iar
Artr'..,^Physics48 (Fail 20II)
Chapter24: Electric potential
"There is nothing either good
or bad, but thinking makes it so." - w'liam
uo
:::"':ft:;irto
Shakespeare
vou thatduermineshowfaryou witt so in tife) it is howyou
handtewhat
Reading: pages628_646
Outline:
= electricpotentialenergyofa system
ofpoint charees
:= electncpotentialandpotential
difference
definition
electronvolt
potentialdueto point charses
potential from an electric dipole
lread on your own)
surfaces
(powerpoinrandLab)
]::,Tp:T"ti"j
= calcutatingthepotentialfrom the
electricfield
+ potential dueto continuouscharge
distributions
= calculatingthe electricfield fio_
,fr" por"rria
> potential ofa chargeisolatedconductor
Problem SolvingTechniques
To solvethe probremsof this chapter,
you shourdknow how to calculate
the electricpotentialof
point chargesandcontinuousdi-;il""*
= ql+rr"o,ro, singiepoint
usethe sum ofthe potentiarsdue to
chargeand
"?"it*g"lur"'i
individuJriruig"J;;
or
point
charges.
use
v,=(l/4ze)[(r/r)dqfor
a continuous
distribution'of " "o,recuon
Repracedq with ].dsfor a line of
"h*g".
chargeor with odA for a surfaceof
charge.you shouldknow how to compute
the work doneby the
;,iffi;fifJ"t;:fl:l,and
thewo'k('-*l;;;;;;;temar
:H:XT*$:\,:::#Jff:f_tl.re
agent
whenacharge
qismoved
energy
qyrenfar
oracorrection
orcharges
: t) = qrq2
/ 4nesr
j"?l,i,?,".ffi ,i|ffi:?JJl,xt;*l*:y$j#il.;h
ortn.poi*nriui";*,i
compute.the
ii:*::ff
changein rhepotenti; ."".cy
ffi;;;;;J"-ou"r.
you shouldknowiooi
to
int..p."t
f il%"T:T1,"#'i;;:f"*l."li$:*:g"n"il"'".t1,
$-qTH''i,l:l'fl
'r energyto computechanges
kinetic
energyofa movingcharg".
in the
askyouto skerchequipotenrial
surfaces.
It i
*I:1-rl.::
easv1odrawelectric
fieldlinesandrhensk"tcb,,rrfa"".thar
areperpend,.r,*,"irffilll.fairly
Questionsand ExampleProblemsfrom Chapter 24
Question 1
The figure below showsthreepatls alongwhich we canmovepositively chargedsphereA closerto
positivcly chargedsphereB, which is fixed in place. (a) Would sphereA be movedto a higher or
lower electricpotential?Is the work done(b) by our forceand (c) by the electric field (dueto the
secondsphere)positive,negative, or zmo?(d) Rank the pathsaccordingto the work our force does,
geatest first.
(q)
(b)
(.)
(r) cltA.l4p11y,.J.,
;J.
,/- td.\/U
'
l-.
Qt/r\44/V\D-<k ^p
^A
Question2
Thefigure below showsa thin, uniformly chargedrod andthreepoints at the samedistanced from
the rod. Without calculation,rank the magnitudeof the electricpotentialthe rod producesat these
threepoints, greatestfirst.
v-h"f +
-t1** olll '\'aLLo4'^h(tn']A
F"*., p*^oi p
\_/
V^
' '-1 tV*
"
tt z---{-
r.1z----f'- a--l
Ap'.
|
\)L-,/)
VU,
V,
tu't". J*^+...r 8 "4-tk0-,..,JB
,po." p-;J C_
Problem I
A particle of chargeq is fixed at point P, anda secondparticleof massm andthe samechargeq is
initially held a distancerl from P. The secondpanicle is thenreleased.Determineits speedwhenit
is a distance12from P. kt q = 3.1 pC, m = 20 mg, 11= 0.90mm, and 12: 2.5 rrun.
rl
P, c,
Vi =o
Ca= {-, = O"90*10-3r"r
ot_------>
q
f7
.^
tyr-n rtrr'*l"vd"--r4
a
K,
L
t)
t'*llfV
i
|
S
*, ) --* ri =
= 3.l"lo-oc
D , A ot t A - b K g
i
?'t' -
V;r,rvr!=
t"(t
il-= f.^= ).'5nlO-3,"r
l(9 = U;- t-?9
U.": l(+",10 *
:
V+=?
'.
.\ ^J
k
(*,
- -.-)
4 tt.f\
I
vi=
{ s'*!' ioqN'.'yc
a (3"1*!o-bc)?'
(!.c',,o-6 (i)
t
c.i,,\ - lit rr
^
F
d,>i
,:
-r-
t'J ) .
Problem 2
Two electronsarefixed 2.0 cn apart.Another electronis shotfrom infinity and stopsmidway
betweenthe two. What is its initial speed?
ta.
-----)
/m//,ttJ:
I
I
$t;.X
I
----->
V,; =V'6; =7;
'./
-
vlg'
\y/
a.{
--
,)
- 3L
2
\/
V".
/, ^yr'
E-,
a^'tt"r-fr^^
4"^'\
.
u,
o$ ttwq.^-4 ---e
oo
,
Art
'I
t/
,a
: +=
r,rv-'
,/.r.
z
1flt"
)(; . rs,
tVll \
U._=
) \)12,* U13+i^-
U': + U"r = /a mui
t o"
'\
t/C -
lr/
=
%'=%^=$z:-€
( SrlL * S]$z)
\
C,3
f^. ,/
[-,":1-;-=
- !,bo|x l5'1c
l,Ocr,r
m = q e l l r ,! o ^ z t l l 7 ,
------>
Vt = {, = 3.e "toa nfs
Problem 3
The electric potentialdifferencebetweenthe groundand a cloud in a particular thunderstomris
1.2 x l0' Y . What is the mnsnitudeof the changein the electricpotentialenergy(in multiples of the
elechon-volt)ofan electronthat movesbetweenthe groundandthe cloud?
-2 -a,Aoc,}^'**t/"r$r', >- +w\
/\V = l,?..lDlV
rr"D-
/fu
- l . 6 o ? . xi D ' ' c
_lq
=a
PhAta k*"4*
F"*^"trt!
.Lg Xfu clb,^,oa,)/ -'4+4etAil,rq
*
'&v +*r'"b^!
e- anJ
Jh V) -frr,n v,e,frt,ra
'.>
AV: ou/"tu
'
AD=9,AV
= (-")(t.a*toqv)
= - I"l* l91e-/
laulj@
\/-i
v)
AU = (-r.r,ot"iorqc)(t.e..loq
= - l.9lv 1p- rr:-
\
(,.--+"u
lA'-'l- l.ga,'!o-"=f..-":;FF;.J
:
rn4.
I n
i
l.a.Y:U
\v. / --
| a {
i.o':'..
-\:
eV
,^r..,
Problem 4
Considera point chargeq = 1.0 pC, point A at distanced1:2.0 m from q, andpoint B at distance
dz: 1.0m. (a) If A andB arediametically oppositeeachother,asin Fig. a below, what is the
electricpotentialdifferenceVa - Vg? O) What is that electricpotentialdifferenceif points A andB
are locatedas in Fig. b?
B
i
"
I
L
6 i
:-4-g.-.r.---.
il.fAqJ"t
\ cYare*
^wft'i
"
"
rr=--l%
-T
u 'ire-'
W
./-------.._:
(r)
(a)
-i")
=
(1.^
t)
,k
(^) Vo-V*= ;i('.^
Vo- Vo = (t.eer lorxr*;i"; ( l.o'.id"c) ( ;:^
= r,r't "5'. lolV J
(b) V*-Vo =\l[5";*;l
rUy'- ^'Y- P'p'
a--ma
-
*,.)
o 'r'Ld\"-'t)
v"' a'b'eA' oha
7a )fi'
V^-Ve
Problem 5
In the figure below, point P is at the centerof the rectangle.With V = 0 at infinity, Qr: 5.00 fC'
=
ez= 2.00 fC, 9: = 3.00fC, andd 2.54 cm, what is the net electricpotentialat P dueto the six
chargedparticles?
aE-
I
d
tl
l*-'J
-Itt
-42
+lt
'S
4g'L q
d ---4
rl =-i-a|{-
r= 1\g %)'
P;'t
&a^Aa "-
""49te^^
u^i ---;"1
q%'
9i
=
=
x
v 7.,78
ok
l1"
d
L
==.*o-It
+,1*
+?L
r_
F.
rf
leiFa^)qWry,'-\
L
^ n_L
-P-e-Yc'E^q d-td*A
*-*2,7=qi.f+ +l
U=(z,el*roqr'rV.")
| Hffi*
@
^
-&'tr-?-Y's.Y'1
L+
4nt4r(-
r ---1
t (A-oorlo'5c')
f D-oa54 h\)
ro^-QgpttU.
t
koblem 6
A long wire'hasa uniform lirrer darge dcnsityX..Wh is the potentialdifferenceVu5betvreentwo
points a andb locatedradial dishncesrumd rb = 2r" fiom the wire?
#/dL
h,/:6 Y,c$lg rrJ'JV
---=
L
zl
vt
1-
)
\
X
--
t^
?,,ti €- C
ft4
--t
^,I
e.]
F
--)
5
VO- V,
//
-)(.t
dr lt ta
t
i
q
fx
Vo- \', =
I
J.l€.,
N1rt.
---)- t^.
do
'x1rto
t
a
^ ) J.
,
/
"n
(9..o-'s" q ) =>
\/
Vq--
r/
\
Vb
-
'--
-lJtlv4
4r€o
gn1ft/r^)
\t
9tl {
Problea 7
The figine below, a plastic rod havinga uriformly distributedchargeQ = -25.6 pC hasbeenbelrt
into a circulr ac ofradius R = 3.71 cm andcentralangle120". With V = 0 at infnity, vihat is the
electricpotentialat P, the centerof curvatureofthe rod?
Y= 4r-',t5+
@
^4i,
: r = A: 3.?l c. $.".,"44 Fd''^l o.,4
)+ ux @l p,.ri{ r .r* 4 frt ^h?r*A
V=
|
/r
4ft.A ) *t'
5ft=Q=-As,{"pc
r r= -!V
-9
lTU4
(.o.o: rl * )
Problen 8
The figure below showsa thin pla$ic rod of lengthL : 12.0cm anduniform positive&ge
Q : 56.1pC lylng on an x axis.With V : 0 at infinity, fnd the electricpotentialat poid Psoa the
axis,at distanced: 2.50cm from oneentlofthe rod.
!
L-- l!'-or lo'r'.
,
/
r^ tr lY
6 = gb. lu iO-,tC
d = ].5onlo-'.
x
---
=
(-
..
!ru=u_.}-1
rf- /
qTL
/L+d
\
(-
q Tt-
X dr
x*J
.
d- r
i
=
r
4rt"
) dr
=
x*d
o
v= (7" ^
:>
qTt"
a ,/ = --l4 FL"
-d,
di= \d*
do
,
ln(x.J)1
-u\
IJ Tt"
o
(+)
-p^,
.= ( t.t: ^toqN.y'') (
\-{)
lL
A=*y't
t l'/.5*lotr"
56. l* lc-'lc \\n
t
Y
\
o
"
l X. {o-}- /
L ?.5qo'-
V=?-39V
Problem9
Thethin plastic rod of lengthL = 10.0cm in the figure below basa mtmiform linear chargedensity
L = cx, wherec = 49.9 pC/m. (a) With V : 0 at infinity, find tbe electricpotential at point P2on the y
axis, a distancey : D = 3.56 cm. (b) Find the electricfield componentF. xPr. (c) Why cannotthe \J
field componentE" at Pzbe fomd usingthe resultof (a)?
[-= J y'*y"
A=cx
c
+lrt"
,.
(
/
,
^
. r /*
.
(x'*y')
u
-v.
1
xdx
d u = /s(x1y')
{ y"*r,\
-=
L
rr=-9-(Ju,
v
+Tt" )
C
(r.".r=)"/
4rC
=
'v = (rti.l^ro'" y'^ (z.Zl, )DqN.nZ)
)
[
-+(
tTttu
f t ^p' E^, pun Mrc'lA l*Ah
(a,
X,/.=-'
/(*t-
{r'"y^ -) /
r-+t
,, .-?-.ra
,.err..-\a
(O.lori / J ( e'o32or'\)
\
-(D,a39v-
= (,-&^)
(b)trr=-,4t: hfr (f,r-r'-))
ft"
(.) ,utu+r'^*a V^^.M*.,1a,l) .
''
C, A7TN/c
V(x,r)+ frp,)
\',-
-
-- l3"rr""to'v
'\-/
\/ =-)y
aX
Problem 10
The electri^o
potential I/ n the spacebetweentwo flat parallelplates 1 and2 is given (in volts) by
V: 15001, whercx (in meters)is ttreperpendiculardistancefrom plate 1. At r : 1.3 cm, (a) what
is the magnitudeof the electric field and(b) is the field directedtoward or awayfrom plate 1?
V
= / Fooxa -a
a
-L$.r,"d!* ttru&u a4 V = ( tSoo Vmor)*
-%, (lsoc{/n,,';
- Jnl,
=
=
v'
ts
|
'o f
L-yE *=
- {3*r'/.+''r) X
(
o.t x = i.3cnq, JE'l = f 3 ocov/sr^,;
,} t" =,.:o-'*. )
(")
I
r_
lEtl-
( b) V -.-,-.h
*ra*4{
Problem 1l
^A -+
l\)JA-\g
lr-,.=,=--
t-
3.l r lo,v/m
t^-" p1o11 lr. p**1
-Ab^4 | '.,/.
.-D., $rrd.r,
v
-)
4
J-
I -a^;r*,F+.";*
S^"h)
/
.\
|' -l
l
"'
+r-,"
f'^-",4,t
Ja^""'r-*",1
_---
what is themagnitude
of theelectricfield at tb point 1r.ooi- z.ooj + 4.00$n if theelectric
potential is given by V : 2.00xy*, where Z is in volts andx, y, and,z arc in meters?
-JY"^ E-./- -)uA1 E_=-JYez
E^=
= -/^^ ( !.ooxy =^) -- -a"oo) e'
L-
- t,;"oxzl
Ey=-y'rr().Doxvz"):
Eu= -%= (a"**7.') = -?. ooxyz
X:3.oo
- -aaoo
)
7 = l"oo
.-/
|
E,
= -].oo(-f
.co)(?"oo)*
L,
-- )"oo ( 3.oo)(*4.o")a = * Q$.ovf p
E*
= - 4.oo (z-co)(-r'oo)(+"a>;,= 16.ov/n
(tq.ou/r)x + (q6"ov/^)e* (e6.ov7^rlt
lEovfn