Exam 1 Solutions. Prof. Darin Acosta Prof. Selman Hershfield February 6, 2007

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1 PHY049 Spring 008 Prf. Darin Acta Prf. Selman Herhfiel Februar 6, 007 Nte: Mt prblem have mre than ne verin with ifferent anwer. Be careful that u check ur eam againt ur verin f the prblem. 1. Tw charge, q 1 = -1 C an q = -4 C (r vice vera), are place alng the -ai a itance L apart with charge q 1 at the rigin an q at =L (ee figure). A thir charge, q 3 = +4/9 C, i al place alng the -ai uch that there i n net Culmb frce n an f the charge. What i pitin f thi charge alng the ai in unit f L, i.e. what i /L? (1) 1/3 () /3 (3) 1/ (4) 4/3 (5) -/3 The nl place t place a thir charge that i pitive i between the tw negative charge. If there i n net Culmb frce, we mut atif fr charge 1 the fllwing: q1q3 q1q F k k 0 L q3 4 / 9 1 L q L 4 3 If q 1 an q are wappe, then the anwer wul be /3. 1

2 PHY049 Spring 008. Three charge frm an equilateral triangle f ie length = 0cm a hwn in the figure. If q A = -1 nc, q B =+nc, an q C =+1nC what i the hrizntal cmpnent (r cmpnent) f the net electrtatic frce n particle A? (1) N () N (3) N (4) N (5) 0 N A B Wrk ut the an cmpnent f the frce n particle A: qaqb qaqc F K in 30 K in N q q q q F K c30 K N A B A C c30 3. Tw electrn each with ma m e = kg are pace 1mm apart. What i the magnitue f the acceleratin fr ne f the electrn? (1) m/ () m/ (3) m/ (4) m/ (5) 50 m/ Set the culmb frce equal t m*a an lve fr acceleratin: F e 8 a k.5 10 m/ 31 m (0.001 m) kg C

3 PHY049 Spring Cnier electric charge Q = +3nC itribute unifrml alng a ne-imeninal path in the hape f a emicircle f raiu R=0.1m. Fin the cmpnent f the electric fiel alng the -ai at the rigin (0,0). (1) N/C () -700 N/C (3) 0 (4) N/C (5) +700 N/C R φ +Q B mmetr, the cmpnent f the fiel cancel at the rigin an we nl have the cmpnent, which pint in the negative irectin ince the fiel pint awa frm pitive charge. The linear charge enit i λ = +Q/ πr fr the emicircle. S cmputing the cmpnent f the fiel: q R E E k in k in 0 R R E k k kq c 1700 N/C R R R 0 3

4 PHY049 Spring What i the cmpnent (r the cmpnent) f the electric fiel at the rigin (center) f the heagnal arra f charge particle. The ie length =0cm an q= C. (1) 1130 N/C () 50 N/C (3) 3380 N/C (4) N/C (5) 0 +4q +q q E q Nte that the fiel cancel fr all pping pair f particle ecept fr +q in the tp left crner. The net fiel pint wn an t the right. The itance frm that charge t the rigin i. S the cmpnent f the fiel are: 9 q E k c N/C 0. E +q +q 9 q k in N/C 0. 4

5 6. A cube f ie m ha ne crner at the rigin a hwn in the figure. If E = (1+ )î+(+ )ĵ+(3+3z )ˆk V/m when,, an z are meaure in meter, what i the flu thrugh the tp face? Becaue the electric fiel i cntant n each face f the cube, the flu i Φ = ˆn E A: Φ tp Φ bttm Φ left Φ right = ˆk E(z = ) A = 60 V m = ˆk E(z = 0) A = 1 V m = ĵ E( = 0) A = 8 V m = ĵ E( = ) A = 40 V m 7. A cnucting phere i inie a cnucting hell a hwn in the figure. The net charge n the phere i -3µC, an the net charge n the hell i 5µC. If a = 1 m, b = m, an c =.5 m, what i the magnitue an irectin f the electric fiel at r = 1.5 m? Appl Gau Law, Φ = ˆn EA = Q enc /ǫ, t a phere f raiu r = 1.5m, which i in between the phere an the utie hell (A = 4πr ). The charge encle i thu the charge n the phere. Fr 3µC n the phere, ˆn E = 1000 V/m, an fr µc n the phere, ˆn E = 8000 V/m. The fact that thee are negative implie that the fiel i irecte inwar. 8. The electric fiel in the irectin i pltte in the figure. If V (0) = V an E = 6 V/m, what i the vltage at = m? The vltage i relate t the area uner the electric fiel graph. V () = V (0) 0 E. The lpe f the graph i E /3, an cnequentl the electric fiel at = m i E /3. The integral i the area uner the graph, which i negative. V () = V (0) + 1 E 3 Fr E = 6V/m, V () = 6V, an fr E = 3V/m, V () = 4V.

6 9. An infinite plane with charge enit σ = C/m lie in the = 0 plane (ee figure). A particle with charge C i mve frm (,, z) = (1, 1, 0) t (,, z) = (,, 0), where itance are meaure in meter. What i the change in ptential energ f the particle, U f U i? The electric fiel n the right i σ/ǫ t the right. If u nt remember thi, it can eail be erive frm Gau Law. The frce n the particle i F = qσ/ǫ t the right. The change in the ptential energ i U f U i = F = F (f i ) = 0.68J. 10. Tw prtn are initiall at ret an eparate b a itance f 1 cm. The are bth releae an mve awa frm each ther. What i the pee f ne f the prtn when the are infinitel far awa? Thi i a cnervatin f energ prblem, K i + U i = K f + U f, which implie that 0 + e 4πǫ = 1 m pv f + 1 m pv f, v f = e. 4πǫ m p Fr = 1 cm, v f = 3.7 m/, an fr = 1 mm, v f = 11.7 m/. Becaue thee anwer are nt eactl n the anwer heet, fr = 1 cm we are cunting bth 5. m/ an 3.3 m/ a crrect, an fr = 1 mm we are cunting bth 16.6 m/ an 1.4m/ a crrect. Thi al accunt fr tuent wh thught ne f the prtn wa fie.