Question

The electric field $E$,current density$J$,and conductivity $\sigma$of a conductor are related as:
A. $\sigma = \dfrac{E}{J}$
B. $\sigma = \dfrac{J}{E}$
C. $\sigma = JE$
D. $\sigma = \dfrac{1}{E}$

Answer 1

Electrical resistivity $\rho$ is the resistance of a conductor of unit cross sectional area$A$ and unit length $l$. Electrical resistivity $\rho$of a material is defined as how strongly it resists the current flow. Contrastingly, the term that defines the conductivity of a material is called electrical conductivity $\sigma$. Therefore, the inverse of electrical resistivity is called electrical conductivity. $\therefore \sigma = \dfrac{1}{\rho }$.
Formula used:
(i)Ohm's law,$V = IR$
(ii)Current density,$J = \dfrac{I}{A}$
(iii)Electric field, $E = \dfrac{V}{l}$
(iv) Electrical resistivity,
$\rho = \dfrac{{RA}}{l}$
$R = \dfrac{{\rho l}}{A}$
(v)Electrical conductivity,$\sigma = \dfrac{1}{\rho }$
Where,
$V$= potential difference
$I$= Current
$R$= Resistance of the wire
$A$= Cross sectional area of the wire
$l$= length of the wire

Complete step-by-step answer:
Let us consider ohm’s law. The states that the electric current is directly proportional to the voltage and inversely proportional to the resistance. That is,
$V = IR$
$\Rightarrow I = \dfrac{V}{R}$
$R$is known as the electrical resistivity. It is the reciprocal of the electrical conductance.
Substituting the value of $R$in the above equation,
$I = \dfrac{V}{R}$
$I = \dfrac{V}{{\dfrac{{\rho l}}{A}}}$
Transforming $A$from Right hand side to left hand side
$\dfrac{I}{A} = \dfrac{V}{{\rho l}}$
Separating the terms for better understanding,
$\left( {\dfrac{I}{A}} \right) = \dfrac{1}{\rho } \times \dfrac{V}{l}$
$J$ is known as the current density. It is defined as the current flowing through the given unit area. It is measured by amperes per meter square. It is given as,
$J = \dfrac{I}{A}$,
Considering the electric field and conductance
$E = \dfrac{V}{l}$
$\sigma = \dfrac{1}{\rho }$.
Therefore, the equation can be changed into,
$J = \sigma E$
We need the relation for electrical conductivity$\sigma$,
$\sigma = \dfrac{1}{E}$
Therefore, the correct option is option B.
Additional information:
The unit of electrical resistivity $\rho$is ohm metre$\left( {\Omega m} \right)$.The unit of electrical conductivity$\sigma$is${\Omega ^{ - 1}}{m^{ - 1}}$
Electrical conductivity can also be called as specific conductance.
Metals are the good conductors of electricity. Non-metals and insulators like glass, plastics are the poor conductors of electricity. Semiconductors like germanium are behaving intermediate between conductors and insulators.
Pure water is the bad conductor of electricity whereas salt water is conductive.
Current density $J$is the amount of electric current $I$passing through the unit cross sectional area $A$.

Note: Electrical conductivity does not have a specific formula for it. It can be derived from Electrical resistivity $\rho$. By knowing the conductivity of a material, we can understand the applications of a material. While approaching this type of question in the problematic type units are must.