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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: VaVb 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 rdependences 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 rdependence 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 rdependence (or the zdependence) 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 parallelplate 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  —IIIIV— ? 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 (11!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=UU’= (11/K) U=(11/1<) Q2/(2C). 26.7.1 1c & Qinduced ofa conductor Given a p1atecapacitor, with an airgap, 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 = QQind Explanaﬁon: Inside the conducting slab the ﬁeld '20.
The effective plate charge, which creates E', is therefore 0,
i.e. Qeff= QQind = 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.
 Fall '08
 Turner
 Physics

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