X along copper wire is given.

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï„}\mathbf{Ï„}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

The electric field is also defined as the rate of change of potential difference per unit length.

The current density is electric current per unit cross-section area at a given point.

We use the definition of the electrical field in terms of potential to rank the three sections according to the magnitude of the electric field, the greatest first. Then, we use the relation between the electric field and the current density to rank the three sections according to the magnitude of the current density.

Formulae:

The electric field relates to the differential form of potential,$E=-\frac{dV}{dx}$ …(¾±)

Here,E is the electric field,V is the potential difference.

The electric field due to uniform current density,$E=\mathrm{ÒÏ}J$ …(¾±¾±)

Here,E is the electric field,$\mathrm{ÒÏ}$ is the charge density,J and is the current density

From the value of the electric field of equation (i),we can say that the electric field is a slope of the V vs X graph.

From the given graph, we can infer that,

$$\begin{gathered} E_A=1, \\ {E}_B=2, \\ {E}_C=\frac{1}{4} \end{gathered}$$

Therefore, the rank of the three sections according to the magnitude of the electric field, the greatest first is$E_B>E_A>E_C$

From the relation of electric field and density of equation (ii), we can say thatfor the same wire,

$J\mathrm{Î\pm }E$

Therefore, the rank of the three sections according to the magnitude of the current density, the greatest first is$J_B>J_A>J_C$