Question

Figure 3.6 shows a copper conductor of resistivity ρ = 1.72 x 10 -8 Ωm having a current density J = 2.54 x 106 Am-2. Calculate the electric field in the copper.
What is the potential difference between the two points a and b, 100m apart?

Answer 1

$J=\sigma E$ where $\sigma$ is the conductivity of the material $E$ is the electric field and $J$ is the current density

$\sigma=1/\rho$ where $\rho$ is the resistivity of the material

$2.54 \times 10^6=\dfrac{1}{1.72\times10^{-8}}\times E \newline E=2.54\times1.72\times10^{-2}V/m$ $=4.37\times10^{-2}V/m$

In the next figure has not been provided so I am unable to answer it

but if the electric field is constant you can apply the formula

$E=-\dfrac{ \Delta V}{\Delta r}$ and if the electric field is variable apply $dE=\dfrac{dV}{dr}$