Answer to Question #104237 in Electricity and Magnetism for mon aldossari

Question #104237
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?
1
Expert's answer
2020-03-04T10:02:10-0500

J=σEJ=\sigma E where σ\sigma is the conductivity of the material EE is the electric field and JJ is the current density

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

2.54×106=11.72×108×EE=2.54×1.72×102V/m2.54 \times 10^6=\dfrac{1}{1.72\times10^{-8}}\times E \newline E=2.54\times1.72\times10^{-2}V/m =4.37×102V/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=ΔVΔrE=-\dfrac{ \Delta V}{\Delta r} and if the electric field is variable apply dE=dVdrdE=\dfrac{dV}{dr}


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