jangyasinniTeacher

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To compute the conductivity ((sigma)), we use the relation given by Ohm’s law at the microscopic level, which is:

[

sigma = frac{J}{E}

]

where:

– (sigma) is the conductivity,

– (J) is the current density, and

– (E) is the electric field.

Given that the current density (J = 12) units and the electric field (E = 20) units, we substitute these values into the formula:

[

sigma = frac{12}{20} = 0.6

]

Thus, the conductivity of the material is (0.6) units.

Regarding the nature of the material, conductivity values give us insight into whether a material is a good conductor, semiconductor, or insulator:

– Conductors have high conductivity (e.g., metals).

– Semiconductors have intermediate conductivity, which can vary with doping, temperature, and other factors.

– Insulators have very low conductivity.

Since the conductivity is (0.6) units without a specific reference scale or comparison to known materials’ conductivities, it’s difficult to definitively categorize the material as a conductor, semiconductor, or insulator. However, here are general pointers:

– If this value is compared to known conductors (like copper, for which conductivity is in the order of (10^7) S/m under standard conditions), (0.6) units might suggest it

c

Explanation: The current density is the product of conductivity and electric field intensity.

J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less

than unity, the material is a dielectric.

c

Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.