I\'m not sure where to even start with this. Can you lease show the formulas, st

Question

I'm not sure where to even start with this. Can you lease show the formulas, steps and solutions for this problem. WileyPLUS × _CC's Coffee House ea MasterngEngineeringh -1Ger ral Prioples General Pringles gen Your answer is partially correct. Try again. Figure (a) gives the magnitude E(x) of the electric fields that have been set up by a battery along a resistive rod of length L = 9.30 mm (see Figure (b) The vertical scale is set by E = 4.00 x 103 V/m. The rod consists of three sections of the same material but with scale is set by E -4.00 x10 V/m. The rod consists of three sections of the same material but with different not indicate the different radi.)The radus of section 3 is 2.27 mm. What is the sche radius of Co) section l and (n ) ection radius of (a) section 1 and (b) section 2? E, x (mm) (a) Number (b) Number Click if you would like to Show Work for this question: Open Show Work Version 4.16.1.7 sy Policy L 2000-2015 John Wiley & Sons Inc, All Rights Reserved. A Division of 2ohn Wiley& Sens, Inc. est/aglist.uni N10148 0@

Explanation / Answer

V = Ex

from ohms law v = i*R

i*R = E*L

R = resistance R = rho*L/A

rho = resistivity

i*rho*L/A = E*L

i*rho/A = E

i*rho = E*A

here i & rho and same

E*A = constant

A = area of cross section = pi*r^2
(a)

E1*A1 = E3*A3

E1*r1^2 = E3*r3^2

5*4*10^3*r1^2 = 3*4*10^3*2.27^2

r1 = 1.76 mm   <<-answer

(b)

E2*r2^2 = E3*r3^2

E2*r2^2 = E3*r3^2

8*4*10^3*r2^2 = 3*4*10^3*2.27^2

r2 = 1.4 mm   <<-answer

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