This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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; = gem1" 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'lefttoplane: (frighttoplane < 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 xaxis. 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 xaxis, 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 = qzqin= 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 etenmne 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: = qzqm= 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 lplate 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. ...
View
Full
Document
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.
 Fall '08
 Turner
 Physics

Click to edit the document details