# Phys2-T1 - Pl-IY 2426 Engineering Physics 2 10 f l...

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Unformatted text preview: Pl-IY 2426: Engineering Physics 2 10'; f l 1 '3‘“: Exam i-A: ll February 2002 Name: LDC D‘si SSN: 52 Li 5’ [Last 4 Digits} —___—.. ____— _— _— ——-_.—, l __ 1.] Twc ting.»r ccnelucting balls cf identical mass, in, and identical charge, q, hang vertically frctn ncncnndncting massiess strings cf identical length L. The},I are in equilibrium, 2 F=ﬂ. See ﬁgure belcw. M, replacing taut-ii E B for small angles, Ute value of the fcllcwing ratic 2.) There are three (3) charges located on a straight line at points I, 2, and 3 reapectively. See ﬁgure below. The charges are separated by d = iﬂﬂ mm, and magnitude of |Q| = I p.113. What is the magnitude and direction of the net Egg {F3} on charge {+Q) located at Point 3 due to the other charges? Shoe.r direction to Leg or ﬂ'gﬁr on Point 3 of the picture. Hint, draw a ﬂea body diagram of forces on charge at Point 3 ONLY. Hint, this is not equilibrium, F3 31E 0! Use SI units. {1: rmmm: U.im 'Q +4Q +Q {it 1 Ii d —0——.—O—~ x P11 PtE P13/ g; 3' ‘6 —’5 4am! / F35] : “GEE—IL: 1{ 3-43 3H0 FE! Pt 15 JTL lie-3&1! irirfﬂxgﬁqu _ e QQLHI'E N / :- GMQWL #1 .4. r “3 a1 “Mathew 1{ m j, H J 11M“); 1 ______________ H gqa 3.) A cube of edge, d = 10 nun, has a charge, :1 = 2t] m; at the geometric center. - Caution, use S] units. : as s 10— EC. d 3a] What is the net electric ﬂux. (1), through the entire surface of the cube? (genders: JD “trail 1'- le / E : : I1 ——~ Z 12 Q 1 ID —-— 5e 2 s 5 all It:— .51 1. C’- 3.b] What is the net electric ﬂux. 111, through the. bottom face only for the same cube‘? r__ L/DW glgexlﬂg g 1 t at 2 F: f: be: "I LE 6 6 a lo} What would be the net electric ﬂux, ED, through the entire surface if the dimension of all the cubc‘s edges were now suddenly doubled, d = 2t] nun? Th? Its! electric fit” Lula“ i234" The (Leer-r1111 _ Pb ‘II- {ﬁtmh Hm ﬁght? Slum-film’— :1; Tm: L6H? CBW - / 4.) A hollow metal cylinder {pipe} of radius Ra, has a net positive charge unifonnly distributed on it with a charge density,r of o. It has a very,r long length L :33:- R0. _+ Ru- f L se- RG [1 mf-—_____———‘ '1. M f? 7/3 4e) Derive an expression for the value of the electric [CREE region 1 {HRH}. .J —=- (:1ch -"'] @315 51h: I E 5 0 % E33- 0 um Ta FM ml: 4.1:!) Derive an expression for the value of the electric field, E”, in region II (9RD). QngE=-———Q%U 613:4 q:6A I 6:. 6:: F 4.c} Give the expression for the value of the electric ﬁeldc E, on the surface of the metal pipe at r = R“. Hint, concept question. 5a} What is the magnitude of the electric field Ep(x=c} at point x = c, if an insulating thin red of length L has a net tetal charge, +q, tmifennly spread ever its length L? Hint, use linear charge density, 1,:qu and it=cqudx. to change variable tc dq=ldx. 5.1:} Next, chew the directicn cf Er. at pcint x = c by placing an arrcw en the pcint x = c in the correct directicn (Le. Painting to the Leg er right). 3’ lt=x+¢.} vb x / / 30 Cl): ﬁ—H ————-— = x dq a Epfc) 1——————r§u———_+ L c E L .= I _ E 2' l4- — 1, cit: :1”- T 7 cl Lift) , 519 d -Aa'k A- if": I} 61 Let U :1 +6.2. L —? {if} = l. a. ﬂuid-H: E F El Aer in J c GHQ") .5 x20 _7. u:c with — l 1 E I = '— -’ L” =UL , ,_ HA E : ‘— l (I _ l >1¢2A :hléI—i- (Cl+L—-:) E : F r" +l— Tr a “‘ MC l M . . . +L. ic) Next, give an expression fer HP at point c=xbieL. Hint, cencept intestacy Q (c 3 _ EFL BIC-F‘ t. q / “2'4 5 LHT éc @z W H ...
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Phys2-T1 - Pl-IY 2426 Engineering Physics 2 10 f l...

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