halliday hw ch25

8/11/2019 Halliday HW Ch25

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Fundamentals of PhysicsHalliday & Resnic

( ) ( )2

12 221

3

4(45 1 F m 4(2 1 m1(!! 1 F 1!!pF(

1(3 1 m

RC

d

= = = =

*"+ $he char'e on the positi)e plate is 'i)en "y q CV= , %here Vis the potential differenceacross the plates( $hus, 1 4*1(!! 1 +*12 + 1(73 1 17(3q F V C nC = = =

Pro"lem 25#7

hat is the capacitance of a drop that results %hen t%o mercury spheres, each of radius 2R mm= ,

mer'e-

56789:;


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\]^>_`ab4aV 1cd$%"V 1e&

25#29

0'(*a+ and *"+ $he ori'inal potential difference 1V across 1C is

( ) ( )e61

1 2

3(1 F 1( /21(1/(

1( F 5( F

C VV

C C

= = =

+ +

$hus 1 1 21(1 74(9V V V V = = and

!

1 1 1*1 +*74(9 + 7(49 1q C V F V C = = =

Pro"lem 25#15

1 pF capacitor is char'ed to a potential difference of 5 /, and the char'in' "attery isdisconnected( $he capacitor is then connected in parallel %ith a second *initially unchar'ed+

capacitor( .f the potential difference across the first capacitor drops to 35 /, %hat is the capacitance

of this second capacitor-

15 1pFHO 5/

0'($he char'e initially on the char'ed capacitor is 'i)en "y 1 q C V= , %here 1 1C pF= is the

capacitance and 5V V= is the initial potential difference( fter the "attery is disconnectedand the second capacitor %ired in parallel to the first, the char'e on the first capacitor is

1 1q C V= , %here 35V V= is the ne% potential difference( ince char'e is conser)ed in the

process, the char'e on the second capacitor is 2 1q q q= , %here 2C is the capacitance of thesecond capacitor( u"stitutin' 1 C V for q and 1C Vfor 1q , %e o"tain 2 1 * +q C V V = ( $he

potential difference across the second capacitor is also V, so the capacitance is

( )22 15 / 35/

1 pF !3pF(35/

V VqC C

V V

= = = =

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Pro"lem 25#2

.n Fi'( 25#3, 1V V= , 1 1C F= , and 2 3 2C C F= = ( %itch is first thro%n to the left sideuntil capacitor 1 reaches e6uili"rium( $hen the s%itch is thro%n to the ri'ht( hen e6uili"rium is

a'ain reached, ho% much char'e is on capacitor 1-

25#3

0'(e do not employ ener'y conser)ation since, in reachin' e6uili"rium, some ener'y is

dissipated either as heat or radio %a)es( Char'e is conser)ed therefore, if

11

batQ C V C = = , and 1q , 2q and 3q are the char'es on 1C , 2C and 3C after the s%itch isthro%n to the ri'ht and e6uili"rium is reached, then 1 2 3Q q q q= + + (

ince the parallel pair 2C and 3C are identical, it is clear that 2 3q q= ( $hey are in parallel%ith 1C so that 1 3V V= , or

31

1 3

qq

C C=

%hich leads to 312

qq = ( $herefore, 3 33 3

5* +

2 2

q qQ q q= + + =

%hich yields 32 2*1 C+

! C5 5

Qq

= = = and conse6uently 31 2

2

qq C= =

Pro"lem 25#27

Fi'ure 25#!3 sho%s a 12( / "attery and four unchar'ed capacitors of capacitances 1 1C F= ,

22C F= , 3 3C F= , and ! !C F= ( .f only s%itch 1 is closed, %hat is the char'e on *a+

capacitor 1, *"+ capacitor 2, *c+ capacitor 3, and *d+ capacitor !- .f "oth s%itches are closed, %hat is

the char'e on *e+ capacitor 1, *f+ capacitor 2, *'+ capacitor 3, and *h+ capacitor !-

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25#!3

0'(*a+ .n this situation, capacitors 1 and 3 are in series, %hich means their char'es are

necessarily the same;

( ) ( ) ( )1 31 3

1 3

1( F 3( F 12(/ 9( C(1( F


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to o"tain 71 3( 1kW h J = ( $hus,

C U

V

= =

=2 2 3 1

1

722

7

2

( =

/

F(c

b gPro"lem 25#33

ssume that a stationary electron is a point of char'e( hat is the ener'y density u of its electric

field at radial distances *a+ 1r mm= , *"+ 1r m= , *c+ 1r nm= , and *d+ 1r pm= - *e+ hat is u inthe limit as r -

0'($he ener'y per unit )olume is

u E e

r

e

r= = HG KJ =

1

2

1

2 ! 32

2

2

22

!

02 (

*a+ t 31( 1 mr = , %ith 191( 1 Ce = and 12 2 2 4(45 1 C >? m = , %e ha)e

14 39(1 1 =>mu = (

*"+ imilarly, at

1( 1 mr

= , 3

9(1 1 =>mu

= (*c+ t 91( 1 mr = , 39(1 1 =>mu= (*d+ t 121( 1 mr = , 14 39(1 1 =>mu= (*e+ From the e@pression a"o)e !u r ( $hus, for r , the ener'y density u (

Pro"lem 25#35

$he parallel plates in a capacitor, %ith a plate area of 24(5cm and an air#filled separation of 3(

mm, are char'ed "y a ( / "attery( $hey are then disconnected from the "attery and pulled apart

*%ithout dischar'e+ to a separation of 4( mm( ?e'lectin' frin'in', find *a+ the potential difference

"et%een the plates, *"+ the initial stored ener'y, *c+ the final stored ener'y, and *d+ the %or: re6uired

to separate the plates(

0'(*a+ Aet q"e the char'e on the positi)e plate( ince the capacitance of a parallel#plate

capacitor is 'i)en "y

i

A

d

, the char'e is

i

i

AVq CV

d

= = ( fter the plates are pulled

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apart, their separation is fd and the potential difference is Vf( $hen f

f

AVq

d

= and

(f f f

f i i

i i

d d dAV q V V

A A d d

= = =

ith33( 1 m

id = , ( /iV= and

34( 1 mf

d = , %e ha)e 1( /fV = (

*"+ $he initial ener'y stored in the capacitor is2 12 2 2 ! 2 2

2 11

3

1 *4(45 1 C >? m +*4(5 1 m +*( /+!(51 1 =(

2 2 2*3( 1 m+

ii i

i

AVU CV

d

= = = =

*c+ $he final ener'y stored is

22

2 1 1

(2 2

f f fif f i i

f f i i i i

d d dA A AVU V V U

d d d d d d

= = = =

ith4

3

f

i

d

d= , %e ha)e 11(2 1 =(fU

=

*d+ $he %or: done to pull the plates apart is the difference in the ener'y;

117(52 1 =(f i

W U U = =

Pro"lem 25#37

.n Fi'( 25#!5, 1 1C F= , 2 2C F= , and 3 25C F= ( .f no capacitor can %ithstand a potentialdifference of more than 1 / %ithout failure, %hat are *a+ the ma'nitude of the ma@imum potential

difference that can e@ist "et%een points and B and *"+ the ma@imum ener'y that can "e stored in

the three#capacitor arran'ement-

25#!5

0'(*a+ $hey each store the same char'e, so the ma@imum )olta'e is across the smallest

capacitor( ith 1 / across 1 F, then the )olta'e across the 2 F capacitor is 5 /

and the )olta'e across the 25 F capacitor is ! /( $herefore, the )olta'e across the

arran'ement is 19 /(

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*"+ sin' 86( 25#21 or 86( 25#22, %e sum the ener'ies on the capacitors and o"tain

(95total

U J= (

Pro"lem 25#!1

Di)en a 7(! pF air#filled capacitor, you are as:ed to con)ert it to a capacitor that can store up to

7(! J %ith a ma@imum potential difference of 52 /( hich dielectric in $a"le 25#1 should you

use to fill the 'ap in the capacitor if you do not allo% for a mar'in of error-

$a"le 25#1

0'($he capacitance %ith the dielectric in place is 'i)en "y C C= , %here C is thecapacitance "efore the dielectric is inserted( $he ener'y stored is 'i)en "y

U CV C V = =12

2 12

2 , so

2 12 2

2 2*7(! 1 =+!(7(

*7(! 1 F+*52/+

U

C V

= = =

ccordin' to $a"le 25#1, you should use Pyre@(

Pro"lem 25#!3

coa@ial ca"le used in a transmission line has an inner radius of (1 mm and an outer radius of

( mm( Calculate the capacitance per meter for the ca"le( ssume that the space "et%een the

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conductors is filled %ith polystyrene(

0'($he capacitance of a cylindrical capacitor is 'i)en "y

C C L

b a= =

2ln* > +

,

%here C is the capacitance %ithout the dielectric, is the dielectric constant, L is the

len'th, a is the inner radius, and b is the outer radius( $he capacitance per unit len'th of

the ca"le is12

112 2 F>m+

4(1 1 F>m 41 pF>m(

ln* > + lnE*( mm+>*(1 mm+

C

L b a

(2.6)(8.8510

= = = =

Pro"lem 25#!5

certain su"stance has a dielectric constant of 2(4 and a dielectric stren'th of 14 />m( .f it is

used as the dielectric material in a parallel#plate capacitor, %hat minimum area should the plates of

the capacitor ha)e to o"tain a capacitance of 27 1 F and to ensure that the capacitor %ill "ea"le to %ithstand a potential difference of !( :/-

0'($he capacitance is 'i)en "y A

C Cd

= = , %here C is the capacitance %ithout the

dielectric, is the dielectric constant, A is the plate area, and dis the plate separation( $he

electric field "et%een the plates is 'i)en "yV

Ed

= , %here Vis the potential difference

"et%een the plates( $hus,V

dE

= and AE

CV

= ( $hus, A

CV

E=

(

For the area to "e a minimum, the electric field must "e the 'reatest it can "e %ithout

"rea:do%n occurrin'( $hat is, A=

=

* (

( * (( (

7 1

2 4 4 45 1 3

4

12

2F+*!( 1 /+

F > m+*14 1 / > m+m

3

Pro"lem 25#!9

Fi'ure 25#!9 sho%s a parallelplate capacitor %ith a plate area 27(49A cm= and plate separation!(2d mm= ($he top half of the 'ap is filled %ith material of dielectric constant 1 11 = the "ottom

half is filled %ith material of dielectric constant 2 12 = ( hat is the capacitance-

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plates, and *c+ the ma'nitude of the induced surface char'e on the mica(

0'(*a+ $he electric field in the re'ion "et%een the plates is 'i)en "yV

Ed

= , %here Vis the

potential difference "et%een the plates and dis the plate separation( $he capacitance is

'i)en "y A

Cd

= , %hereAis the plate area and is the dielectric constant, so

A

dC

= and

E VC

A= =

=

12

12 ! 2

!5 1 1

5 ! 4 45 1 1 11 1

/ F

F m m/ mgcc hc h( ( ( (

*"+ $he free char'e on the plates is12 9*1 1 +*5 + 5 1

fq CV F V C = = = (

*c+ $he electric field is produced "y "oth the free and induced char'e( ince the field of a

lar'e uniform layer of char'e is

2

q

A, the field "et%een the plates is

E qA

qA

qA

qA

f f i i= + 2 2 2 2 ,

%here the first term is due to the positi)e free char'e on one plate, the second is due to

the ne'ati)e free char'e on the other plate, the third is due to the positi)e induced char'e

on one dielectric surface, and the fourth is due to the ne'ati)e induced char'e on the

other dielectric surface( ?ote that the field due to the induced char'e is opposite the field

due to the free char'e, so they tend to cancel( $he induced char'e is therefore

( ) ( ) ( )9 12 ! 2 !9

5( 1 C 4(45 1 F m 1 1 m 1( 1 / m

!(1 1 C !(1nC(

i fq q AE

= =

= =

Pro"lem 25#53

$he space "et%een t%o concentric conductin' spherical shells of radii 1(7b cm= and 1(2a cm= is

filled %ith a su"stance of dielectric constant 23(5 = ( potential difference 73V V= is applied

across the inner and outer shells( etermine *a+ the capacitance of the de)ice, *"+ the free char'e q

on the inner shell, and *c+ the char'e q induced alon' the surface of the inner shell(

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0'(*a+ ccordin' to 86( 25#17 the capacitance of an air#filled spherical capacitor is 'i)en "y

! (

ab

C b a

= p

hen the dielectric is inserted "et%een the plates the capacitance is 'reater "y a factor

of the dielectric constant ( Conse6uently, the ne% capacitance is

9 2 2

23(5 *(12 m+*(17 m+! (17 nF(

4(99 1 ? m C (17 m (12 m

abC

b a

= = =

*"+ $he char'e on the positi)e plate is *(17 nF+*73( /+ 7(79 nC(q CV= = =

*c+ Aet the char'e on the inner conductor "e Gq( .mmediately adacent to it is the induced

char'e q'( ince the electric field is less "y a factor 1>than the field %hen no dielectric

is present, then Gq ! q' " #q>( $hus,

( ) 1 23(5 1(

! 1 *7(79 nC+ 7(!5 nC(23(5

abq q V

b a

= = = =

Pro"lem 25#55

parallel#plate capacitor has plates of area 2(12m and a separation of 1(2 cm( "attery char'es

the plates to a potential difference of 12 / and is then disconnected( dielectric sla" of thic:ness

!( mm and dielectric constant !(4 is then placed symmetrically "et%een the plates( *a+ hat is the

capacitance "efore the sla" is inserted- *"+ hat is the capacitance %ith the sla" in place- hat isthe free char'e 6 *c+ "efore and *d+ after the sla" is inserted- hat is the ma'nitude of the electric

field *e+ in the space "et%een the plates and dielectric and *f+ in the dielectric itself- *'+ ith the

sla" in place, %hat is the potential difference across the plates- *h+ Ho% much e@ternal %or: is

in)ol)ed in insertin' the sla"-

0'(*a+ .nitially, the capacitance is ( )12 2 2 2

2

4(45 1 C >? m *(12 m + 49 pF(1(2 1 m

ACd

= = =

*"+ or:in' throu'h ample Pro"lem 25#7 al'e"raically, %e find;

( )12 2 2 22

2 3

4(45 1 C >? m *(12m +*!(4+1(2 1 pF(

* + *!(4+*1(2 (!+*1 m+ *!( 1 m+

AC

d b b

= = =

+ +

*c+ Before the insertion, *49 +*12 + 11q C V pF V nC = = = (*d+ ince the "attery is disconnected, q%ill remain the same after the insertion of the sla",

%ith 11q nC= (

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*e+

9

12 2 2 2

11 11 >

*4(45 1 > +*(12 +

q CE kV m

A C $ m m

= = =

*f+

1 >

2(1 >!(4

E kV m

E kV m= = =*'+ $he potential difference across the plates is

3* + *1 > +*(12 (! + *2(1 > +*(! 1 + 44V E d b E b kV m m m kV m m V = + = + =*h+ $he %or: done is

2 9 27

e@t 12 12

1 1 *11 1 C+ 1 11(7 1 =(

2 2 49 1 F 12 1 F

qW U

C C

= = = =

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halliday hw ch25

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