Question

If the magnitude of electric field in a conductor is $200\ \text{NC}^{-1}$ and the current density is $10^9\ \text{Am}^{-2}$, then the resistivity of the material of the conductor is:

• A. $5\times10^6\ \Omega\text{m}$
• B. $2\times10^{11}\ \Omega\text{m}$
• C. $2\times10^{-7}\ \Omega\text{m}$
• D. 1800 $\Omega\text{m}$

Hint:

Resistivity $\rho$ relates current density $J$ and electric field $E$.
Always check units: NC$^{-1}$ for E, A/m$^2$ for J, result in $\Omega$m.
Use simple formula $\rho = E/J$.

Answer 1

The Correct Option is C

• Ohm's law: $J = \dfrac{E}{\rho} \Rightarrow \rho = \dfrac{E}{J}$.
• Given: $E = 200\ \text{N/C}$, $J = 10^9\ \text{A/m}^2$.
• $\rho = \dfrac{200}{10^9} = 2 \times 10^{-7}\ \Omega\text{m}$.
• Hence, correct answer: $2\times10^{-7\ \Omega\text{m}$}.