(5:56:46 PM) PbG 5000: ok 2 b and c are kQ/(A(A-L)) (5:57:22 PM) PbG 5000: 6 is Q / (Eo * 6) (5:57:42 PM) PbG 5000: 7 is finite to zero (5:58:16 PM) PbG 5000: 9 is 0, k*q1/r^2, k/r^2*(q1+q2), -, -1 (5:59:22 PM) PbG 5000: 10 a is p * (4/3) *(pi) r^3 (6:00:13 PM) PbG 5000: bc are kqr/R^3 (6:00:52 PM) PbG 5000: de are just gausses law (using a sphere of radius Rgiven) (6:01:25 PM) PbG 5000: for 11 2*pi*C*r (6:01:48 PM) PbG 5000: then b2 and b4 are C*r/(Eo) (6:01:52 PM) PbG 5000: b3 = 0 (6:02:01 PM) PbG 5000: b1 = C/(Eo) (6:02:37 PM) PbG 5000: 12 a3 and a4 are -Q and Q respectively (6:03:08 PM) PbG 5000: then a1 is -Q/4pi(Rinner)^2 (6:03:28 PM) PbG 5000: a2 is Q/(4pi(Router)^2) (6:05:37 PM) PbG 5000: b2 = 0 (6:06:13 PM) PbG 5000: b1 = b3 = Qenc/ (Eo*4*pi) (6:06:35 PM) PbG 5000: c3 = a3 (6:06:50 PM) PbG 5000: c4 = a4 + the new Q (6:07:02 PM) Doug: ty a lot (6:07:04 PM) Doug: quick question (6:07:14 PM) Doug: for 2 what is A? (6:07:36 PM) PbG 5000: c1 and c2 do the same emthod as for a1 and a2 with the new c3 and c4 being used as the -Q and Q (6:07:51 PM) PbG 5000: part a of question 2 (6:08:04 PM) PbG 5000: lamda * delta X (6:08:24 PM) PbG 5000: oh sorry forgot to mention (6:08:32 PM) PbG 5000: A is the x given in the part (6:08:33 PM) Doug: nvm figured it out (6:08:49 PM) Doug: which Rs are which for 9? (6:08:51 PM) PbG 5000: and X is distance between A and the second x (6:09:06 PM) PbG 5000: for 9? (6:09:10 PM) PbG 5000: i gave you the answers in order (6:09:11 PM) Doug: yea (6:09:19 PM) Doug: k*q1/r^2 (6:09:22 PM) PbG 5000: 9 is 0, k*q1/r^2, k/r^2*(q1+q2), -, -1 (6:09:27 PM) Doug: no r1 or r2? (6:09:32 PM) PbG 5000: nope (6:09:38 PM) PbG 5000: lower case r (6:09:42 PM) PbG 5000: not upper case r (6:09:48 PM) Doug: k (6:09:59 PM) PbG 5000: ok (6:10:19 PM) PbG 5000: 12 c5 = b1 (6:10:23 PM) PbG 5000: c6 = 0 (6:10:46 PM) PbG 5000: and c7 use the same method you did to solve for b1 but use c4 as your charge (6:10:58 PM) PbG 5000: then 13 (6:11:31 PM) PbG 5000: a = (delta X) * lamda (6:12:28 PM) PbG 5000: b c d e = 2*k*(lamda)*sin(theta) / y (6:12:39 PM) PbG 5000: and you'll ahve to use some geometry to find sintheta (6:12:44 PM) PbG 5000: (it is the angle (6:13:07 PM) PbG 5000: for the diagram of the angle (6:13:14 PM) PbG 5000: Jan 31 E fields.pdf (6:13:17 PM) PbG 5000: on eber's site (6:13:25 PM) PbG 5000: page 7 (6:13:28 PM) PbG 5000: and thats it (6:13:31 PM) Doug: wow major ty (6:13:35 PM) Doug: for 11 (6:13:39 PM) Doug: which r do i use for a (6:13:41 PM) PbG 5000: damn i just closed it (6:13:43 PM) PbG 5000: lemme check (6:13:45 PM) Doug: lol k (6:13:46 PM) Doug: sorry (6:13:57 PM) Doug: dont forget to forget about english (6:14:08 PM) PbG 5000: oh yeah (6:14:10 PM) PbG 5000: 11 (6:14:14 PM) PbG 5000: its 1.5 cm (6:14:23 PM) Doug: there we go (6:14:25 PM) PbG 5000: (3cm diameter / 2) (6:14:31 PM) PbG 5000: done? (6:14:47 PM) PbG 5000: if you have any questions ask xiao he's done too i'm going downstairs for dinner (6:14:49 PM) PbG 5000: ttyl (6:15:17 PM) Doug: ttyl