Chapter 24 IQ Questions - 24.1.1 A cubic box Consider the...

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Unformatted text preview: 24.1.1 A cubic box Consider the space of a cubic box. Electric field is parallel to X—axis. Flux from left: ICDLl = 2Nm2 I'C. Flux leaving right: l‘I’Rl = le2 IC. Find Qencl, , the net charge enclosed. Choose one: _1- w Him: (I) s=¢ L+¢ PHD 31229.5 Explanation: Gauss Law states that ‘13; = gem-1" 5‘0 . Here (133 is the flux leaving the cubic region. So ¢S='|¢L|+I¢R|= _2+1:chlflcou SO Qenclz —£D Ans = 3. 24.1.3 Comparison of fluxes (A) Consider two Gaussian surfaces above. Each has a shaded vertical flat surface connected to an unshaded round surface. The shaded surface and its location relative to q for both cases are the same. The two unshaded surfaces as shown are different. Define the flux through the shaded surface due to q to be 11:1 and the magnitude of the flux through the unshaded surface to be |¢z|. Compare 1111 & |¢2|' 4 case A ¢1=|¢z| ¢1=|¢2| ¢1<I¢zl ¢1<|¢2| case B ¢1=|¢2| ¢1<|¢2| ¢1=|¢2| ¢1<l¢2| Hint: For case 2, sketch a plane through the charge q with the plane being perpendicular to the horizontal line. The portion of the flux emanating from (1 to the left of this plane is the same as the portion to the right of this plane. Explanation: For case 1, since q is outside, ¢1+ 132:0, so $1: — ID2=| ¢2|. For case 2, the comibination of the hint and the inspection leads to: “’1 < 'I'left-to-plane: (fright-to-plane < Ans:2- 24.1.2 Electric flux Given a constant electric field E along the x direction. The first rectangle abcd has an area Al It is perpendicualr to E. The second rectangle abc’d’ has an area AZ and it is inclined with an angle Adad’ze. Find (1)2, the electric flux due to the field E through the second rectangle: abc’d’. - EAzcosm Hint: Flux “(DzEAl’K For abcd, ¢1=EA1. Extra: What would be the flux through the second rectangle abc’d’, if the rectangle is now H to the x-axis. Explanation: For the second retangle, the projection of its area which is perpendicular to E is “A i” : areaabc. c1.cos9 :Azcose. Ans:2. Extra: When abc’d’ is parallel to the x-axis, 9:900. So (1)220. This is expected, since now there is no flux through the area. 24.3.1 Conducting shell & a point charge Consider an electrostatic situation. A point charge ql is located at the center of a thick spherical conducting shell. Net charge on the shell is qz. Let S be a concentric spherical surface (Gaussian surface) with a radius r. Hint: For static case, what is E inside of a conductor? Extra: Show that the charge on the outer surface of the spherical shell is q/2. Explanation: For an electrostatic case, inside of a conductor, or in a conducting medium, E=0. This implies that cps = fsEvdA= 0 Extra: For an electrostatic case, there must not be charge(s) inside of a conductor (otherwise Einside would not be 0, Why? and it violates the static condition). So the charges can only reside on the inner surface & outer surface of the conducting shell. We denote them by qm & qout. Since (D3 is 0, the charge enclosed q1+qin=0. But qin+qout=q2. This leads to qout = qz-qin= C12+Ch= q+(-q/2) = q+(-q/2) = q/2. 24.3.3 Charged spherical shell A spherical shell (insulator) has inner radius a and outer 24.3.2 Uniformly charged sphere radius b_ e“ a Let the Gaussian surface S be a concentric spherical surface Posmve Charge is unifomfly dismbmed throughout the with radius 0A 2 I. D etenm-ne the charge enclosed by S' shell. Consider the field vector E at A, where OA=af2. | | 1 | 2 1 2 3 4 | |E| | > 0 | >0 4 3 / 3 I, 3 I 3 / Direction of radially radially -- Qencl 7rr p 3 ‘ 47rR p 3 271T p 3 27rR p 3 E inward outward Hint. Flux: = flows outward from the r (D5 Qencl / 80 Extra: Show the magnitude of the field at A: E=:—. GaUSSian surface S 0 Explanation: Since the charge distribution is spherically symmetric, |E| must be the same everywhere on S. And by dezpvzpmmyn Ans:1_ symmetry E must be directed radially either out (+) or in(-). , _ 2 _ Qencl _ _ Extra. Apply Gauss,s law. (D8 2 £1sz 2 fl . We may write: COS i4Ttr E — 80 . Qencl — 0, or E—O. 80 Q end 47: 3 2 r Digression: For an spherically symmetric charge Wehave 8 ZPEI :47“ E, orE=%. : éeflcl ’ 0 0 0 distribution, 4m280 where Qencl is defined by a A sphere (insulator) has a uniform charge Q and radius R. b Its charge density p: wherev : 43113 . Explanation: For the definition of charge density, spherical Guassian surface with a radius r0 . 24.3.4 Attraction between Plate: 24.3.5 Conducting shell & a point charge E Consider an electrostatic situation. A point charge q1 is :Li._£39;~0 located at the center of a thick spherical conducting shell. . a ' Net charge on the shell is q2. A parallel plate system has a plate charge Q. Within the gap: -- " E _ C"plate _ Q gap _ e _ e A' is O 0 ; 3 Determine the magnitude of the electric field 0 within the gap contributed by the bottom plate alone. ‘ [4,112 Choose one: Find Ep, the magnitude of the radial electric field vector at P, which is at a distance r from the center. Extra: Determine the electric force which the (ll/(4719011) q2[(47rgor2) (ql+q2)/(4jf80r2) bottom Plate Pulls on the to? Plate- Extra: Show that the charge on the outer surface of the Explanation: Electric field due to the bottom plate Sphefical _Shefl is (Ill-(h- _ . alone contributes to one—half of the total field in the Explanatlon: Due to Sphencal Symmeuy ©S’ emanatlng gap, i_e_ through the Gaussian surface S Which passes through P, is E Q en C1 E gap given by <I>s =47rrlE. Applying Gauss’s law, it gives 4nr’E 2 EOA 2 —(q1+q2)/80, which leads to Ans—3. QEgap Extra: For an electrostatic case, inside of a conductor, or in Exam The Pulling force F = QEl = a conducting medium, E=0. The charges can only reside on ' " ‘41? I? -'~' '-‘—'~"-' '-‘« ‘—‘-’ ‘1‘ to; the inner surface & outer surface ofthe conducting shell. it. .l .. .. . . .-_.. I ‘ We denote them by an & qm. an=-q1. But an+qm=q2. So "‘9 1 qua: = qz-qm= qz+q1. 24.4.6 Field near an uniformly charge plate Given an uniformly charged conducting plate total charge Q (Q>0) and the plate area A. To determine the electric field near the surface of this plate we use a pill box to define the Gaussian surface, Where the area of the top and that of the bottom are AA. Find the total electric flux coming out of the pill box. E1 Choose one: 1. E1 AA 2. 2E1AA 3. 0 4. E1*(areas of the side + 2AA) Extra: Show E1= Q/(280A) Explanation: (1) = (PM, + (Dbottom + CDSide = 2E AA Extra: Use Gauss law ZEAA = AQ/SO 24.4.7 One plate vs parallel plates Q Given 1—plate pattern, E = : 1 250A +A I ++++++++ II_ — Q, A 111 Find electric fields of parallel plate system in I, II and III. 24.4.8 Electric field due to 3 charged plates +Q -Q +Q +Q -Q A- <— —> —> <— El El El E1 ‘11 (a) (b) (0) Consider the case (a), based on the superposition principle determine the electric field vector at A, E A. Here Q>0. 1 | 2 | 3 4 | EA=|EA| omen) ‘Q/(SUA) iQ/(ZsDA) ("z/(220A) l dirofEA <— i —> i <— —> i Hint: For one plate alone, the +Q case is shown in (b), and the -Q case in (c). EI=|E1|= Q/(ZSDA). Extra: Find q], the charge on the left surface of the center plate in (a). Explanation: Taking into account the direction of the field vector contributed by each of the three plates (from left to right), one obtains EA =+ E1 + E]- E1=+ E1= Q/(ZSUA). as Extra: Near the surface of a conductor, “El=0' surface 0 ‘ Since E A is pointing toward the left surface of the center plate, the magnitude E A=|E Al: — q1/( 80A). But from above, EA: Q/(ZeUA). Equating the two, one arrives at q1=-Q/2. This agrees with the answer based on a symmetry argument. 24.4.9 Fields & surface charges of H plate capacitor Re call the electric field pattern for l-plate with a positive charge +Q, and the pattern for with a negative charge -Q. For each case, the magnitude of the electric field above and that below the plate are given by E1=Q/(283A). I / qupp er +Q qluwer I—J_Q 111 Find the electric fields of in regions: I, II and III, with - sign denotes downward. Hint: Use the supersposition principle. Extra: Determine the surface charge on the lower surface of the top plate, qlwm, and the surface charge on the upper surface of the top plate, qupper. Explanation: See 24.4.7 Ans = 1. Extra: At the surface of a conductor “E l=qsurface/(SDA)”, where Elis positive if it is outgoing from the surface. In region 1, E1: quppa/(SUAFO. This implies that qupper=0. Eu = (how/(80A) = 2E1 = Q/(SUA) , so that qlower=Q. ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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Chapter 24 IQ Questions - 24.1.1 A cubic box Consider the...

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