# Sol-Ch23 - CHAPTER 23 E xercises 1 Use mg = eE(a 5.58 1 0...

This preview shows pages 1–3. Sign up to view the full content.

Exercises 1. Use mg = eE : (a) 5.58 × 10 –11 N/C; (b) 1.02 × 10 –7 N/C 2. (a) F = eE = 1.92 × 10 – 17 N; (b) a = F/m = 1.15 × 10 10 m/s 2 3. (a) E = F/q 1 = 2500i N/C ; (b) F = q 2 E = –1.6 × 10 –5 i N 4. (a) 4/d 2 = 9/(1 + d) 2 , thus d = 2m, or x = –2 m. (b) 4/x 2 = 1/(1 – x) 2 , thus x = 2/3 m 5. (a) E y = –12kQ/(2) 1/2 L 2 = –7.64 × 10 10 Q/L 2 ; (b) E x = –8kQ/5(5) 1/2 L 2 ; E y = –12kQ[(1 + 1/5(5) 1/2 ]/L 2 E = (–6.44i – 11 j   )   × 10 9 Q/L 2 6. k(Q 1 + Q 2 ) = 10.8, k(Q 1 /9 – Q 2 /1) = –0.8 Find Q 1 = 1 nC; Q 2 = 0.2 nC 7. 2eE = mg, E = 3.06 × 10 6 N/C 8. (a) E 1 = (kq 1 /r 2 )i = 4.22 × 10 3 i N/C (b) E 2 = (kq 2 /r 2 )i = 9.84 × 10 3 i N/C (c) F 2 = q 2 E 1 = –2.95 × 10 –5 i N (d) F 1 = q 1 E 2 = +2.95 × 10 –5 i N 9. (a) E x = –Ecos θ , Ey = Esin θ , where E = 9000 N/C E = –8050i + 4020 j   N/C ; (b) E = 3460 N/C, E = 1920i – 2880 j   N/C 10. E 1 = 2.12 × 10 3 N/C, E 2 = 27 × 10 3 N/C. E x = –E 1 /(17) 1/2 + 2E 2 /(5) 1/2 = 23.6 × 10 3 N/C E y = –4E 1 /(17) 1/2 + E 2 /(5) 1/2 = 10.0 × 10 3 N/C 11. (a) k(2 µC)/(5 cm) 2 = 7.2 × 10 6 N/C E x = –14.4 ×   10 6 + 7.2 × 10 6 cos 60 ° = –10.8 × 10 6 N/C E y = 7.2 × 10 6 sin60 ° = 6.2 × 10 6 N/C (b) F = qE = 32.4i – 18.7 j   N; CHAPTER 23

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(c) Not at all 12. (a) Q 1 = Q 2 ; (b) Q 1 = –4Q 2 ; (c) Q 2 = –9Q 1 13. E = kQr /r 3 , E x = E .i = kQx/r 3 , etc. 14. E = kq/r 2 : (a) 2.25 × 10 2 1 N/C ; (b) 5.13 × 10 11 N/C 15. (x < 0) E = kq[–1/x 2 + 1/(6 – x) 2 ] (0 < x < 6) E = kq[1/x 2 + 1/(6 – x) 2 ] (x > 6) E = kq[1/x 2 + 1/(x – 6) 2 ] 16. (a) (x < 0) E = kq[–2/x 2 + 1/(6 – x) 2 ] (0 < x < 6) E = kq[2/x 2 + 1/(6 – x) 2 ] (x > 6) E =kq[2/x 2 – 1/(x – 6) 2 ] (b) 2/x 2 = 1/(x – 6) 2 , so x = 20.5 m 17 E 1 = 500i N/C; E 2 = 720(0.6i + 0.8 j   )N/C E
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern