Chapter 26 IQ Questions

Chapter 26 IQ Questions - 26.2.2 Spherical Capacitor 26.2.1...

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Unformatted text preview: 26.2.2 Spherical Capacitor 26.2.1 Potential of a conductor Given a spherical capacitor, see sketch. It consists of an inner conducting sphere with a radius “a”, and a concentric . . i . conducting shell with an inner radius “b” and an outer COHSIder a conducung Sphere Wlth a radlus R, and Chalge radius “0”. The shell is grounded. There is a positive Q. It is in electrostatic equilibrium. charge +Q on the inner sphere. _ . _ Determine the magnitude of the field at the point A, which Fmd the potential at A, OA= r<R, and the potentlal at O. is located on the outm- surface of the shell, Choose one: 1 3 kQ/r kQ/R 00 kQ/R Hint: The shell is grounded, potential of the shell is 0. Extra: Find the charge on the outer surface of the shell. Explanation: Suppose there were charges at the surface _ I of the shell, there would be an electric field perpendicular Hmt: compare the POtemlal at A and that at B' to the surface, since E gammy EU. In turn, there would be - . - - - ~ - a charge flow between the surface of the shell and the Explanation. Bemg m51de of an equtpotentlal body, ground‘ This is con w the fact am the shell is VOZVAZVszQ/R- grounded, i.e. there is no potential difference between the Ans = 3. shell and the ground. So there can be no field at surface. Extra: From the explanation we see that there can be no charge on the surface. 26.3.3 Potential at O ofan llltgl‘OlllldCd 931cm An ungrtmnded spherical capacitor has a sphere and a concentric shell. Both are conductors. The charge on the sphere is -Q. The net charge on the shell is zero. ‘3 Find the potential at O. llint: V at 0 tluc to one sphere ol‘ R and Q is H) R. Use the superposition principle. Explanation: 'I here are 3 concentric spherical charge distributions: ‘ The superposition principle implies that at O: V,, :V,,+Vp V: le| ‘a -l b *l ‘cl. Ans‘ : 3 26.2.5 One plate vs parallel plates T +Q E1 ++++++ H Al» l+++++l+ E Q Given 1- late attem, E = : P p 1 250A 1 +A I ++++++++ II_ -Q, A In Find electric fields of parallel plate system in I, II and III. 2 Extra: Find the potential difference across the plates. I ly the suerposition principle Extra: V:Ed:Qd:’(£0A), where cl is the gap width. 26.2.4 A parallel plate capacitor +Q Two plates a and b are separated by a distance d. The “plate charge” is Q. Between the gap E is constant. Find the potential difference. Choose one: Va-Vb Ed -Ed Ed/2 -Ed/2 1‘23‘4 Hint: AV=_E.As=_EASCQSQ, G is the angle defined by E and As. Notice E is down, As is up. Explanation: From the sketch, Va _Vb = AV: — EAs c03180°=Ed. a b MI W. An independent check: We recall that the natural tendency for a positive charge is to move along E. So Va>Vb. The sign is therefore correct. 26.2.6 Attraction between plates +Q : i : Egap -Q A parallel plate system has a plate charge Q. Within the gap: = Uplate = £ gap 80 E Determine electric force with which the bottom plate pulls the to late. - QEgap 0'5QEgap Hint: The electric force “F:qE“. ° "q" is the charge of "test charge" 0 "E" is that portion of the field due to the "source charge“ only. Explanation: The electric field due to the bottom plate as shown is = Qencl = 1 230A 2 ' This leadsto F=QE1= 26.2.7 Various symmetries Syrnm. CD5 E V l Dipole axial -- N r—3 ? Point ~i N 1 charge spher. 4M2E r2 r Long ~ 1 wire cylind. ZMLE r ? EA const. ~z Plane planar {2EA N ZO Find the r-dependences of V for the two cases with “?’s” Choose one: Denote C=constant | 1 2 3 4 Dipole N; ~ 1 Ni ~ 1 1'2 I' 1'2 I' L 911g ‘ -lnr +C -lnr+ c C C Wife Hint: E = — dV/dr. Explanation: To solve for the r-dependence of V, one needs to perform the appropriate integration. For the dipole V=—I§0Edr~— case, 1.2. For the long wire case: V=—J;Edr~—lnr+const. Conversely, one may take the derivative of V to verify the expected r-dependence (or the z-dependence) of E. The reader should make this check for all four cases tabulated. Ans = 1. 26.3.2 Capacitor network II C C “C A AI A" —C/2 —C/2 _C B T, T» T Determine the resultant CAB for the above network. | 1 | 2 3 | C | 0/2 20 i CAB Hint: Parallel C=C1+C2, Series lzi+_1 C_ CC1 C2 Explanation: we Tom C 3c, B"— C1C2 _ C1+C2 B” By inspection, C AB '= C A"B " = C. This leads to: CAB = sketch = C/2, Ans = 2. 26.3.1 Two capacitors in series Given Cl=1uf, C2=2pf, VAB=3V. A I9 CI B M ind: The ratios QI/QZ. Choose one: l 2 3 4 Ql/Q2 1 2 1/2 1/3 Hint: "Q=VC". Series: V=V1+V2 Extra: Show the ratio Vl/V2=2. Explanation: Notice each capacitor is a neutral system, so for C1 and C2 the corresponding charges must be (Q1,-Q2) and (Q2,-Q2). From the figure it follows that Q1=Q2. So Ql/Q2=1. Extra: Since V1=Q1/C1=Q/Cl Vl/V2=C2/Cl=2/ 1=2. Water beaker analogy: The correspondence of the relationship Q=VC is Volume=Height*Area. The present case corresponds to having a fixed volume. When the area is doubled, the height is reduced to 1/2 of its original value. The situation is illustrated by the sketch: Il—J and V2=Q/C2, so 26.3.3 A capacitor network Which one of following diagrams represents the same network as one above? Choose one: C C raj H :H l 4 C I I C Fl: I—I :h B 1 4 C C rl :N: l— B r 4 Hint: Deformation rule A wire segment may be shrunk down to a point and a point may be expanded to a segment. Explanation: Using the deformation rule, one finds that "1" is the match. 26'5'1 Energy and Energy density 26 5 2 Capacitor: and dielectric: Two parallel-plate capacitors: +Q +Q C1 C2 d F - Q -Q V Case 1: Air—gap. Energy density: I lF—U Illl Ad 2 C 2 Given capacitances of #1 and #2: C2>C1. Compare V1 Case 2: Gap with dielectrics of K. and V2. Corresponding energy density: U 2 u =—K=ls (9)E 2 withU =Q— K Ad 2 0 ' ’ " 2CK V1> v2 v1 =v2 v1 <v2 Determine (?), i.e. the K factor. Hint: Q: VC Ch : °°'° °”° 1 2 3 Extra: If c2 is filled with a deielctric, show that V2’<V2 0) K0 K1 K2 Explanation: Since it is a series connection Q:Q1=Q2. Q=V1C1=V2C2. So V11V2=C2fC1 >1. Hint: Etc = E / K. Explanation: Extra: The ratio of the potentials after the insertion is U & U 1 £2 1 2 given by V1 ’/V2’=C2’/C1=KC2/C1. Out of the total If . . ’=—=—=—=— =— {5" i _ 7‘ Ad 2AM: I" 2ft?) 2 éb A I So 0):“ potential drop of V, V2 gets a lesser portion thaan 26.5.3 A fixed V case Ic— 26.5.4 Capacitors and dielectrics V -Q Q - —IIII|V|— ? Consider atypical capacitor, such as a ||-plate capacitor or a spherical capacitor. For each case, the capacitance is defined by A capacitor “rub a capammce C Is connecwd w a battery C=QN. In the presence of the dielectric with the dielectric wnh a Vglggfi It has ,Thpla‘etecfifl’gemgmad'f’? m t constant K(>l), while keeping Q fixed, the electric field between energy ' 1 9 gap “n m n w 0 1e ec c the gap will be reduced to E’=E/K. The primed quantity is the constant K' The corresponding new quantities are QI and corresponding quantity after the insertion ofthe dielectric. Choose U'. one: 2 Him: VEV, ..Q:VC.., ..U = Q_ =1CV2.. 2 2C 2 Hint:V=EdandU=Q_ Extra: Show that the ratio of total energies is U’/U=K. 2C Ex lmfion_ Extra: For fixed Q, show that the ratio ofthe total energies stored p Q: I Q Q: CI is U’fU=1/1q. V'=—=V=—,or—=—=K. _ C' C Q C Explanation: V=Ed, V’=E’d=Ed/1C=V/1C. Extra: Thf ratiozof total energies is: c=QN’ C,=QN,=KQN=KC_ U' Extra: The ratio ofthe energies is given by: — = — = K. U 10,2 U’/U=[Q2/(2C’)]/[QZI(ZC)]=CIC ’=1/1c, 2 26.5.5 Insert a dielectric slab _____ Insert a dielectric slab half way into a charged capacitor. Find the direction of the force on the slab. Choose one: 1 d' . r fire: toleft Hint: Natural tendency for the slab is to move from high potential to low potential. Extra: Show the magnitude of the work in inserting the slab into the system is given by (1-1!1<)Q2/(2C). Explanation: Going from no slab to the presence of a slab corresponds to going from a high potential to a low potential. The natural tendency is to have the slab moving toward filling the gap. In other words, there is an attractive force to the right. Extra: Work=U-U’= (1-1/K) U=(1-1/1<) Q2/(2C). 26.7.1 1c & Qinduced ofa conductor Given a ||-p1ate-capacitor, with an air-gap, the plate charge is Q. Now fill the gap with a conducting slab. What are the dielectric constant of the conducting slab, and the induced charge on this slab? Choose one: 3 1c 00 Qind oo 0 Q Hint: Inside a conductor, the electric field E': 0. For a parallel plate setup, the effective source charge for E' is Qeff= Q/K = Q-Qind Explanafion: Inside the conducting slab the field '20. The effective plate charge, which creates E', is therefore 0, i.e. Qeff= Q-Qind = 0 or the induced charge Qind=Q. One sees that K=E/ E‘=oo. Ans = 3. 26.6.1 Flip a dipole -Q II A dipole of :: q with a separation d is placed in a uniform field E. Determine the net force on the dipole, and the potential energy released in flipping it from I to II. Choose one: 1 2 3 4 F 0 0 2qE 2qE |AU| qu 2qu 2qu 4qu Hint: First find the potential energy difference for the charge +q in going from I to 11. Then do the same for iq. Explanation: Electric forces on the two charges asserted by the electric field are equal in magnitude and opposite in direction. So the net electric force is zero. For +q, the potential energy released is AU=q(V1 —V11)=qu. Taking into account the potential energy released for -q, one finds that the total potential energy released is 2qu. ...
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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.

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Chapter 26 IQ Questions - 26.2.2 Spherical Capacitor 26.2.1...

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