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EECE261_PS3_solution

# EECE261_PS3_solution - E€CE 2.1'PRoBl—EM 515T 2...

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Unformatted text preview: E€CE 2.1,! 'PRoBl—EM 515T 2 SDLUTI'DNS " Determine the work done in carrying a Z—MC charge from (2,1,—1) to (8,2,—1) in the ﬁeld E : ya, + \$3,, / "B ‘8 .2 2 1 For the straight line a: 2 6y — 4: Here, y = m/6 + 2/3, and the work is .8 '2 W- :—2 1‘6 E 3 d X 0 [/1 (6+3 dm+.1 (6y—4)dy 2—28/LJ M0 Otlanat (A rtSu/f ‘ef (114(95er #«MS ACC‘KK E if Consermvfwe. 2. In spherical coordinates, E = 2r/(r2 + a2)2 a, V / m. Find the potential at any point, using the reference a) V : U at inﬁnity: We write in general 27‘ dr V(r)=— .———+C= C , (r2 + a2)2 r2 + a2 + With a zero reference at 7‘ —> 00, C : 0 and therefore V(r) : 1/(7‘2 + a2). b) V : 0 at r = 0: Using the general expression, we ﬁnd 1 1 Therefore 2 1 1 —r V“) ' r2 + a2 — 35 — az(r2 + a2) 3. A certain current density is given in cylindrical coordinates as J = 1006—2z (pap + az) A/ 1112. Find the total current passing through each of these surfaces: a) z :7 0, 0 S p S 1, in the az direction: 27r 1 Ia = / J.dS =/ / 1006_2(0)(pap+a2)-azpdpd¢= 100% S O 0 where ap - az = 0. b) 2 = 1, 0 S p S 1, in the az direction: 27r l 1b = / J - dS 2/ / 1006—2(1)(pap +az) -azpdpdq§ = loom—2 s 0 0 c) closed cylinder deﬁned by 0 S z S 1, 0 S p S l, in an outward direction: >1 ‘27r IT 2 Ib—Ia+/ / 1006-22( (l)a,,+az)~ap (1) dgbdz = 10071'(e_2—-1)+1007r(1—e_2) : Q . 0 . 0 The region between plates is ﬁlled with material having conductivity 0(53) 2 age“\$/“ where 00 is a constant. Voltage V5 is applied to the plate at z = d; the plate at z = 0 is at zero a) the electric ﬁeld. intensity E within the material: We know that E will be z—directed but the conduct1v1ty varies With 1‘. We therefore expect no 2 variation in E7 and alsd note that the line integral of E between the bottom andt l t ‘ ‘ ' Therefore E : _V0/daz V/m- op p a es must always give VI). b) the total current ﬂowing between plates: We have _008~\$/a% J = a(m)E = d az Using this, we ﬁnd b a _ —x/aV I : J' : 006 0 owa _ 0.63 b / dS _/0/0 d az'(—az)d\$dy: d 0(1_e ”2%01/014 c) the resistance of the material: We use I 0.63ab00 , A dipole for which p = 1060 az C - m is located at the origin. What is the equation of the surface on which Ez : 0 but E 7E 0? First we ﬁnd the 2 component: 5. 10 . 5 2 — ‘_ 2 E2 = E - a2 = 47rr3 [200st] (aT ~az) +s1n(9(a9 -az)] = W [200s 0 sin 6] This will be zero when [2 cos2 6 — sin2 0] = 0. Using identities, we write 1 2 cos2 6 — sin2 0 : E [1 + 3 cos(29)] ‘ v _. r O ‘ _ x r : o The above becomes zero on the cone surfaces, 6 — o4.7 and 6 — 120.5 . ...
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