Question

Electric field $\left(E\right)$ and current density $\left(J\right)$ have relation
(A) $E\propto J^{-1}$
(B) $E\propto J$
(C) $E\propto {\displaystyle \frac{1}{J^2}}$
(D) $E^2\propto {\displaystyle \frac{1}{J}}$

Answer 1

Hint: With the use of Ohm’s law we can establish the relationship between the current density $J$ and electric field $E$ . The current density is the amount of current flowing through a particular charged area or flowing through the cross-sectional area. Once we establish an expression for current density in terms of the electric field, we can find the relation between them.

Formula used:
In this question, we will use the following formulas:
 $V=IR$
 $J={\displaystyle \frac{I}{A}}$
 $J=\sigma E$.

Complete step by step solution:
We know that according to ohm’s law
 $V=IR$
That is, the voltage drop across two conductors is directly proportional to the flow of current where $R$ is the proportionality constant and the resistance of the conductor which opposes the current flow.
Now, the resistance of the conductor depends on the distance between the two plates of the conductor and the area of the plates. Let say, the distance between two plates is $l$ and the area of the plates is $A$ , then
 $\Rightarrow R\propto {\displaystyle \frac{l}{A}}$
 $\Rightarrow R={\displaystyle \frac{\rho l}{A}}$
where $\rho$ is proportionality constant and known as resistivity.
Now, the above value of $R$ can be substituted in the ohm’s law
 $V={\displaystyle \frac{I\rho l}{A}}$
Now here we can again substitute $J$ in place of $I/A$ as $J$ is the current density which gives us the current per area.
 $\Rightarrow V=J\rho l$
Let the electric field across the conductor be $E$ , then the potential difference $V$ can be given as
 $V=El$
We can substitute this value of $V$ in the above equation.
 $\Rightarrow El=J\rho l$
 $l$ from both sides get cancelled and we get
 $\Rightarrow E=J\rho$
Current density hence can also be given as
 $J=\sigma E$
where $\sigma$ is the conductivity and reciprocal of resistivity.
Hence, the current density is directly proportional to the electric field so we can write it as
 $\Rightarrow J\propto E$
Therefore the correct answer is option (B).

Note:
Always remember that the ratio of current density to electric field is conductivity but it's vice versa that is the ratio of the electric field to current density is none other than the resistivity. Conductivity tells us the capacity to conduct whereas resistivity tells us the capacity to resist the flow of charges. Hence conductivity is a reciprocal of resistivity.