Chapte 9 Webaign Help Poblem 4 5 6 7 8 9 0 Poblem 4: The pictue fo thi poblem i a bit mileading. They eally jut give you the pictue fo Pat b. So let fix that. Hee i the pictue fo Pat (a): Pat (a) imply involve finding the electic field at the oigin ceated by the two paticle which I have labeled q and q. The poblem ak fo you to epaate the component of the electic field in the x and y diection. But you can ee that by the way the paticle ae aange, the E field fom q, which I have labeled E, i only along the x axi (in the -x diection) and the E field fom q, which I have labeled E, i only along the y axi (in the -y diection.) Which mean... x component of electic field i E. Ue k q / b to calculate it. y component of electic field i E. Ue k q / a to calculate it. Make ue you ente both of you anwe a negative, becaue that the diection of both component. Fo Pat (b)... Same idea... the foce on q 3 along the x axi i only fom q, and the foce along the y-axi i only fom q. You can calculate the foce uing k Q q /, but it much eaie to jut ue foce on q 3 fo each dimenion = q 3 E whee E i each electic field you found in Pat (a).
Poblem 5: Thi poblem decibe a continuou ditibution of chage, evenly pead ove the length of a od (i.e. in one dimenion.) We can imagine that the chage along thi od i an infinite numbe of infinitely mall chaged paticle, each with chage dq. Then the electic field due to each bit of chage can be witten: de k dq Eentially, to get the total electic field due to all of the chage we will need to add all the bit of de. Thi will, of coue, equie an integal. But we need to fit the piece of infomation togethe to give u an expeion that we can integate. Conide thi pictue: In the poblem you ae given, the length of the od, and D, the ditance fom the cente of the od to the point on the x axi. I have alo labeled ditance in thi pictue, which i fom the end of the od to the point on the x axi. You ae told that the od ha a total chage Q, which i unifomly ditibuted along the length of the od. We can ue thi to wite and equate two imple atio: o: chage fo entie od / length of entie od = chage of tiny bit / length of tiny bit Q dq Q o dq dx dx Alo, fo any bit of chage on the od at a poition on the x-axi of x, the ditance fom that bit of chage to the point whee we want the electic field i = + - x We can ue thee two eult to modify ou expeion fo de above: de Q dx / dx x x k We can now ue an integal to add all thee tiny bit of E field and get the total E field. It i woth noting that ince you ae give the total chage i negative, evey bit of chage i negative, which mean evey bit of E field will point to the left (i.e. towad the chage.)
And the total E field i... x u u du x dx E 0 0 0 O... E Afte all of that, we have thi imple expeion. Note that you ae given D and, but you need and fo thi calculation. So ue D and to get the value of. Then calculate E. Watch you unit, and do not include the ign of Q with you calculation, a you only want to calculate the magnitude of the electic field. Poblem 6: Thi poblem i imila to Poblem 5, in that it decibe a continuou chage ditibution. In thi poblem, howeve, the chage i aanged in a ing intead of along a one-dimenional od. Conide the ing of chage in the diagam hee: You ae told that the ing ha a total chage Q. Uing the ame atio a in Poblem 5, we can wite:
o: chage fo entie ing / length of entie ing = chage of tiny bit / length of tiny bit Q R dq d o dq Q R d We can then wite the electic field fo one bit of chage: de k Q d / R d R Now conide the ed and blue bit of chage at oppoite ide of the ing. The electic field ceated by thee two bit each have a component paallel to the x-axi... and they each have a component pependicula to the x-axi. Due to the ymmety of the ituation, the pependicula component will um to zeo. Which mean... The total electic field will be the um of the component of de that ae paallel to the x-axi. Thi i tue of the electic field ceated by evey bit of chage and it coeponding bit of chage on the othe ide of the ing. We can wite the component of the electic field that i paallel to the x-axi a: de de co x d R co Note that fo a given value of x, i.e. a pecific point on the x-axi, the value of R and ae contant. Which mean eveything in thi expeion can be factoed out of the integal! E x d co co d co R co R R R We can eplace co in tem of x and, and then eplace in tem of x and R: E x ( x / ) x 3 x x 3 x R x R 3/ O: E x x x R 3/
It not the pettiet expeion, but at leat you can now ue it. Note that each of the fou pat to thi poblem imply peent you with a diffeent value of x. Be ue to convet it to mete, and convet you value of R to mete alo. Then caefully calculate E x fo each of the given value of x. Watch you unit and facto of 0. Poblem 7: Fo thi poblem, you hould daw a poton (poitively chaged paticle) and an electic field (I would ecommend dawing the E field pointing to the ight... o it will acceleate the poton to the ight.) abel the initial poition and peed of the poton a x = 0 and v = 0. The poton will acceleate ome ditance, and at a point to the ight omewhee, which you can label x =?, it will have peed v =... (i.e. the given value.) abel a =? and t =? in the inteval. And ecognize that thi i a motion in one dimenion poblem. Wite the two equation fom Chapte : v f = v i + at and x f = x i + v i t + ½ a t You can alo wite F = ma and F = qe. Combine thee two to calculate the acceleation of the poton. Fo a poton, m =.67 x 0-7 kg and q =.60 x 0-9 C Once you have the acceleation, ue the two equation fom Chapte to calculate the anwe to Pat (b) and (c). Then fo Pat (d), ue the definition of kinetic enegy K = ½ m v. Poblem 8: Thi i a imple electic flux... quetion. It not even a poblem. You ae given the diamete of a loop, o you ae eentially given the aea of the loop. You ae told the maximum poible flux though the loop. And you know, by definition: = E A co whee i the angle between E and A. It hould be obviou that fo diffeent angle, co vaie fom - to. So the maximum poible flux will happen when co i equal to (i.e. when the E field pae diectly though the aea.) So fo thi poblem, you have = E A. You ae given and A, and aked to find E. Poblem 9: Thi poblem elie on the pinciple behind Gau aw, that the total flux though a cloed uface i popotional to the amount of chage encloed by the uface. Specifically: total = qinide / 0 whee 0 = 8.85 x 0 - C / N-m Alo fo thi poblem, you ae told that the uface i a cube and the chaged paticle i at the cente of the cube. Thi peent a ymmetical ituation in which all ix face of the cube ae the ame ditance fom and oientation to the chaged paticle.
Calculate the anwe to Pat (b) fit, uing the imple equation above. Watch you unit and facto of 0. Then anwe Pat (a), uing the ymmety decibed in the poblem. That i, each face hould have one-ixth of the total flux pa though it aea. To anwe Pat (c)... if the chage wa not at the cente, Gau aw fo the total flux would till be the ame. That i, the anwe to Pat (b) would not change. But the ymmety of the ix ide would no longe exit; ome face of the cube would have le flux and ome moe (but the total would emain the ame.) Poblem 0: We did thee kind of poblem back in A... foce poblem with omething hanging fom a ting. Daw a fee body diagam. You hould have mg down qe to the ight tenion up and left Split the tenion into two component uing the given angle. Ue the x and y axe a hown in the given pictue. Wite the foce equation fo each of the two dimenion. Divide the equation to eliminate T. You hould then be able to olve fo q in tem of m g E and. Poblem : Thi poblem i much like Poblem. You ae peented with fou chaged paticle and aked to find the total foce on one paticle due to the othe thee. The tep ae exactly the ame a thoe fo Poblem. (The only light diffeence between the poblem i that Poblem gave you jut thee paticle intead of fou.) If we label the fou paticle with q q q 3 q 4, we can then define foce F F and F 4 acting on paticle 3: Now you can: Calculate the magnitude of F F and F 4 Split F into x and y component; you will need an angle fom the oiginal pictue. Calculate the total foce in the x-diection, and the total foce in the y-diection. Ceate a total foce tiangle, with x-diection y-diection and hypotenue. The hypotenue of the total foce tiangle i the magnitude of the total foce.