Answer:
(a) 1.3 V/m
(b) [tex]R\approx 1.6*10^{-2}[/tex] Ω
(c) 0.22 V
Explanation:
From Ohms law
[tex]J=\sigma E[/tex] or [tex]J=\frac {E}{\rho}[/tex] or [tex]\frac {I}{A}=\frac {E}{\rho}[/tex] hence [tex]E=\rho \frac {I}{A}[/tex]
At [tex]120^{\circ}[/tex]
[tex]\rho=5.25*10^{-8}+5.25*10^{-8}(0.0045)*(120-20)=7.61*10^{-8}[/tex] Ω⋅m
Area of cylinder is given by
[tex]A=\pi (\frac {d}{2})^{2}=\pi (\frac {1*10^{-3}}{2})^{2}=7.85398*10^{-7}[/tex]
Maximum electric field is given by
[tex]E=\frac {I\rho}{A}=\frac {13.5*(7.6125*10^{-8})}{7.85398*10^{-7}}=1.308492365 V/m[/tex]
[tex]E\approx 1.3 V/m[/tex]
(b)
Resistance of electric field
[tex]R=\frac {\rho_{120}L}{A}=\frac {7.6125*10^{-8}*(16*10^{-2})}{7.85398*10^{-7}}=0.015508058[/tex]
[tex]R\approx 1.6*10^{-2}[/tex] Ω
(c
Maximum potential drop
[tex]\triangle V=IR=13.5*0.016=0.216 V [/tex]
[tex]\triangle V\approx 0.22 V[/tex]
