Figure 26-23 gives, for three wires of radius R, the current density J(r) versus radius r, as measured from the center of a circular cross section through the wire. The wires are all made from the same material. Rank the wires according to the magnitude of the electric field (a) at the center, (b) halfway to the surface, and (c) at the surface, greatest first.

The figure of the current density J(r) versus radius r for three wires of radius R is given.

The electric field is defined as the potential difference per unit length. From Ohm’s law, the potential difference is directly proportional to the current through the conductor. The constant of proportionality is called the resistance of that conductor. The current density is the current across the unit area at a given point in the conductor. We can use the relation between the electrical field and the current density and resistivity to rank the wires according to the magnitude of the electric field at the center, halfway to the surface, and at the surface of the wire.

Formula:

The electric field produced by a uniform charge density, $E=pJ$ …(i)

E is the electric field, p is charge density and J is the current density

Since all the wires are made from the same material, resistivity p will be same for all the wires. Thus, using equation (i), we can get the following electric field relation as follows:

$E\alpha J$

From the figure at the center, wireand wirehave maximum value of the current density than wire. So electric field at center is,$E_a=E_c>E_b$

Therefore, the rank of the wires according to the magnitude of the electric field at the center, greatest first is$E_a=E_c>E_b$

As all the wires are made from the same material, conductivity p will be same for all wires. Thus, using equation (i), we can get the following electric field relation as follows:

$E\alpha J$

From the figure at halfway to the surface, wirehas maximum value of the current density than wireand wirehas maximum value of the current density than wire. So, the electric field at halfway to the surface is,

$E_a>E_b>E_C$

Therefore, the rank of the wires according to the magnitude of the electric field halfway to the surface, greatest first is$E_a>E_b>E_C$

As all the wires are made from the same material, conductivity p will be the same for all the wires. Thus, using equation (i), we can get the following electric field relation as follows:

$E\alpha J$

From the figure at the surface of wire, wireand wirehave maximum value of the current density than wire. So, electric field at the surface of wire is,$E_a=E_b>E_C$

Therefore, the rank of the wires according to the magnitude of the electric field at the surface, the greatest first is$E_a=E_b>E_C$