Tht reminds me must downlod the test prep HW. dpted from http://www.neringzero.net (nz118.jpg)
Em 1: Tuesdy, Feb 14, 5:00-6:00 PM Test rooms: Instructor Sections Room Dr. Hle F, H 104 Physics Dr. Kurter B, N 15 BCH Dr. Mdison K, M 199 Toomey Dr. Prris J, L St Pt s Bllroom* Mr. Upshw A, C, E, G G-3 Schrenk Dr. Wddill D 10 BCH Specil Accommodtions Testing Center *em 1 only If t 5:00 on test dy you re lost, go to 104 Physics nd check the em room schedule, then go to the pproprite room nd tke the em there.
Em Reminders 5 multiple choice questions, 4 worked problems bring clcultor (ny clcultor tht does not communicte with the outside world is OK) no eternl communictions, ny use of cell phone, tblet, smrtwtch etc. will be considered cheting no hedphones be on time, you will not be dmitted fter 5:15pm
Em Reminders grde spredsheets will be posted the dy fter the em you will need your PIN to find your grde (PINs were emiled by the recittion instructors) test preprtion homework 1 is posted on course website, will be discussed in recittion tomorrow problems on the test preprtion home work re NOT gurnteed to cover ll topics on the em!!! LEAD review session Mondy from 7 to 9 pm in BCH 10
Em 1 topics Electric chrge nd electric force, Coulomb s Lw Electric field (clculting electric fields, motion of chrged prticle in n electric field, dipoles) Guss Lw (electric flu, clculting electric fields vi Gussin surfces, fields nd surfce chrges of conductors) Electric potentil nd potentil energy (clculting work, potentil energy nd potentil, clculting fields from potentils, equipotentils, potentils of conductors) Cpcitors (clculting cpcitnce, equivlent cpcitnce of cpcitor network, chrges nd voltges in cpcitor network)
Em 1 topics don t forget the Physics 1135 concepts look t old tests (014 to 016 tests re on course website) em problems my come from topics not covered in test preprtion homework or test review lecture
Three chrges +Q, +Q, nd Q, re locted t the corners of n equilterl tringle with sides of length. Wht is the force on the chrge locted t point P (see digrm)? y F 1 ˆ F F F i F F ˆj 1 1y y +Q P =60 +Q F -Q F F cos F cos ˆi 1 F sin F sin ˆj 1 Q Q Q F k k 1 Q Q Q F k k Note: if there is not problem like this on Em 1, there will be one on the Finl!
Three chrges +Q, +Q, nd Q, re locted t the corners of n equilterl tringle with sides of length. Wht is the force on the chrge locted t point P (see digrm)? y P +Q F 1 F F cos 60 cos 60 i kq kq ˆ kq kq sin 60 sin 60 ˆj +Q -Q I could hve stted tht F y =0 nd F =F 1 by symmetry, but I decided to do the full clcultion here.
Three chrges +Q, +Q, nd Q, re locted t the corners of n equilterl tringle with sides of length. Wht is the force on the chrge locted t point P (see digrm)? y P +Q F 1 F F kq F cos 60 ˆi kq F ˆ i +Q -Q
Wht is the electric field t P due to the two chrges t the bse of the tringle? y P You cn repet the bove clcultion, replcing F by E (nd using Coulomb s Lw). Or you cn be smrt F qe E kq î F kq q Q ˆi +Q -Q This is the chrge which hd been t point P, feeling the force F. Cution: never write q E F. Why?
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? y dq de = k r
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? y dq de = k = k r dq de dq chrge dq = ds = ds length dq = Q length of rc ds Q 4 Q dq = ds = ds 8
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? y d ds ds = d de
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? y de = k 4 Q d de dq de = - de cos de y = - desin E = - E = - y 4 0 4 0 decos desin 1 = 8 4
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? de = 4k Q d 44k Q 4k Q 4 E = cos d = cos d 0 0 4k Q 4k Q E = sin = sin sin 0 0 4 4 4k Q k Q E = 0 =
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? de = 4k Q d 44k Q 4k Q 4 E y = sin d = sin d 0 0 4k Q 4k Q E y = cos = cos cos 0 0 4 4 4k Q 4k Q E y = 1 = 1
A rod is bent into n eighth of circle of rdius, s shown. The rod crries totl positive chrge +Q uniformly distributed over its length. Wht is the electric field t the origin? kq ˆ 4kQ E = i 1 ˆ j E = kq ˆ i ˆ j You should provide resonbly simplified nswers on ems, but remember, ech lgebr step is chnce to mke mistke.
Wht would be different if the chrge were negtive? Wht would you do differently if you were sked to clculte the potentil rther thn the electric field? How would you find the force on test chrge -q t the origin?
An insulting sphericl shell hs n inner rdius b nd outer rdius c. The shell hs uniformly distributed totl chrge +Q. Concentric with the shell is solid conducting sphere of totl chrge +Q nd rdius <b. Find the mgnitude of the electric field for r<. Use first nd lst slide for in-person lecture; delete for video lecture This looks like test preprtion homework problem, but it is different!
An insulting sphericl shell hs n inner rdius b nd outer rdius c. The shell hs uniformly distributed totl chrge +Q. Concentric with the shell is solid conducting sphere of totl chrge +Q nd rdius <b. Find the mgnitude of the electric field for r<. For 0<r<, we re inside the conductor, so E=0. If E=0 there is no need to specify direction (nd the problem doesn t sk for one nywy). b c +Q +Q
An insulting sphericl shell hs n inner rdius b nd outer rdius c. The shell hs uniformly distributed totl chrge +Q. Concentric with the shell is solid conducting sphere of totl chrge +Q nd rdius <b. Use Guss Lw to find the mgnitude of the electric field for <r<b. E da E 4r q enclosed o o Q b c r +Q +Q E Q r o Be ble to do this: begin with sttement of Guss s Lw. Drw n pproprite Gussin surfce on the digrm nd lbel its rdius r. Justify the steps leding to your nswer.
An insulting sphericl shell hs n inner rdius b nd outer rdius c. The shell hs uniformly distributed totl chrge +Q. Concentric with the shell is solid conducting sphere of totl chrge +Q nd rdius <b. Use Guss Lw to find the mgnitude of the electric field for b<r<c. E da q q enclosed o q shell,enclosed conductor,enclosed E 4 r o b c +Q +Q r qconductor,enclosed Q Q q V V shell shell,enclosed shell shell,enclosed shell,enclosed Vshell
q shell,enclosed Q V shell shell V shell,enclosed b +Q +Q r Q 4 4 q shell,enclosed r b c b 3 3 3 3 4 3 4 3 3 3 c q c Q r b 3 3 shell,enclosed 3 3 b E 4r c Q r b 3 3 b 3 3 o Q The direction of E is shown in the digrm. Solving for the mgnitude E (do it!) is just mth.
E 4r c Q r b 3 3 b 3 3 o Q b c +Q +Q r Wht would be different if we hd concentric cylinders insted of concentric spheres? Wht would be different if the outer shell were conductor insted of n insultor?
An insulting sphericl shell hs n inner rdius b nd outer rdius c. The shell hs uniformly distributed totl chrge +Q. Concentric with the shell is solid conducting sphere of totl chrge +Q nd rdius <b. Find the mgnitude of the electric field for b<r<c. E 4r 4 4 Qshell c b 3 3 E da 3 3 q enclosed 4 4 3 3 o 3 3 r b Q Wht would be different if we hd concentric cylinders insted of concentric spheres? Wht would be different if the outer shell were conductor insted of n insultor? o
A ring with rdius R hs uniform positive chrge density. Clculte the potentil difference between the point t the center of the ring nd point on the is of the ring tht is distnce of 3R from the center of the ring. R 3R Begin by deriving the eqution for the potentil long the centrl is of ring of chrge. We did this bck in prt of lecture 6. I m going to be lzy err, efficient nd just copy the pproprite slides.
dq R r P Every dq of chrge on the ring is the sme distnce from the point P. Q dv dq k r dq k R kdq k V dv dq ring ring ring R R
dq R r P Q V k R ring dq V kq R Q R V kr R
A ring with rdius R hs uniform positive chrge density. Clculte the potentil difference between the point t the center of the ring nd point on the is of the ring tht is distnce of 3R from the center of the ring. R 3R V() kr R kr kr 1 1 0 R 3R R V(0) V(3R) kr R R 10
A ring with rdius R hs uniform positive chrge density. Clculte the potentil difference between the point t the center of the ring nd point on the is of the ring tht is distnce of 3R from the center of the ring. R 3R 10 1 V(0) V(3R) k 10
If proton is relesed from rest t the center of the ring, how fst will it be t point P?
For the cpcitor system shown, C 1 =6.0 F, C =.0 F, nd C 3 =10.0 F. () Find the equivlent cpcitnce. C 1 =6F V 0 C =F C 3 =10F C 3 = C + C 3 = + 10 = 1μF
For the cpcitor system shown, C 1 =6.0 F, C =.0 F, nd C 3 =10.0 F. () Find the equivlent cpcitnce. C 1 =6F V 0 C 3 =1F 1 1 1 1 1 1 3 1 = + = + = + = = C C C 6 1 1 1 1 4 eq C eq = 4μF 1 3 Don t epect the equivlent cpcitnce to lwys be n integer!
For the cpcitor system shown, C 1 =6.0 F, C =.0 F, nd C 3 =10.0 F. (b) The chrge on cpcitor C 3 is found to be 30.0 C. Find V 0. C 1 =6F V 0 C =F C 3 =10F Q = CV C 3 =10F Q 3 = 30C V 3 =? There re severl correct wys to solve this. Shown here is just one. Q Q3 30 V = V 3 = V = V 3 = = = 3 V C C 10 3
For the cpcitor system shown, C 1 =6.0 F, C =.0 F, nd C 3 =10.0 F. (b) The chrge on cpcitor C 3 is found to be 30.0 C. Find V 0. C 1 =6F V 0 C 3 =1F Q 3 =? V 3 = 3V Q = C V = 1 3 = 36 μc = Q = Q = C V 3 3 3 1 eq eq 0 36 36 V 0 = = = 9 V C 4 eq