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 DocumentThis 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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (a} The values ofthe four capacitanees show in the diagram are: C1 = 15 pF, C2 = l pF,
C3 = 16 pF, and C4 = 6 pie". Calculate the effective capacitance between terminals :1 and b. A l2 V battery is connected across terminals :1 and h and remains in place for the
remainderol" this problem. [b] How much charge is stored on the positive plate of CI when the network is Fully
charged? [c] How much charge is stored on the positive plate of C2 when fully charged? (a) How much charge is stored on the positive plate ol't'L';1 when the network is Fully
charged? J— ; 1. Fe ,
5%) Ca. Ca CH (3"! ' Spherical Capacitors The two sphedealrshell capacitors shown
in the diagram have inner radii nfn = 15 em
and outer radii oft: = '? em. The space
between the spherical plates of capacitor
{1. is ﬁlled with air and in capacitor C9 the
space is ﬁlled with a dielectric with rr= 2.9. {a} Caienlatc the two capacitanees. The two capacitors are connected to an E = 2.5 V battery using die circuit shown in the
diagram. (h) Calculate the charge {including the appropriate algebraic sign] that appears on the
inner plate of CA when ﬁillj,»r charged. to] Calculate the charge that appears on the outer plate of CD when it is fully' charged. “:3 E r: 5"":
b 31"— Capaeitnr Newark V= 12V I l :1 C1= 3le Y C] I" It} C ﬁzjzm:
2 C4 5 3215 c..=1luF " Cs=4uF In this circuit? what is the potential difference across {34? I
C'BH : HE : r I
__L___ Ci at
cm”. .r. C1 + (‘3, + C1— .—. E, .35} 99E " e.
F.— ,L ._4
C1 G193 (3% L (Dam: : CGBJJV $3.059le (ax/)H‘Zkrfaﬁuc Q . 1
Vain = V—A/l = V! —*— ; jluﬁzigi’ﬂcjqulv
C! 3/1? 82L; 3 (9‘3, 1 @131” l "' Battery and Capacitors An E = l? V battery is used to charge two identical capacitors Q. = CH = 3 uF, as shown in this diagram. J:_ I . C C
{a} Caieulate the resulting charges Q; and 9;; on the left I' R
and right capacitor respectively.
The battery is now removed as indicated in this diagram. l
{b} Calculate the total energy stored in the capacitor CL c”
network.
Next a dielectric With rr= L4 is insETIEd between the plates of CR.
Fully ﬁlling the available space as shown. (T he dielectric cam'es zero
net charge} EL CR [c] Caiculate the new 1“values of the charges on each capacitor. til Calculate the subsequent voltage across Cg.
{c} 1ulnl'hat is the total energy.r stored in the circuit in its ﬁnal conﬁguration? (ﬂ Compare the total energy stored in the capacitor network before and after the
dielectric is inserted1 and from this determine whether the dielectric is patted into C; or
must be pushed into C”. 69 {TIQILi QR ‘ 1?) U : ULtng. = a %(3::JF>(DV)l [email protected]§ﬂ
no} Choeacn £6110 mdcttdbm m UL: Vt;
agate : I I P 63 + 9'1 2‘5; _.
@‘Z‘ L r = Liqffig.
ma 1+
5 Hr
@Lq .®L+ (31E ﬁﬂ I
1 r ——
‘9 VPL=V .. $511: /w @L (49V
HCR L CL.
1' l I
EL) U : UL+ '1.. r1
1 ELLA/L + 51mm? : g mﬁrYgPF) (m. ml: 511m 7 UHCU ,: ED 11"»? a‘blrcm “mg h) 31.;
ﬁhkﬂJE— EMS—mﬁl ._ M ML..ML Ln EMA ﬁg, (RH 51 m, admit. {mat Dielectric A. parallel plate capacitor is shown in the diagram in cross I
section. A slab of dielectric be: 4.55} is positioned midway
between the plates. The area of the plates and dielectric slab is A I:
= 2.9 mg. The spacings area =c I =l.4cmartdh= l.4crn. {a} Find the capacitance. {b} A battery with E = 7 V is now connected across the capacitor. How much energ},r is
stored in the capacitor when fully charged? [c] The dielectric slab is removed and replaced by a slab chonducting material with identical dimensions. The spacings a. £1, and c are unchanged. Calculate the capacitance
new. {d} If slabs ofdielectrie with a’ = 2.4 are next inserted. ﬁlling the ernpt}r space between the top plate and conducting slab and the empty space between the lower plate and the
conducting slab, what is the capacitance of the resulting conﬁguration? JD?) ILL %CU
J, ’L "I
= (1510,4812; xlD ﬂax/J ...
View
Full Document
 Fall '08
 HerreraSiklody
 Work

Click to edit the document details