38% 7:50 swyp Illi C edugen.wileyplus.com Chapter 23, Problem 029 The figure is

Question

38% 7:50 swyp Illi C edugen.wileyplus.com Chapter 23, Problem 029 The figure is a section of a conducting rod of radius R1 1.30 mm and length L-14.40 m inside a thin-walled coaxial conducting cylindrical shell of radius R2 10. 8R1 and the (same) length L. The net charge on the rod is Q1 = +3.71 × 10-12 C; that on the shell is Q2--224Q1" What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 216R2? what are (c) E and (d) the direction at r = 5.19R? what is the charge on the (e) interior and (f) exterior surface of the shell? (a) Number (e) Number (e) Number ) Number Question Attempts:00f 3 used sa PO·LA -

Explanation / Answer

Given,

R1= 1.30 mm = 1.30 × 10 -3 m

R2= 10.8 R1= 10.8 × 1.30 × 10 -3 m =14.04 × 10 -3 m

L= 14.40 m

Q1= +3.71× 10 -12 C

Q2= -2.2 Q1= -2.2 × (+3.71× 10 -12) C= -8.3104 × 10 -12 C

a) In this case we have to find out the magnitude of E at the point

r = 2.16 R2 =2.16× 14.04 × 10 -3 m = 30.3264 × 10 -3 m

To find out the electric field at the given point we have to construct a cylindrical Gaussian surface of radius r and length L and the electric filed at surface of this cylinder gives required value of electric field. Thus applying Gauss’s law we have

                                          2 rL E= Qenc/o

              In this case Qenc= (-8.3104 + 3.71) × 10 -12 C= -4.6004 × 10 -12C

            Thus,

                                                E= 2Qenc /(4orL)

              Substituting the values of Qenc , r, and L we have

                  E = (9×10 9× 2×4.6004 ×10-12) / (30.3264×10-3× 14.40) N/C (taken only the magnitudes of charges)                       

                           E= 0.1896 N/C

b) The direction of the electric field is normal to the curved surface of the cylinder and in inward direction. Since the total enclosed charge is negative.

c) In this case we have to find out the magnitude of the electric field E at

r= 5.19 R1= 5.19× 1.30 mm = 6.747 mm= 6.747 × 10 -3 m

Thus, we have to construct the cylindrical Gaussian surface of radius r= 6.747 × 10 -3 m      and by applying Gauss’s law we have

                                                  2 rLE= Qenc/o   or E= 2Qenc /(4orL)

              In this case Qenc= + 3.71 × 10 -12 C

             Substituting the values of r, L and Qenc in the above equation, we have

                                                 E= (9×10 9× 2×3.71 ×10-12) / (6.747×10-3× 14.40) N/C

                                                  E= 0.6873 N/C

d) The direction of this electric field is normal to the curved surface of the cylinder and in the outward direction. Since the charge enclosed by the cylinder is positive.

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