(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