A rod of semiconducting material of length L = 10 m and cross-sectional area A = 4.5 mm² lies along L. The material obeys Ohm's law, and its resistivity varies along the rod where po = 5 x 10 4N · m. The end of the rod at z = 0 is at a potential the x-axis between z = 0 and æ = according to p = po (1 L² Vo = 30 V greater than the end at a = L. D) What is the electric-field magnitude E, in units of V /m, in the rod at z = L/2? Answer:

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Chapter1: Units, Trigonometry. And Vectors
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(1) THE FOLLOWING QUESTIONS ARE BASED ON THE INFORMATION GIVEN HERE.
A rod of semiconducting material of length L = 10 m and cross-sectional area A = 4.5 mm² lies along
the x-axis between r = 0 and z = L. The material obeys Ohm's law, and its resistivity varies along the rod
where po = 5 x 10 4N · m. The end of the rod at z = 0 is at a potential
according to p = po (1
L²
Vo = 30 V greater than the end at z = L.
D) What is the electric-field magnitude E, in units of V /m, in the rod at z = L/2?
Answer:
Transcribed Image Text:(1) THE FOLLOWING QUESTIONS ARE BASED ON THE INFORMATION GIVEN HERE. A rod of semiconducting material of length L = 10 m and cross-sectional area A = 4.5 mm² lies along the x-axis between r = 0 and z = L. The material obeys Ohm's law, and its resistivity varies along the rod where po = 5 x 10 4N · m. The end of the rod at z = 0 is at a potential according to p = po (1 L² Vo = 30 V greater than the end at z = L. D) What is the electric-field magnitude E, in units of V /m, in the rod at z = L/2? Answer:
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