ragsdale (zdr82) HW2 ditmire (58335) 1

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1 rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering points A chrge of 8. µc is t the geometric center of cube. Wht is the electric flux through one of the fces? Correct nswer: N m 2 /C. Let : q 8. µc C. By Guss lw, Φ E d A q The totl flux through the cube is given by Φ tot q C C 2 /N m N m 2 /C. 5. A correct (b ) A Bsic Concepts: Φ Surfce E d A For uniform electric field nd flt surfce, this simplifies to Φ E A. Solution: The electric flux through surfce is given by Φ E A, where E is the electric field nd A is the vector which is directed perpendiculr to the surfce nd hs mgnitude equl to the re of the surfce. When the surfce is in the yz plne, it is perpendiculr to the x-direction. Therefore, the resulting flux is due to the x-component of the electric field only. This cn lso be written s Φ E A (î bĵ) Aî A. The flux through one side of the cube is then y y y E E E Φ 6 Φ tot N m 2 /C. z x x z z fig fig 2 fig 3 x 002 (prt of 3) 0.0 points A uniform electric field î b ĵ, intersects surfce of re A. Wht is the flux through this re if the surfce lies in the yz plne?. b A 2. A 2 b 2 3. ( b) A. ( b) A 003 (prt 2 of 3) 0.0 points Wht is the flux if the surfce lies in the xz plne?. b A correct A. ( b) A 5. ( b) A

2 rgsdle (zdr82) HW2 ditmire (58335) 2 6. A 2 b 2 7. (b ) A When the surfce is in the xz plne, it is perpendiculr to the y-direction. Therefore Φ (î bĵ) Aĵ b A. 00 (prt 3 of 3) 0.0 points Wht is the flux if the surfce lies in the xy plne?. (b ) A 2. 0 correct 3. b A. ( b) A 5. A 6. ( b) A 7. A 2 b 2 When the surfce is in the xy plne, it is perpendiculr to the z-direction. There is no component of the electric field in the z- direction, so there is no component of the electric field perpendiculr to the surfce. This mens tht the electric flux through the surfce is zero. Mthemticlly, this is written Φ (î bĵ) Aˆk (prt of ) 0.0 points A solid sphere of rdius 59 cm hs totl positive chrge of 2. µc uniformly distributed throughout its volume. Clculte the mgnitude of the electric field t the center of the sphere. Correct nswer: 0 N/C. Let : r 59 cm By Guss lw, nd 2. µc C. E d A q The enclosed chrge is zero, so 0 N/C. 006 (prt 2 of ) 0.0 points Clculte the mgnitude of the electric field.75 cm from the center of the sphere. Correct nswer: N/C. Let : r.75 cm. E r 2 3 r3 ρ r ρ, 3 where the chrge density is ρ 3 r C (59 cm) C/m 3 nd R is the rdius of the sphere. Thus, r ρ 3 (0.75 m) ( C/m 3 ) 3 ( C 2 /N m 2 ) N/C. 007 (prt 3 of ) 0.0 points Clculte the mgnitude of the electric field

3 rgsdle (zdr82) HW2 ditmire (58335) 3 59 cm from the center of the sphere. Correct nswer: N/C. E Let : r 59 cm. The field outside the sphere is the sme s tht for point chrge. Thus r 2 () C (0.59 m) C 2 /N m N/C. 008 (prt of ) 0.0 points Clculte the mgnitude of the electric field 69.7 cm from the center of the sphere. Correct nswer: N/C. Using Eq., Let : r 69.7 cm. r C (0.697 m) C 2 /N m N/C points A cubic box of side, oriented s shown, contins n unknown chrge. The verticlly directed electric field hs uniform mgnitude E t the top surfce nd 2 E t the bottom surfce. 2 E How much chrge is inside the box?. encl E 2 correct 2. encl 3 E 2 3. encl 2 E 2. encl 0 5. encl 2 E 2 6. insufficient informtion Electric flux through surfce S is, by convention, positive for electric field lines going out of the surfce S nd negtive for lines going in. Here the surfce is cube nd no flux goes through the verticl sides. The top receives Φ top E 2 (inwrd is negtive) nd the bottom Φ bottom 2 E 2. The totl electric flux is Φ E 2 2 E 2 E 2. Using Guss s Lw, the chrge inside the box is encl Φ E points A uniformly chrged conducting plte with re A hs totl chrge which is positive. The figure below shows cross-sectionl view of the plne nd the electric field lines due to the chrge on the plne. The figure is not drwn to scle.

4 rgsdle (zdr82) HW2 ditmire (58335) E P Find the mgnitude of the field t point P, which is distnce from the plte. Assume tht is very smll when compred to the dimensions of the plte, such tht edge effects cn be ignored.. E 2 A E Due to the symmetry of the problem, there is n electric flux only through the right nd left surfces nd these two re equl. If the cross section of the surfce is S, then Guss Lw sttes tht Φ TOTAL 2 E S A S, so 2 A. 0 (prt of 3) 0.0 points Consider solid insulting sphere of rdius b with nonuniform chrge density ρ r, where is constnt. 2. E A 3. E 2. E b O r dr 5. E 6. E 2 A correct 7. E A 8. E 2 9. E 2 0. E A Bsic Concepts Guss Lw, electrosttic properties of conductors. Solution: Let us consider the Gussin surfce shown in the figure. E E S Find the chrge r contined within the rdius r, when r < b s in the figure. Note: The volume element dv for sphericl shell of rdius r nd thickness dr is equl to r 2 dr.). r r2 2. r r3 3. r r correct. r r 5. r r 6. r r3 7. r r 3 8. r r 2 9. r r 2 0. r 0

5 rgsdle (zdr82) HW2 ditmire (58335) 5 Bsic Concepts: d ρ dv ; Guss lw. Solution: A chrge element is given by dq ρ dv ( r)( r 2 dr) r 3 dr For r < b, we integrte to find the totl chrge within rdius r: dq r 0 r r. r 3 dr 02 (prt 2 of 3) 0.0 points If C/m nd b m, find E t r 0.7 m. Correct nswer: N/C. Pick sphere of rdius r, concentric with the first, s the Gussin surfce (since the problem exhibits sphericl symmetry). Over this surfce, the flux is constnt due to symmetry, so Guss Lw gives Since encl r r 2 encl ( C/m ) (0.7 m) C, we cn substitute this into Guss Lw to find encl r C C 2 /N m 2 (0.7 m) N/C t the rdius r 0.7 m, directed outwrd. 03 (prt 3 of 3) 0.0 points Find the chrge b contined within the rdius r, when r > b.. b b 2. b b3 3. b b. b b 3 5. b b3 6. b b 2 7. b b2 8. b b correct 9. b b2 0. b 0 For r > b, dq b 0 b b r 3 dr Obviously, the chrge enclosed within rdius r does not depend on r when we go outside the sphere; it is just the chrge of the entire sphere. Therefore, we could hve used the result r from the Prt nd just inserted r b insted of crrying out the integrtion gin.

6 While the chrge does not chnge ny more s we go outside the sphere, the electric field flls off s due to Guss Lw. r2 0 (prt of 5) 0.0 points Consider cylindricl distribution [drk gry] extending from r 0 to r 8.2 cm, of chrge ρ r 0, where 0 27 cm m 3 /µc. This cylindricl chrge distribution [drk gry] is surrounded by dielectric shell [light gry] whose inner rdius is 7. cm nd outer rdius is 2.6 cm s shown in the figure. 2.6 cm 7. cm 8.2 cm Wht is the electric field t.3 cm? Assume the length L is very long compred to the dimeter of the dielectric shell. Neglect edge effects. Correct nswer: V/m. Guss lw where rgsdle (zdr82) HW2 ditmire (58335) 6 L Let : r.3 cm. E d enc r 0 dv d ( r 2 L ) 2 r L dr. In our cse, Guss lw gives 2 r L 0 r2 3 0 r 0 ρ(r) dv r 2 r L dr, C 2 / N m 2 (.3 cm) 2 3 (27 cm m 3 /µc) m 00 cm C 0 6 µc V/m. 05 (prt 2 of 5) 0.0 points Wht is the electric field t 2. cm? Correct nswer: V/m. Let : r 2. cm. Now the integrtion in the RHS of Guss lw goes up to the boundry of our chrge distribution 2 r R r 2 r dr, 0 0 giving R3 3 0 r C 2 / N m 2 (8.2 cm) 3 3 (27 cm m 3 /µc)(2. cm) m 00 cm C 0 6 µc V/m. 06 (prt 3 of 5) 0.0 points Wht is the electric field t 20.2 cm? Correct nswer: 63.6 V/m. Let : r 20.2 cm κ nd This cse is equivlent to the previous one except tht we re inside the dielectric which

7 rgsdle (zdr82) HW2 ditmire (58335) 7 will decrese the electric field by fctor of κ : R κ r C 2 / N m 2 (8.2 cm) 3 3 (27 cm m 3 /µc)(6.03) (20.2 cm) m 00 cm C 0 6 µc 63.6 V/m. 07 (prt of 5) 0.0 points Wht is the electric field t 29.2 cm? Correct nswer: V/m. Let : r 29.2 cm. This cse is completely equivlent to tht of Prt 2, becuse lthough there is induced surfce chrge on the shell, the net chrge on it is zero, nd we cn disregrd the shell when using Guss lw: R3 3 0 r C 2 / N m 2 (8.2 cm) 3 3 (27 cm m 3 /µc) (29.2 cm) m 00 cm C 0 6 µc V/m. 08 (prt 5 of 5) 0.0 points Wht is the surfce chrge density on the inner surfce of the dielectric? Correct nswer: µc/m 2. The electric field jumps from E(r in ) just outside the dielectric to E(r in) inside the κ dielectric t its boundry. This shows us tht the surfce chrge density induced t the boundry is ( σ E(r in ) ), κ where E(r in ) Therefore R r in. (κ ) σ R3 3 κ 0 r in (6.03) (8.2 cm) 3 (27 cm m 3 /µc)(7. cm) m 00 cm µc/m (prt of 3) 0.0 points A solid conducting sphere of rdius is plced inside of conducting shell which hs n inner rdius b nd n outer rdius c. There is chrge q on the sphere nd chrge q 2 on the shell. S r A O q q 2 Find the electric field t point A, where the distnce from the center O to A is d, such tht < d < b.. E A k q 2 d 2 2. E A k (q q 2 ) d 2 3. E A k q 2 b c

8 rgsdle (zdr82) HW2 ditmire (58335) 8. E A q k d 2 5. E A 0 6. E A k q d 2 correct 7. E A k (q q 2 ) d 8. E A k q d 9. E A q 2 k d 2 0. E A k q Tking Gussin sphere of rdius d from the center nd pplying Guss Lw, E A d 2 enclosed q, E A q d 2 k q d (prt 2 of 3) 0.0 points Find the totl flux emnting through Gussin surfce S which hs rdius r (see the figure).. Φ S q 2 q 2. Φ S q q 2 r 3. Φ S q q 2. Φ S q 5. Φ S q 2 6. Φ S q q 2 correct 7. Φ S (q q 2 ) 8. Φ S q q 2 r 9. Φ S q 0. Φ S q 2 Bsic Concepts: Guss Lw, electrosttic properties of conductors. Solution: According to Guss Lw, the flux on the surfce S is equl to the enclosed chrge over. In this cse, the chrge enclosed is q q 2, so the flux is Φ S q q (prt 3 of 3) 0.0 points Find the surfce chrge density on the outer surfce of the shell.. σ outer q 2 c 2 2. σ outer q 2 q c 2 q 2 3. σ outer c 2 q 2. σ outer b 2 5. σ outer q q 2 c 2 correct 6. σ outer q 2 b 2 7. σ outer q 2 q 8. σ outer 2 9. σ outer q q 2 c 2 0. σ outer q 2 q c 2 Using the fct tht inside the conductor the electric field is zero, we cn see tht the chrge on the inner surfce of the shell q inner is equl to q. Since the totl chrge on the shell is q 2, then q outer q 2 q inner q 2 q.

9 rgsdle (zdr82) HW2 ditmire (58335) 9 The chrge density is the chrge per unit re, so on the outer surfce of the shell, it is given by σ outer q outer c 2 q q 2 c points An infinite line chrge λ is locted long the z-xis. A prticle of mss m tht crries chrge q whose sign is opposite to tht of λ is in circulr orbit in the xy plne bout the line chrge. Obtin n expression for the period T of the orbit in terms of m, q, R, nd λ, where R is the rdius of the orbit. (2 ). T 3 m R 2 2 m R 2. T 2 (2 ) 3. T 3 m R (2 ). T 3 m R 2 (2 ) 5. T 3 m R 2 correct 6. T (2 ) 3 m R 2 m R 7. T 2 8. T 2 m R 2 m R 9. T m R 0. T x z λ r m,q Becuse of symmetry, the electric field produced by the line chrge is perpendiculr to the line. We cn choose closed cylindericl surfce nd pply Guss Lw to obtin the electric field: E n da net S E 2 R L λ L λ 2 R. Applying uniform circulr motion, keywords: ( ) 2 2 q E m R T ( ) R m R T (2 ) T 3 m R 2 2 mǫ o 2 R. y

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere