Chapter 23, Problem 029 The figure is a section of a conducting rod of radius R1

Question

Chapter 23, Problem 029 The figure is a section of a conducting rod of radius R1 = 1.50 mm and length L = 12.50 m inside a thin-walled coaxial conducting cylindrical shell of radius R2 = 12.5R 1 and the (same) length L. The net charge on the rod is Q1 = +3.55 × 10-12 C; that on the shell is Q2 =-2.1201. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.06R2? what are (c) E and (d) the direction at r = 5.05R? what is the charge on the (e) interior and (f) exterior surface of the shell? (a) Number Units (c) Number Units (e) Number Units (f) Number Units

Explanation / Answer

Given,

R1 = 1.5 mm ; R2 = 12.5 R1 = 18.75 mm ; L = 12.5 m ;

Q1 = 3.55 x 10-12 C ; Q2 = -2.12 Q1 = - 7.526 x 10-12 C

We know from Gauss law that,

E.dA = Q(enc)/ eo

E (2 pi r L) = Q(enc)/ eo

E = 1/ 2 pi eo x Q(enc)/ r L

a) r = 2.06 R2 = 2.06 x 18.75 mm = 0.038625 m

At this point, Q(enc) = Q1 + Q2 = 3.55 pC - 7.53 pC = -3.98 x 10-12 C

E =  1/ 2 pi eo x Q(enc)/ r L

E = [1/ 2 x 3.14 x 8.85 x 10-12 ] x [ (-3.98 x 10-12)/ 0.038625 x 12.5 ] = -0.1483 N/C

Hene, E = 0.148 N/C

b) The direction would be radially inwards since Q(enc) < 0

c) Now, r = 5.05 R1 = 5.05 x 1.5 mm = 7.575 mm = 0.007575 m

Q(enc) = q1 = 3.55 x 10-12 C

E =  1/ 2 pi o x Q(enc)/ r L

E =  [1/ (2 x 3.14 x 8.85 x 10-12 )] x [ 3.55x 10-12/ 0.007575 x 12.5] = -0.675 N/C

Hence, E = 0.675 N/C

d) Direction would be radially outwards since Q(enc) > 0 or positive.

e) Qin = -Q1 = -3.55 x 10-12 C

f) Qex = Q2 - Qin = -7.526 x 10-12 - (- 3.55x 10-12 )= -3.98 x 10-12 C

Q ex = -3.98 x 10-12 C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.