Physics, 6th Edition
Chapter 25. Electric Potential
Chapter 25. Electric Potential Work and Electric Potential Energy 25-1. An positively charged plate is 30 mm aove a negatively charged plate, and the electric !ield intensity has a magnit"de o! 6 # 10$ %&C. 'o( m"ch (or) is done *+ the electric !ield (hen a $-µC charge is moved !rom the negative plate to the positive plate
Work = Fd = qEd; F opposite to d makes work negative. 30 mm
E
Fe
-
or) / $ # 10-6 C6 # 10$ %&C0.030 m Work = -.20 -.20 # 10-3 4
Work = -.20 -.20 m
25-2. n Prolem 25-1, ho( m"ch (or) is done % or against the electric !ield hat is the the
electric potential energy at the positive plate 30 mm
E
Fext
The displacement occurs against the electric force, so that
-
work done ! the field due to outside "# force F e$t in same direction as the displacement. e$t in %ince the field is in a position to do positive work when at the positive, plate, the electric potential potential energ& is positive positive at that that point' point'
E m. p / .20 m.
25-3. 7he electric !ield intensity et(een t(o t(o parallel plates, separated y 25 mm is 8000 %&C. 'o( m"ch (or) is done *+ the electric !ield in moving a 92-µC charge !rom the negative plate to the positive positive plate plate hat is the (or) (or) done *+ the !ield !ield in moving moving the the same charge charge
ac) to the the positive positive plate plate (Electric force with motion) 25 mm
E
Fe
Work done *+ field is positive, F e with displacement.
-
Work = qEd / 2 # 10-6 C8000 %&C0.025 m Work done *+ = oss of electric energ&. Work = $.00 # 10-$ Work done *+ field is negative'
%o(, in coming ac) electric !orce opposes motion. Work = - $.00 # 10-$ .
8:
Physics, 6th Edition
Chapter 25. Electric Potential
25-$. n Prolem 25-3, (hat is the potential energy (hen the charge is at a the positive plate and
the negative plate
E
25 mm
Fe
ememer, potential energ& represents the work that the
-
electric field can do when electric forces are free to act. When the /0 µC charge is at the positive plate, the E field can do no work, thus with reference to that point, E p = 1. (a) 2t 3 plate' E p = 0 !o work can e done & the electric field. -$ () 2t / plate, the field can do 3work' E p = $.00 # 10
25-5. hat is the potential energy o! a 6 nC charge located 50 mm a(ay !rom a 80-µC charge hat is the potential energy i! the same charge is 50 mm !rom a 980-µC charge #.E . =
k4q r
=
P
: # 10: % ⋅ m2 &C2 +80 # 10-6 C6 # 10−: C 50 mm
0.050 m
;80 µC
#.E . / 86.$ m k4q
#.E . =
r
=
: # 10: % ⋅ m2 &C2 −80 # 10-6 C6 # 10−: C 0.050 m
4
#.E. = -86.$ m
25-6. At (hat distance !rom a 9-µC charge (ill a 93-nC charge have a potential energy o! 60 m hat initial !orce (ill the 93-nC charge e#perience #.E. =
k4q
F =
k4q
r
2
r
4 r =
=
k4q #.E .
=
: # 10: % ⋅ m2 &C2 − # 10-6 C-3 # 10-6 C 0.060
: # 10: % ⋅ m2 &C2 − # 10-6 C-3 # 10-: C -3 2
3.15 # 10
4
r = 3.15 mm
F = 1:.0 %, rep"lsion
!ote' This value can also e otained from' F =
:0
4
#.E . r
Physics, 6th Edition
Chapter 25. Electric Potential
25-. A 8-nC charge is placed at a point #, $0 mm !rom a 12-µC charge. hat is the potential energy per "nit charge at point # in
is removed #.E . =
k4q r
=
$0 mm
: # 10: % ⋅ m2 &C2 +12 # 10-6 C8 # 10−: C 0.0$0 m
12 µC
#.E . / 0.0216 4 5 =
#.E . q
=
0.020 -:
8 # 10 C
4
5 / 2.0 # 106 &C 4
%o
The #.E.6q is a propert& of space.
7f another charge were placed there or if no charge were there, the #.E.6q is the same.
25-8. A charge o! 6 µC is 30 mm a(ay !rom another charge o! 16 µC. hat is the potential energy o! the system #.E . =
k4q r
=
: # 10: % ⋅ m2 &C2 +6 # 10-6 C16 # 10−6 C 0.030 m
4
#.E . / 28.8
25-:. n Prolem 25-8, (hat is the change in potential energy i! the 6-µC charge is moved to a distance o! only 5 mm s this an increase or decrease in potential energy P.E.30 / 28.8 from previous e$ample. !ow assume charge is moved. #.E .5
=
k4q r
=
: # 10: % ⋅ m2 &C2 +6 # 10-6 C16 # 10−6 C 0.005 m
Change in P.E. / 12. 9 28.8 4
4
P.E.5 / 13
Change / 1$$ , increase
25-10. A 93-µC charge is placed 6 mm a(ay !rom a -:-µC charge. hat is the potential energy s it negative or positive #.E . =
k4q r
=
: # 10: % ⋅ m2 &C2 −3 # 10-6 C-: # 10−6 C 0.006 m
:1
4
#.E . / $0.5
Physics, 6th Edition
Chapter 25. Electric Potential
25-11. hat is the change in potential energy (hen a 3-nC charge is moved !rom a point 8 cm a(ay !rom a 96-µC charge to a point that is 20 cm a(ay
s this an increase or decrease o! *
potential energy (8oves from 2 to * on figure.) #.E .8
k4q
=
r
=
: # 10: % ⋅ m2 &C2 −6 # 10-6 C3 # 10−6 C
P.E.8 / -2.025 , #.E . 20
8 cm
(!egative potential energ&)
: # 10: % ⋅ m2 &C2 −6 # 10-6 C3 # 10−6 C
=
A
0.08 m
0.20 m
9hange = final / initial / -0.810 9 -2.025 4
20 cm -6 µC
P.E.20 / -0.810 ,
Change in P.E. / 1.22 , increase
25-12. At (hat distance !rom a 9-µC charge m"st a charge o! 912 nC e placed i! the potential energy is to e : # 10-5 #.E. =
k4q r
4 r =
k4q #.E .
=
: # 10: % ⋅ m2 &C2 − # 10-6 C-12 # 10-: C -5
: # 10
4
r = 8.$0 m
25-13. 7he potential energy o! a system consisting o! t(o identical charges is $.50 m (hen their separation is 38 mm. hat is the magnit"de o! each charge #.E. =
k4q r
=
kq 2 r
4 q=
r #.E.
0.038 m0.00$5
k
: # 10: % ⋅ m2 &C2
=
4
q = 138 nC
Electric Potential and Potential Difference 25-1$. hat is the electric potential at a point that is 6 cm !rom a 8.$0-µC charge hat is the potential energy o! a 2 nC charge placed at that point 5 =
k4 r
=
: # 10: % ⋅ m2 &C2 8.$0 # 10-6 C 0.06 m
#.E. = q5 = 2 # 10-: C1.26 # 106 =4
:2
4
5 = 1.26 # 106 = P.E. / 2.52 m
Physics, 6th Edition
Chapter 25. Electric Potential
25-15. Calc"late the potential at point 2 that is 50 mm !rom a 9$0- µC charge. hat is the potential energy i! a 3-µC charge is placed at point A 5 =
k4 r
=
: # 10: % ⋅ m2 &C2 −$0 # 10-6 C 0.050 m
P.E. / >= / 3 # 10-6 C-.2 # 1064
4
5 / -.20 # 106 = P.E. / -21.6
25-16. hat is the potential at the midpoint o! a line
∑ k4r
(!et potential is algeraic sum)
5 =
k4
: # 10: % ⋅ m 2 &C2 −12 # 10-6 C
∑
r
=
0.0$0 m
5 / -2.0 # 106 = 0.65 # 106 =4
3 µC
-12 µC $0 mm
+
$0 mm
: # 10: % ⋅ m2 &C2 + 3 # 10-6 C 0.0$0 m
5 / -2.025 # 106 =4
5 = -0.10 ?=
25-1. A $5-nC charge is 68 mm to the le!t o! a 9:-nC charge. hat is the potential at a located $0 mm to the le!t o! the 9:-nC charge
5 =
∑
k4 r
=
: # 10: % ⋅ m2 &C2 $5 # 10-: C 0.028 m
-: nC
$5 nC
Find potential due to each charge, then add'
+
point
28 mm
$0 mm
: # 10: % ⋅ m2 &C2 − : # 10-: C
5 / 1$.5 # 103 = -2.025 # 103 =4
0.0$0 m 5 = 312.$ )=
@25-18. Points A and * are 68 mm and 26 mm a(ay !rom a :0-µC charge. Calc"late the potential di!!erence et(een points A and * 'o( m"ch (or) is done *+ the electric !ield as a A
µC charge moves !rom A to * 5 *
=
: # 10: % ⋅ m 2 &C 2 :0 # 10 -6C 0.026 m
;
=* / 3.115 # 10 =
:3
*
68 mm
26 mm
:0 µC
-5-
Physics, 6th Edition
Chapter 25. Electric Potential
@25-18. Cont.
5 2
=
: # 10: % ⋅ m2 &C2 :0 # 10-6 C 0.068 m
=A / 1.1: # 10 =4
-5 µC
;
=* / 3.115 # 10 =
5 * / 5 2 = 3.115 # 10 = 9 1.1: # 10 =4
∆= / 1.:2 # 10 =
A
*
68 mm
26 mm
:0 µC
!ote that the potential 7!9E2%E% ecause * is at a higher potential than 2 !ow for the field' or)A* / >=A - =* / -5 # 10-6 C1.1: # 10 = 9 3.11: # 10 =4 or) A* / :6.2 m4
The field does positive work on a negative charge.
@25-1:. Points A and * are $0 mm and 25 mm a(ay !rom a 6-µC charge. 'o( m"ch (or) m"st e done against the electric !ield y e#ternal !orces in moving a 5-µC charge !rom point A 5 µC
to point * 5 2
5 *
= =
: # 10: % ⋅ m2 &C2 6 # 10-6 C 0.0$0 m : # 10 % ⋅ m &C 6 # 10 C :
2
2
; 5 2 / 1.35 # 106 =
*
$0 mm
25 mm
6 µC
-6
0.025 m
A
6
; 5 * / 2.16 # 10 =
or)A* / >=A - =* / 5 # 10-6 C1.35 # 106 = 9 2.16 # 106 =4
Work 2* = $.05
!ote' The work *+ the field is negative, ecause the motion is against the field forces.
@25-20. A 6 µC charge is located at # / 0 on the #-a#is, and a -2-µC charge is located at # / 8 cm. 'o( m"ch (or) is done *+ the electric !ield in moving a 93-µC charge !rom the point # / 6 µC
10 cm to the point # / 3 cm 5 2
5 2
=
=Σ
k4 r
4
5 *
=Σ
k4
: # 10: % ⋅ m 2 &C 2 6 # 10 -6C 0.10 m
*
-2 nC
A
3 cm
$ = 1
5 cm
2 cm
r
+
: # 10 : % ⋅ m 2&C 2 −2 # 10 -6C 0.020 m
:$
; 5 2 = -360 )=
Physics, 6th Edition
Chapter 25. Electric Potential @25-20. Cont.
5 2 / -360 = 5 *
5 *
=
=Σ
6 µC
k4
* 3 cm 5 cm
$ = 1
r
: # 10: % ⋅ m2 &C2 6 # 10-6 C 0.030 m
+
A
-2 nC 2 cm
: # 10: % ⋅ m2 &C2 − 2 # 10-6 C 0.050 m
or)A* / >=A - =* / -3 # 10-6 C-360 )= 9 1$$0 )=4
; =* / 1$$0 )=
Work 2* = 5.$0
Challenge Problems 25-21. Point A is $0 mm aove a 9:-µC charge and point * is located 60 mm elo( the same charge. A 93-nC charge is moved !rom point * to point A. hat is the change in potential 2
energy
$0 mm #.E . 2
#.E . *
= =
k4q r k4q r
= =
−6
-:
−6
-:
: # 10 % ⋅ m &C −: # 10 C-3 # 10 C :
2
2
-: n9
0.0$0 m
60 mm
: # 10 % ⋅ m &C −: # 10 C-3 # 10 C :
2
2
*
0.060 m
P.E.A / 6.05 # 10-3 4
P.E.* / $.05 # 10-3 4
∆E p / 2.02 m4
-: µC
∆E p / 6.05 m 9 $.05 m
The potential energ& increases.
25-22. 7(o parallel plates are separated y 50 mm in air. ! the electric !ield intensity et(een the plates is 2 # 10$ %&C, (hat is the potential di!!erence et(een the plates E
=
5 d
4
5
= Ed = 2 # 10$ %&C0.05 m E 5 = 1000 =
:5
2 # 10$ %&C 50 mm
Physics, 6th Edition
Chapter 25. Electric Potential
25-23. 7he potential di!!erence et(een t(o parallel plates 60 mm apart is $000 volts. hat is the electric !ield intensity et(een the plates E =
5 d
=
$000 = 0.060 m
;
E = 66. )=&m
25-2$. ! an electron is located at the plate o! lo(er potential in Prolem 25-23, (hat (ill e its )inetic energy (hen it reaches the plate o! higher potential. hat is the energy e#pressed in electronvolts
Work done on electron equals its change in kinetic energ&.)
mv0 = qEd = 1.6 # 10-1: 66,00 =&m0.060 m / 6.$0 # 10-16 :.11 # 10-31 )gv0 = 6.$0 # 10-16 4
v / 3.5 # 10 m&s
25-25. Bho( that the potential gradient =&m is e>"ivalent to the "nit %&C !or electric !ield. 1
= 1 &C 1 % m
m
1=
1
% = 1 C
25-26. hat is the di!!erence in potential et(een t(o points 30 and 60 cm a(ay !rom a 950-µC charge 5 2*
=
*
∆E p / =A 9 =*
: # 10: % ⋅ m 2 &C2 −50 # 10-6 0.030 m
6
=A* / -1.50 # 10 = 9 -.50 # 10 4
−
: # 10: % ⋅ m2 &C2 − 50 # 10-6 0.060 m
∆=A* / -.50 # 10 6
A 30 cm
60 cm -50 µC
25-2. 7he potential gradient et(een t(o parallel plates $ mm apart is 6000 =&m. hat is the potential di!!erence et(een the plates 5 = Ed = 6000 =&m0.00$ m4
:6
5 = 2$.0 =
Physics, 6th Edition
Chapter 25. Electric Potential
25-28. 7he electric !ield et(een t(o plates separated y 50 mm is 6 # 105 =&m. hat is the potential di!!erence et(een the plates 5 = Ed = 600,000 =&m0.005 m4
5 = 3 )=
25-2:. hat m"st e the separation o! t(o parallel plates i! the !ield intensity is 5 # 10$ =&m and the potential di!!erence is $00 = 5
=
Ed 4
d=
5 E
=
$00= 50,000 =&m
4
d = 8.00 mm
25-30. 7he potential di!!erence et(een t(o parallel plates is 600 =. A 6-µC charge is accelerated thro"gh the entire potential di!!erence. hat is the )inetic energy given to the charge ∆ E k = Work = q5;
∆ E k = 6 # 10-6 C600 = );
∆ E k / 3.60 m
25-31. etermine the )inetic energy o! an alpha particle 2e that is accelerated thro"gh a potential di!!erence o! 800 )=. Dive the ans(er in oth electronvolts and
∆E) / or) / >=4 ∆E) / 2e8 # 105 =4
∆ E p = 1.60 ?e=
∆E) / or) / >=4 ∆E) / 21.6 # 10-1: C8 # 105 =4
∆ E p =2.56 # 10-13
25-32. A linear accelerator accelerates an electron thro"gh a potential di!!erence o! $ ?=, hat is the energy o! an emergent electron in electronvolts and in
∆ E k = q5= 1 e$ # 106 =4
∆ E k / $.00 ?e=
∆ E k = q5 / 1.6 # 10-1: $ # 106 =4 ∆E) / 6.$0 # 10-13 25-33. An electron ac>"ires an energy o! 2.8 # 10-15 as it passes !rom point A to point *. hat is the potential di!!erence et(een these points in volts
∆E) / >=4
1.6 # 10-1: C5 / 2.8 # 10-15 4
:
5 / 1.5 )=
Physics, 6th Edition
Chapter 25. Electric Potential
@25-3$. Bho( that the total potential energy o! the three charges placed at the corners o! the 3q
e>"ilateral triangle sho(n in ig. 25-11 is given yF
−kq 2
d
d
d -0q
First find the work required to ring the two 3qs together.
3q
d
Then add the e$tra work to ring the /0q charge to each q' #.E . =
∑
k4q r
=
kqq d
+
k −2 q q d
+
k −2 q q
4
d #.E . =
#.E . =
kq 2 d
−
2kq 2 d
−
2kq 2 d
=
3kq 2
−
d
−3q 2 d
@25-35. Ass"me that > / 1 µC and d / 20 mm. hat is the potential energy o! the system o! charges in ig. 25-11. #.E . =
−3q 2 d
=
−31 # 10-6 C2 4
P.E. / 1.50 # 10-10
0.020 m
@25-36. 7he potential at a certain distance !rom a point charge is 1200 = and the electric !ield intensity at that point is $00 %&C. hat is the distance to the charge, and (hat is the magnit"de o! the charge 5
E =
=
k4 r
=
1200 =4
k4 k 2 4 2 &1200 =2 4 =
r =
=
r
=
k4 1200 =
1200 =2 k4
4
4
E =
=
4
$00%&C
1200 =
/ $00 %&C
4 = $00 nC
: # 10: % ⋅ m2 &C2 $.00 # 10-: C
:8
=
: # 10: % ⋅ m2 &C2 4
: # 10: % ⋅ m2 &C2 $00 %&C
1200 =
r 2
1.$$ # 106 =2
1.$$ # 106 %&C
k4
k4
4
r = 3.00 m
Physics, 6th Edition
Chapter 25. Electric Potential
@25-3. 7(o large plates are 80 mm apart and have a potential di!!erence o! 800 )=. hat is the magnit"de o! the !orce that (o"ld act on an electron placed at the midpoint et(een these plates hat (o"ld e the )inetic energy o! the electron moving !rom lo( potential plate
to the high potential plate F
= qE =
q5 d
=
1.6 # 10−1: 8 # 105 = 0.080 m
Work = q5 =
e-
80 mm
-12
; F = 1.6 # 10 %
E
E k = 1.6 # 10-1: C8 # 105 =
∆E) / 1.28 # 10-13 Critical hinking Problems 25-38. Plate A has a potential that is 600 = higher than Plate * (hich is 50 mm elo( A. A 2-µC charge moves !rom plate A to plate * hat is the electric !ield intensity et(een the plates hat are the sign and magnit"de o! the (or) done y the electric !ield oes the potential energy increase or decrease %o( ans(er the same >"estions i! a 92-µC charge is moved !rom A to *
F
First find the field E etween the plates' E =
5 d
=
600 = 0.050 m
;
A
600 = 5 cm
E = 12,000 =&m, downward.
E
F
d
*
0=
When a positive charge moves with the field, the work is positive, since F and d are same. When a negative charge moves with E , the work is negative, since F and d are opposite a Work = q ∆5 = 2 # 10-6 C600 = 9 0 = 4
Work / 1.20 m, positive work
The field does work, so the 2-µC charge loses energ&; Work = q ∆5 = -2 # 10-6 C600 = 9 0 = ;
#.E. decreases
Work / -1.20 m, negative work
The field does negative work, so the -2-µC charge gains energ&;
::
#.E .increases
Physics, 6th Edition
Chapter 25. Electric Potential
25-3:. Point A is a distance $ = 3a to the right o! a $-µC charge. 7he right(ard electric !ield at point A is $000 %&C. hat is the distance a hat is the potential at point A hat are the electric !ield and the potential at the point $ / -a. ind the electric !orce and the electric potential energy (hen a 92-nC charge is placed at each point E 2
=
kq a2
4
a
2
=
kq E 2
=
: # 10: % ⋅ m2 &C2 $ # 10-6 C $000 %&C
a = :.00 m2 5 2
E *
=
=
kq a2
5 *
=
k4 a
=
=
*
=
2
q = 3$ µC
: # 10: % ⋅ m2 &C2 $ # 10-6 C 3.00 m
3.00 m2
a
$ = 3a
a / 3.0 m
4
: # 10: % ⋅ m2 &C2 $ # 10-6 C
k4
$ = 1
$ = -a
;
5 2 = 12.0 )=
E * = $000 %, to the left
: # 10: % ⋅ m2 &C2 $ # 10-6 C 3.00 m
F 2 = qE 2 = -2 # 10-: C$000 %&C4
4
F 2 / -8 # 10-6 %, leftward
(#.E.) 2 = q5 2 = -2 # 10-: C12,000 =4 F * = qE * = -2 # 10-: C-$000 %&C4
5 2 = 12.0 )=
P.E.A / 8.00 # 10-6
F 2 / 8 # 10-6 %, rightward
(#.E.) * = q5 * = -2 # 10-: C12,000 =4
P.E.A / 8.00 # 10-6
@25-$0. Points A, *, and C are at the corners o! an e>"ilateral triangle that is 100 mm on each side. At the ase o! the triangle, a 8-µC charge is 100 mm to the le!t o! a 98-µC charge. hat is the potential at the ape# C hat is the potential at a point
9 ?1 cm
3 µ 9 100
2
?1 cm
- µ 9
> cm
0 cm
*
Physics, 6th Edition
Chapter 25. Electric Potential
that is 20 mm to the le!t o! the 98-µC charge 'o( m"ch (or) is done y the electric !ield in moving a 2-µC charge !rom point C to point 5 9 =
5 >
=
: # 10: % ⋅ m2 &C2 8 # 10-6 C 0.10 m : # 10: % ⋅ m2 &C2 8 # 10-6 C 0.08 m
: # 10: % ⋅ m2 &C2 − 8 # 10-6 C
+
;
0.10 m : # 10: % ⋅ m2 &C2 − 8 # 10-6 C
+
0.02 m
Work)9> = q5 9> = 2 # 10-6 C G0 9 -2.0 # 106 =H
5 9 = 0
; 5 > = -2.0 ?=
(Work)9> = 5.$0
@25-$1. 7(o charges o! 12 and 96-µC are separated y 160 mm. hat is the potential at the midpoint A o! a line "al to Iero 5 2
=
: # 10: % ⋅ m2 &C2 12 # 10-6 C 0.08 m
5 *
=
+
: # 10: % ⋅ m2 &C2 − 6 # 10-6 C 0.08 m
: # 10: % ⋅ m2 &C2 12 # 10-6 C # 12 # 10-6 C $
=
+
5 2 = 65 )=
;
: # 10: % ⋅ m2 &C2 − 6 # 10-6 C
6 # 10−6 C
16 cm - #
=0
12 µC
16 cm - #
-6 µC 8 cm
8 cm
216 cm 9 # / #4 $ = 10. cm from the 12-µC charge
?@ cm - $
12 µC
-6 µC
$
@25-$2. or the charges and distances sho(n in ig. 25-12, !ind the potential at points A, *, and C 'o( m"ch (or) is done *+ the electric !ield in moving a 2-µC charge !rom C to A 'o( m"ch (or) is done in moving a 92-µC charge !rom * to A 5 2
5 *
=
=
: # 10: % ⋅ m 2 &C 2 −6 # 10 -:C 0.03 m : # 10: % ⋅ m2 &C2 −6 # 10-: C 0.0: m
+
+
: # 10 : % ⋅ m 2 &C 2 +$ # 10 -:C 0.03 m : # 10: % ⋅ m2 &C2 + $ # 10-: C 0.03 m 101
;
5 2 = -600 =
;
5 * = 600 =
Physics, 6th Edition
Chapter 25. Electric Potential
5 *
=
: # 10: % ⋅ m2 &C2 −6 # 10-: C 0.06 m
+
: # 10: % ⋅ m2 &C2 + $ # 10-: C 0.06 m
(Work)92 = q(5 2 / 5 9 ) = 2 # 10-6 CG-300 = 9 -600 =H
;
9
Work = 36 # 10-$
6 cm
(Work)92 = -2 # 10-6 CG600 = 9 -600 =H
5 9 = -300 =
6 cm
3A n9
-@ n9
Work = -2.$ # 10-3
: cm 2
: cm
: cm
@25-$3. 7he horiIontal plates in ?illi)anJs oil-drop e#periment are 20 mm apart. 7he diameter o! a partic"lar drop o! oil is $ µm, and the density o! oil is :00 )g&m3. Ass"ming that t(o electrons attach themselves to the droplet, (hat potential di!!erence m"st e#ist et(een the plates to estalish e>"iliri"m G =ol"me o! a sphere / $πK 3&3 H 5ol. =
$π 0.02 m3 3
= 33.5 # 10-18 m3 4
ρ
=
m 5
4
m = ρ 5
m / :00 )g&m333.5 # 10-18 m3 / 3.016 # 10-1$ )g4 qE = mg 4
5 =
5 = mg4 d
q
5
3.016 # 10-1$ )g:.8 m&s2 0.02 m -1:
21.6 # 10 C
102
mgd
=
;
q
5 / 18.5 )=
*
