{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test1_solution - Problem 1(30 pts(figure not to scale A...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 1 (30 pts) (figure not to scale) A glass rod, 2 m in length, with a uniformly distributed charge of Q = +5.3e —9 C, lies on the x-axis, its center at the origin. A hollow plastic ball, with a charge of q = —l.4e —9 C uniformly distributed on its surface, is above the rod, and its center at < 0.08, 0.06, 0> m. (a, 20 pts) Calculate the net electric field at the observation location < 0, 0.04, 0> In, marked with an “X” on the diagram. A Show all your work. Express your answer as a vector. ’ ,2 , .27 5.3 ea) i‘ER00i:LTnneo (GE/L.) —(7&‘i) ( ( e :_ //‘7'2.5' é! ._\ r‘ (02 )(0.043 —— —— M "Ma. (UN/72.5 E' I 0)’ AngLL : 9716—61.: (1:77.? ) F: <0)0'0L/)0>m~<0.08)0.OG)O>H: ‘ r <~0.08}‘0.02)f0>m [F]: V*o.0069m1 : 0.033% m ,0: i: (-0.9701—0M2L/310v ——- __ (”(#4) <— .Civo)-0.:zq3,0>= —/85‘2.7('-970;‘.373.0> EBRLL— (fife C?) .0098 ‘9 ~—‘ «.4 M t : 7 _ :1 (I747) 4579,0753; ) Eun' EMLL+ERM <17? ) /6 43 t 07c (b, 4 pts) List any assumptions or approximations you made in your calculation. 0 LEA/67H 0F Ron >7 ptsnucg F120”! R00 7‘0 085. LOC‘ 42 PLASTIC BALL IS NOT POLARIZED MUCH {57’ ERG/J (BALL BEHAVFS L//<e' A Pom/7' CHAKG€> (c, 2 pts) On the diagram, draw an arrow representing the net electric field at the observation location. ((1, 4 pts) An alpha particle (charge = +28) is at the observation location. Calculate the net electric force on it, and express it as a vector. .5 F3? E:(¥)(flée~lé)</7?7) /é‘/3)O> N :<5.7{€«/¢) {owe-m, 0 >A/ Problem 2 (25 pts) (a, 9 pts) Locations A, B, and C are all the same distance from the center of the dipole. At each of the locations, draw an arrow representing the net electric field at that location. A longer arrow should represent a field with larger magnitude. (b, 6 pts) Location C is a distance r = 6e-10 m from the center of the dipole. The magnitude of the charge of one end of the dipole is q — 1. 6e-19 C and the separation between the two charges 15 s I 1. 8e— 14 m. Calculate the magnitude of the electric field at location C. L/fléo i. /§/= 4-— T: (1e?) (Me ”Mme-1‘01 (QC-(0)3 /.;285 E c Now, a second dipole, identical to the first (same q, same s) is placed at location D, a distance y directly below (vertically) location C. The second dipole is placed in such a way that the net electric field at C is now zero. (c, 4 pts) On the diagram at the right, draw the dipole at location D, clearly indicating the orientation of its positive and negative charges. (d, 6 pts) Calculate the distance y from the second dipole at D to observation location C. ) (Ema/=0 157/: Ml Problem 3 (20 pts) The VPython program below, when completed, will calculate and display as arrows the electric field of a -3 nC charged particle at two different observation locations, called obs l and osz in the program. In the blank spaces below, fill in any missing lines of code needed to complete the program. Your code should be written in terms of symbols and defined variables, and should not use numeric values. from __future__ import division from visual import * #constants oofpez = 9e9 q = —3e—9 scalefactor = 5e—7 #objects particle = sphere(pos=( 0.03, 0.02, O), radius = 5e—3, color=color.red) obsl = vector( 0.01, 0.04, O) osz = vector(—0.02, —0.0l, 0) # Write the lines of code needed to calculate the electric field at obsl I“? 1' obsl -' Particle./aos .x >992 +017“sz + ripple.) [419‘447/3 53,“:(1‘1 rims: r1 /V‘j_m¢;, E1'=(oofre2 >4< 3 /P{Lm4;,>l<9€l> * Fit/.44: # Write the lines of code needed to Calculate the electric field at osz f2 = 0bs.2——/9artfc (e -,oos era = Sfirt (r2.xx*2 +V‘2.7’a<*2+r‘2.z>t<x2) fthEZF‘J/V‘QW“? E.) = (009‘ch *f/ VJWM}*:F2) >t< (‘2 hq-t # Fill in the blanks below E_arrow_l = arrow(pos = OLSi , axis = E l. *scalefactor) E_arrow_2 = arrow(pos = 01352- , axis = 6—2 *scalefactor) Problem 4 (25 pts) A small glass ball is rubbed all over with a small silk cloth (both initially neutral). The ball acquires a charge of +5 nC. The silk cloth and the glass ball are placed 30 cm apart. If any of the electric fields you are asked to draw is zero, state this explicitly. (a, 4 pts) At the location marked “x” draw two arrows, representing the electric field due to the silk cloth, and the electric field due to the glass ball. Label them E clo ECLOTH E glass ball L silk cloth '3" L Now a positively charged metal block is placed between the two objects. (b, 6 pts) On the diagram below, draw the approximate charge distribution in and/or on the positively charged metal block, using the diagram convention of this course. At the location marked “x” draw and label four arrows, using the same scale as in part (a): (c, 3 pts) The electric field due to the silk cloth cloth, E (d, 3 pts) The electric field due to the glass ball, E (e, 3 pts) The electric field due to the charges in and/or on the metal block, E (f, 3 pts) The net electric field, Em cloth ball metal T .1. + + +- silk cloth +- .l. glass ball + metal block‘l' Now the metal block is removed, and replaced by a neutral plastic block. (g, 3 pts) Show the polarization of a molecule at each location marked “x” below, using the diagram convention of this course. ‘ @@@——© glass ball silk cloth plastic block m and EM“ respectively. The relative lengths of these arrows should be correct. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern