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lu5\ ,{)cc/.:===--'Physics 48 (Fail 2All)

Chapter 22: Electric Fields\./

6t/

"Whether ou thinkyou can or thinkyou can't, you're usually ight. - Henry Ford

"It is our attitude at the beginning of a dfficult task which, more than anything else, will afect ir 'ssuccessful utcone. -William James

"Thefirst and most mportant step owatd success s the eeling thal we can succeed.Nelson Boswell

Reading: pages 80 596

Outline:

= electric ieldintroduction

= electric ield of a point charge= eiectric ield lines (PowerPoint)= electric ield from a dipole

, binomialexPansionS/ = electric ield calcuiations

continuous harge distributionline of chmgecharged isk

> point chmge n an electric ield+ dipole n an electric ieid (Read on your own)

Problem SolvingTechniques

You should know how to compute he electric reld ofa collection of point charges. Remember hatthe sum of the ndividual fields is a vector sum. Add the x components o get the x component f thesum; add he y components o get he y component f the sum. The magaitude of the electric ieldproduced by a single point charge q, at a point a distance away, s given by E = lql/4nesf. Thismust be multiplied by the sine or cosine of an angle o obtain a component. You must also determine

the sign of the component.

Some problems deal with electric ield lines. You should know that the electric ie1d at aay point istangert to the ine through hat poinl that the number of lines per unit area passing t['ough al areaperpendicular o the lines s proportional o the magritude of the field, and hat iines emanate rompositive harge nd end at negative harge.

r.\_-, You will also need o know that he force on a charge n an eiectric ieid is given by the product of'=.'the charge and he field. For a positive charge, t is in the direction ofthe field while for a negativecharge t is directed opposite o the field.

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You should know how to calculate he electric ield ofa continuous distribution of charge. Use helinear or area charge density o write an expression or the charge n an nfinitesimal region, hen ,^carry out an ntegration over lre entire charge distribution. Don't forget hat the field is a vector and Vyou must evaluate an ntegral or each component or, in special cases, se symmetry o show a fie1d \-

component anishes.

Some problems deal with the trajectories of charges n electric relds. f the field is uniform, theacceleration fthe charge s cohstant and you may use he kinematic equations f constant motion.The problems are quite similar to projectile motion problems; however, he acceleration s now dueto the force of electric ield rather han o gravity.

You should know how to calculate he orque ofa uniform electric ieid on an electric dipole and hepotential energy ofa dipole n an electric ield. ln some cases, ou are asked or the work that sdone by the field on a dipole when he dipole tums. This is the negative of the change n thepotential energy.

Questions and Example Problems from Chapter 22

QuestionThe figure below shows wo protons and an electron hat are evenly spaced n an axis. Where on theaxis (other han at an nfinite distance) s there a point at which their net electric ield is zero: o theleft of the particles, o their righ between he wo protons, or between he electron and le nearerproton? ^ "

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d'QuestionThe figure bplow shows hree nonconducting ods, one circular and wo straight. Each has a uniformcharge of magnitude Q alolg i151ephalf and another along ts bottom half. For each od, what ig thedirection ofthe net electric ield at point P?

- / \ | L --) O'J ta^tzh

Question 4The figure below shows wo disks and a flat ring, each with the same :niform charge Q. Rank he

objects according o themagnitude of the electric ield they create at points P, geatest first.

v

- l2 ,e

^ ,,,P,,,-o'Fr,Y)(., f^,,\,r

v'+5e

-) -$-c-. &r".f

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Problem 2In the figure below, the three particles are fixed in place and have charges ql = q2 ; +e and qz: +2e.Distance a: 6.00 pm. What are he (a) magnitude and (b) direction of the net electric ield at point P i.due o the particles? g- /_ __i _-_,

, \ A"tt-) At' *'A E ^ t''n"'Q)^"\ fue' ^J) \ 4"" ) A,

*"AE^'ton"'"Lq"o"s

T+.t \ )^*"#.g-2'^,-. xru a,wvq

,l L.;oio', -b.i/io?t'4'ta-'';t"*'

1Ii\_

- l

t #"4'a-h^q^-*#€; 1-= a' c-oai56

E = l.60, .goaNr/ ./ r \ n I

- (5) )?@^ln.o\l.o."M/\ y}ec;4nAte Zr.t 4-;,a ;tProblem 3 ' O=?5"The figure below shows an electric dipole. What are he (a) magnitude and @) direction (relative othe positive direction of the x axis) of the dipole's electric ieid at point P, located a distance >> d?

F

\42

t-,t3'

j'v"g --

( ^ )

(b)

I

4'rt"

c/.)5ld"r"

E c.oe----J l

\o' l

Etno-->'E+

1- lr\2

l - E +G 4\r.a-X -tau'"y"n@v^+a^Yd UM.,

Ne"fur K^"I^"'Ji

c'*(d/)"

E^rx ffi21,1'./>'f r\ (16;'1:7'

AL JJxn O

^ f | =L,,.'r]. ' L+fied Lca("a/ J

?,. - - f . j 1 t / a : r -

L I J I

I

"* ( t / t )"

t%drr€. 14

&r, F>>J

4rt.

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Problem 4In the figwe below, wo'curved plastic rods, one of charge +q and he other of charge q, form acircle ofradius R = 8.50 cm in an xy plane. The x axis passes }rough both of the connecting oints,and he charge s distributed uniformly on both rods. fq = i 5.9 Oa, what are he (a) magnitude and(b) direction (relative o the positive direction of the x axis) of the electric ield produced t P, thecenter of the circle?

r- r Jq) l - =

-- . . P -

lTQ- r- "

l r- r )dsd . E - - _ - - - -

qfie-

\=%/t = 4/s

3= )s ---)l \ l

dg= aos--:'

o.td.'=:

r\

l i

F = -'t"-t l < ."7 t t ?o

rT

)o

q Tre" A\f -w"ol,/ L u/

o

= _r_-4lrL"

\---p-

- - \ l

R).J' n" l -

-1

/ 5 'o . ldra c -rl(o"or5o r"r) -

ts_*).t;'%l' [-J)= ( ra.r N/.)C

5,6^/ 1i_" c/^'...-...-'

4^"f b"ot "t -q'pa + lo)" :

4 X ,;ry

-[-r- (r)]= i;"(

+ E' 4;t ) +.a,1 rt^-.) "A 4 cAa^4 *g',ra /\4\y\e aa ufi,pnaaA

r f--J- /- L {rrc." \

A= O,o?Sorr. _

\= %/r:+A=

\^-.D"OAJo4^

a

))

t:^rde =

,^^, P'" y ,*,tutrA ( -3)

)^ rcvi*^blta '.

\7 "r :2

o.ogSofr )

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Problem 5In the figure below, positive charge q: 7.81 pC is spread niformly along a thin nonconducting odof iengti L = 14.5 cm. What are he (a) magnitude and (b) direction (relative o the positive directionofthel axis)of the electric ield at poinl P, at distance : 6.00 cm iom the rod along tsperpendicular isector?

r-- -J- dIdtr - 4rrt- r\\,- l/t----:X= L

JX '\J*

\dxr\

F= tJ4 = ;il. 3N ^^^ )-b": X){-)O

.. F=

_/A.l/n C,

J xaB-

= ) - = rR

t-+g14x

( dx\ ---.-.-

) r*1st)%- ,/r,

E

= ^)*L

t - \

4t€-

( dx

i ("'*)L/x

XALln€"

Jx

ffi2" -l-J 4^d*,hfr

/\, -3"r.^ )v'.,3-.*t*Jt.t

-,r^tltffith:a X=qtq\&

S: ?.f x1c,?

,4 :D.ot .or r | = o ' l ' l5 r" r

\R4-nt.

E= )j- |4t€. L

E-I

4^(r'tBL

i{Lz,6h)'^

l , -- .Ldtr

-qnt.

X -lraw'1.^luO c.-. +---& c'axuaA,

Afiv.; t,a...,?-t tb d,Ad

\ l

X dK - &MqO

F

. X ^ ? .

x 1 6

,'l4nt- A

(rr.rN7")'

4" = At-U

L7,r,4^

-->E

=lX."fN/c ^r"fu.rrd"d

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ItProblem 6A disk of radius 2-5 cm has a surface harge density of 5.3 1tC/m2 n its upper ace. What is themagnitude ofthe electric ield produced by the disk at a point on its cenual axis at distancez: 12 cm from he disk?

A= 2-5t, = o.ox 5r 'r

Z= l3 . tc r ' = O. lO. rn

f = 5,3 u' cl*+

r-3--I r---a<| 214'

Problem

I t -/ l - J Z

----r1l

{z'rA" O.39

fr,/e-Q.5o " tD9

( l .uoa' . ' to".) (Z.so. PtN/.)

\ J '

--)

Fy'Fu =

h" = Y)q' t ov

t") / ,5 * lo 'o

E=*e,(,-ft"):

[^= ( s .3"lo "%n') [ , - D, larq - '1 [ ^--- ,(*fic"%_,)' 6ffi J Y'rJProblem 7At what distance along he central perpendicular xis ofa uniforrniy charged lastic disk of radius0.600 m is the magnitude ofthe electric ield equal o one-halftle magrdtude fthe field at thecenter of the surface of the disk.

r l

L +& ", "l*^tqA -l:-z,,&. E= V^,-(r- ra;)=) o* c-Jth at br^5"4'.6 d/4A I Z -o ao E

-- /aL.

fl ,,* *r"^i z zotrl,.l E= t/x(7'..) -' 7ae-(t-A) :'/r(Y^r)

= '/x + o.,z'--nEia-t -. 4 z^= E Ra

z"= B7s ---4 z= Ytrg -=}$lrA charged loud system p.roduces n electric ield in the air near Earth's surface. A particle of charge-2.0 x 10-eC is acted n by a downward lectrostaticorce of 3.0 x 10-6 when placed n this field.(a) What is the magnitude of the electric ield? What are he (b) magrdtude nd (c) direction of theelectrostatic orce Fel exerted on a proton placed n this field? (d) What s the magnitude ofthegravitational orce F, on the proton? e) What is the ratio of F"1/F n this case?

/^\ = ^F F-P^=t3'"ij9't)3aJ /- . - %b --. E - ,/ \ ' _a 'Oio* c

tc/

( J)

4 L i u r n - l b r . r6 . I ' l v l \

i

- / | | r - r . r ^ - ) ^ 7 v . \ t - I ^ \- - ( r.b ' l" lO-- ' fh)11.8o-ls ' ) :

(a"q . o 'nn)

| , r r1* ld"N ),^t t ^

l .6q *l .5 toN

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Problem 9A 10.0 blockwitha charge f +8.00 10-5 is placed n electric ield E=(SOOOi-OO}hN C . -What are he (a) magnitude ard (b) direction (relative o the positive direction of the x axis) of the \Jforce on the block? fthe block is released rom rest at the origin at t = 0, what are ts (c) x and (d) ycoordinates t = 3.00 s?

- O F . ^_

0 r v v , l L ) L _

: f? "oox ( ' -oo.lo 'rVt 1

= ( 7.oo* dE c] f( z.oo*tolub) e- {c.oo, . o}rr/c)i

f]

E

1.; I (u)

{. ' t t-

%E

i n / N

E-

= (o.avo1.r) - (c.o+:,x;i ' - , / - D ' o ' r 6 N \&= r'& \ o.a,roNla = -, l1"3"Ii -----;6-D

X = X"* Vo*f * ' l ac ty i , "a = l^ ( l l " t1 ) ' (3 .oo ' ) a

{ = lD8ro

An electronwith a speed f 5.00 x 108 mls enters n electric ieid of magnitude .00x 103N/C, -ttraveling along he fieid lines in the direction hat retards ts motion. (a) How far will flre electrontravel n the field before stopping momentarily and (b) how much ime will have elapsed? c) If theregion with the elec.tric reld is only 8.00 mm long (too short or the electron o stop within it), rn'hat

fraction of the electron's nitial kinetic energy will be lost in that region?

V : 5"oo^lobh/s

E -- l"oo loa N/// "

y-Xo= - (5.oo"tou.1s) '(- t' ?{,"to'e"s9

Vj* ),a (x-x") ---

tr/^:o

l4 l=

X- Xo: -u]/^o

\ D J

( c)

lql= l .?6*lo 'q 'e /s*-- ,

712 '* lo -e t t t

( i . r lx o-q ' t )

e = -1"?6r lo 'e r / s r

= e.e nx)?.4 nx)

r r f az- | l b Q r / 6

- 5 .oo { loLof5 f - x.gs* o-es- 1.11n"16't 7r'

4-'\t-

V'.-voa

AV._ .' t U , - \ .

A( - l.tr"" Io'l"r/s ") ( t .oo^ oa^)' ( l

tr=m=@c:)J".=*u:;fr=o#,fi'(d) t:;3J"1, at =E/n= o'T:;8;+'1 &'

P"obl"mi6 t- 3"oos \=t/a.aot^ J:e1.

/ l _ - 4

LLlo

DIL-z AfgC = 7a'n(av)"=-KF. 7e'"'1|

'/a tVo'