Electrostatics Answer Key

Transcription

Electrostatics Answer Key
Electrostatics
AP Physics C
- Forces, Fields and Gauss' Law
OBJECTIVES:
Upon the completion of this unit you should:
a) Understand the concept and source of electric charge and be able to calculate the magnitude and direction of the
force on charges due to other specified charges using Coulomb's law and/or electric fields.
c) Construct and interpret an electric field diagram and calculate the magnitude and direction of the electric field
produced by multiple charges s.
d) Analyze the motion of a charged pa,ticle in an electric field
e) Be able to use the principle of superposition to calculate by illtegration; the electric field of a straight, ulliformly
charged wire, at the cellter of a circular arc of charge, on the axis of a thill ring of charge, a uniformly charged disk.
parallel charged planes, coaxial cylinders, and cOllcentric spheres.
j) SWdelllS should understand Gauss's Law, so they call:
(1) State the 1mI' in integralform, and apply il qualitatively to relatejlux and electric charge for a specified
sUlface.
(2) Apply the law to determille the electric fieldfor a planar, spherical, or cylindrically symmetric charge
distribution and the charge dellsity or total charge 011 a surface in terms of the electric field near the sUlface.
TEXTBOOK READING:
Chapter 21 : Electric Charge and Electric Field
Chapter 22: Gauss's Law
II. Al'PROXIMA TE SCHEDULE
M 1/5
Charge and Force Exploration & Coulomb's Law
Tu 1/6
Forces on Charge Particles
W 1/7
Electric Fields of Charge P31ticles
Th 1/8
Life is not a point ch31'ge- Electric Field by Integmtion
F
1/9 Work on Integration Posters
M
Tu
W
Th
F
1/12
1/13
1/14
1/15
1/16
M 1/19
Tu 1/20
W 1/21
Th 1/22
F
1/23
Independent practice
Motion in an electric Field
Gauss' Law
)
Independent practice
Independent practice
7
~~
No School - Martin Luther King Day
FOI'ce and Field Homework Quiz
Gauss Law HomewOl'k Quiz, handout pre-test
Review
50.pt Quiz F01'CCSand Ficlds
III. ZERO HOUR LABS
F 1/9 Force on a pith ball Lab
*** on-line Charges and Field Lab
Ill. ASSIGNMENTS:
Force and Field Homework Quiz
Gauss Law Homework Quiz, handout pre-test
Force and Field Quiz
DUE DATES:
F 1/16
W 1/21
DUE DATES:
Tu
1/20
W
1/21
F
1/23
AP Advanced Physics
Coulomb s Law Written
I
1.
Name ----------Problems
Two small balls, each of mass 10 g, are attached to silk threads 1 m long and hung from a common point.
When the balls are given equal quantities of negative charge, each thread makes an angle of 20. with the
vertical.
(a) Draw a diagram showing all the forces on each ball.
y'Y1::
A:---~
.r( ~1~1
•
\...:= \
0\
I"h
e :-2-D
~\'J'+E
I'-J
("
o
'('-:..VlL.l
(b) Find the magnitude of the charge on each ball. (1.36xlO'6C)
i9f..n
0
1..0
j(
--
ct.:
_----(-2.
..J.
0
, (,OD?-O
C1
~
-:
v
_
2
\<-If,
\~.JAY\
1JJ<> -::
l~V~
fu-r
CZ:=
-J
-\:p..vI.2.c
0 {~l.~lt"l
( .o\) ("I. 'i) \
\
J
"I
-:: \.<6 S )(.\0''2.
OJ )CIe
2. A charge -3 x 10'9 C is placed at the origin of an xy-coordinate
the positive y-axis at y= 4 em.
(,
\
-:2 'f\ •..•~
f-=
q,2. ')
('2.. f"~
MrA
-
<>
0("
J
.l#"'"
1'2.
1.'6'S~IO- C
_'-
1.~Go\(IO
C
L....:(.,~
__
---system, and a charge 2 x 10'9 C is placed on
(J=
)
(a) If a third charge 4 x1O'9 C is now placed at the point (x= 3 em, y= 4 em), find the components of the total
force exerted on2his charge
the other two. (Fx=+5.41xlO'5N, Fy=-3.46xlO,5N)
_
_
~4
•.
2 ..•. ~
!(l1J;
:/:
-.
I
-?'l<.\O C
-'\
9.'2 :: '2. X \ 0 C I ~.
1.,!>:: '-\){\O-~C
\i I -
\
'L-
-
•cH
~e~~ ,0:>
!=',,\'\.L~-:: ~
oJ
I 0"'"
'?,
C-ei,S
= (q)(IO"l)(3YIO
_---_--,
~)(l.j)l\O
• oS
oy",
l.e:o
L.
}Y\CA
L-...:.....
~.l..\4>):IO-SN
(-Ij)l
_
')
~
')(-0,+)
_
3. Point charges of 2 x 10-9 C are situated at each of three corners of a square whose side is 0.20 m. What
9
would be the magnitude and direction of the resultant force on a point charge of -1 x 10- C if it were placed
(a) at the center of the square? (Fnet=9.05x10-7 N, 45°)
_...•
0.1.
-F~-'c= q.05)(IO N
l{
r): -:: {q •-1"'!>l' \ 0 -, N
F'j ~ (" .OS)C
Ie)
-,N
\
APAdvanced Physics
Name.
1. a. What is the strength and directian of the electric field at the position of point P in the
figure to the right?
so
t -= 2 l'-ct CP~e
~,,'
I .
_
(r:'l
5 Oem
f'2.
(
J..C~Xleo' ') (~x,\o' q)
CD.; L}S"":.
"I '515 N
"-I
(Hi
r\
~ 0/.)2..
Ie
.
)
$'"
b. If the 3 nC change on the bottom of the figure is replaced with a -3 nC charge what is the new strength
and direction of the electric field at point P?
~~
-
I
-0
2. A - 1.0 nC charge is located at the origin. Points I, 2, and 3 have (x,y) coordinates of (1 cm, 0 cm), (0 cm, 1 cm)
and (1 cm, 1 cm) respectively. Determine the electric field in vector notation at each of these locations and show
~
the vectors on an electric field diagram on the grid at the right.
_-Li_
--
1. E-:::(qY:\O"l)(ltIO-'i'j
-=:
-'1ojUOt.,t:) oj't-J1c
'10,000
-h~
Ie!!
(. 0\)1.-
.. j~
::i --
--
~_L~_
--
--
--
,
-"' ..
'1
E----
3.
'1...
~L.(.0\))
3. A 10.0 nC charge is located at position (x,y) = (1.0 cm, 2.0 cm). At what (x,y) positions is the electric field:
a. -225,000 i N/C?
-=
&)1:10"')
(Lo)((6'f)
('2-
'\ ~ .0'2. -:: ;;(",
CYv"\
b. (161,000 i + 80,500
(-;:.Je
c. (21,600
'II
, (p00
~'L~I~OO
"'.~
o
0( -
i-
J
-
I( -:..
I
CIa
I¥ 0
~Ol:'00
J ) N/C?
2.. • V-I
I 003
7- • 2. '-t ~
t
(p • £.,)
(2(".
5 ~e;,:> 01)""
Du,O~
.0 S
he 2.
(g ')
~
J ) N/C?
28,800
'f t>
\h
'I::
'?
s
CrY\.
'?:> \'
81""(5~. ,) -.: Y
J
.s
c..•...•.....•
z. C,V\'\.. L'
\
I- ..
y
+J.
~\ i. oL 'j
~
+
+
\
\
+
~
\
~
e
+
P
x
\
\,
c - ~>-tL
-x~,[X"'2.tc-
I
\()-L
x'2...•L:-
)(
s::~
'\ \(>-(-~ -+ 1.\"
~-+e- )<. )J -Q is
Iong-the-+x-oxis-1'raITl"J<=O-to-x-'-rNeg<ltlvec
}Jii+!-.~
J
5". APositive charge + is distributed uniform I
distributed along the -x-axis from x=o to x=-o. A positive point charge q lies on the positive y-axis, a distance y
from the origin. Find the magnitude and dire.-ctionof the force that the charge distribution exerts on q. Show
that this force approaches(~as
~
~'t
':1
"
~
-r=---EH~1!)
\
d-Q -= A cJ.y.
y becomes large.:.-)
t~ -:;
\C.Q"t
{2.
c-t>se -I:!
OJ co'oeJ
r'-
d (')~ Q cA.)'
~
0-
6. A plastic rod with linear charge density Ais bent into a quarter circle as show in the figure.
What is electric field at the origin?
oL'F -= ~
'0
~e
{L
c1t :. /<. A¥4~ MB
fJ
y.
11"".
0 r cesocW-
-S -::
~ J~
.::;..--
1-
r-: '6\,,&,\}
t~
\)- -1)( ~~
1:
Answers
la. 7.8x103 N/C, right
lb. same magnitude as above but down'
2-1. -90,000 i N/C
2-2. -90,000 J N/C
2-3. (-31,829 i -31,829 J ) N/C
3 a. (-I em, 2 em)
3 b. (3 em, 3 em)
3 e. (4 em, -2 em)
KQ
4.
2
i_kQ(l..
2
x.J x + L
2KQ[l..
5
.
a
y
KJ.. (0 0)
6 .-\I
+}
R
L
lx
1]
Ja2 + y2
~h
..--_--==--------------..,
Motion. in-an, Elea1"ric Field
Written Problems
2. An electron is projected into a uniform electric field of 5000 N/C. The direction of the field is vertically
upward. The initial velocity of the electron is 1.0 x 107 mIs, at an angle of 30 degrees above the horizontal.
(a) Find the maximumdistance the electron rises vertically above its initial elevation. (0.014 m)
I
~~::
('Vtme»(\.lFXIO-
')
9 .lIl{/O
V
_
-~
-s.~
y
(1.0 ;.Ie>
~
I
/--=. v,? ~1-0\.6't1
Q -.
«.1g>,IO'''i
[ \1.. ::0.01'+"" \
.j-
1-( - 'is' '25)£,1d"'i') 6}
.-IfLlL..-_--~-
(b) After wnat horizontal distance does the electron return to its original elevation? (0.10 m)
v~-;;.V i- +0-1I't.
V'f..'"-Y0'"
G~
- '2.( I 't.\O' 'lsi ('-aD')
10"
\
em, 1lt» (1.\4lOD-~)
f:,. Y.
-=- Q. 10
'fY"
\ . V-lit
-~
S
10
-8' .1&' )(.lD~
\
(c) Sketch the trajectory of the electron.
2. An electron is launched at a 45° angle and a speed of 5.0x106m/s from the positive plate of a parallel-plate
capacitor. The electron lands 4.0 cm away.
I
_. _. _. _
3
a. What is the electric field strength inside the capacitor? (3.6x10 N/C)
j
Y' ...
o
"
1:=:="
b. What is the smallest possible spacing between the plates? (1.0 cm)
'" •
_. ~•...
j
-l+T=~=/
~r
L
.
eY~
i
VerticaII y
~B
i\====+
d
Up
v
2. An electric field E exists in the region between the two electrically charged parallel plates shown above. A beam of electrons of
mass //1,charge q. and velocity venters the region through a small hole at positionA. The electrons exit the region between the
plates through a small hole at position B. Express your answers to the following questions in terms of the quantities//1. q. E, and
v. Ignore the effects of gravity.
a.
e,
i.
On the diagram of the parallel plates above, draw and label a vector to show the direction of the eectric field E
between the plates.
ii.
On the following diagram, show the direction of the force(s) acting on an electron after it enters the region between the
plates.
iii. On the diagram of the parallel plates above, show the trajectory of an electron that will exit through the small hole at
position B.
b.
Detennine the magnitude of the acceleration of an electron after it has entered the region between the parallel ~ates. (Eq/m)
c.
Detelmine the total time that it takes the electrons to go from position A to position B. (2mvsin8/Eq)
V+- ""'- V ~-I--CAf::. 't
i'V'
~{; = :u-n V'1rl\c\.Gd.
Determine the distance ~etwee~~tions
vt -= ViIt bt-
A and B. (vcos8)(2mvsin8/Eq)]
::: (V C 0,& f:J)
(ZAV....V 'in A
C't
e.
t:> t
- V %i "'8- -:::.
\J~ N27 - ~
Q:. \
)
Now assume that the effects of gmvity cannot be ignored in this problem. How would the distanced change for an electron
entering the region at A and leaving at B? Explain your reasoning.
~
cxGV'tt-cA./\ (.e
~
~.
~
CU:,c.rV..ifR.;h~
~n..,
Law
Name
_
AP Advanced Physics
>.
A conducting spherical shell of inner radius a and outer radius b has a positive point charge Q located at its
center. The total charge on the shell is -3Q, and it is insulated from its surroundings.
\
(a) Derive the expressions for the electric-field magnitude in terms the distance r from the cent~ for the
/
at-
regions:
i)r<a
6 .d..A -=
_Cf.,.f.V\-
,&:::
,
~
t; -::LflTr"2.
ii)
:::~
1~
'Eo
a <r <b
1
liI)r>b
(b) What is the surface charge density on the inner surface of the conducting shell?
-Q
0=
G
V
(c) What is the surface charge density on the outer surface of the conducting shell?
o -:: -LQ
S....fIT
-3
Ov?>
(d) On picture above, draw a sketch showing electric-field lines and the location of the charges.
(e) Draw a graph
4W-
01-.E as a function
of r.
E(r)
.,l;)
a
b
2. A long coaxial cable consists of an inner cylindrical conductor of radius aand outer conducting coaxial cylinder
of inner radius b and outer radius c. The outer cylinder has no net charge. The inner cylinder has a uniform
positive charge +Aper unit length. Calculate the electric field
a) at;y
(~~e~J)he_CYlintl
'Go
b) at any external point.
Cross section of
cylinders
c) Sketch a graph of the magnitude of E as a function of the distance r from the axis of the cable, from r = 0
tor=2c.
E(r)
I'••'*AAoo.,
Ivy? if
a
b
I
c
>
2c
d) Find the charge per unit length on the inner surface of the outer cylinder, and on the outer surface. Explain
your reasoning.
+
A
A
3. (200BE1). A metal sphere of radius a contains a charge +Q and is surrounded by an
'ncharged, concentric, metallic shell of inner radius b and outer radius c, as shown
Jbove. Express all algebraic answers in terms of the given quantities and fundamental
constants.
(a) Determine the induced charge on each of the following and explain your
reasoning in each case.
i.
-Q
The inner surface af the metallic shell
-+
ii. The outer surface of the metallic shell
Q
(b) Determine expressions for the magnitude of the electric field E as a function of r, the distance from
the center of the inner sphere, in each of the following regions.
i.
r<a
ii.
a<r<b
iii.
b<r<c
iv.
c<r
t::
v-=-'O
5~O
-S
(c) On the axes below, sketch a graph of E as a function of r.
/oe.
fcl(
00
E
r
r
r
r
I
r
r
~
=V~«,.
I
o
c
Ii
a
)J':
~
(d) An electron of mass me carrying a charge -e is released from rest at a very large distance from the
spheres. Derive an expression for the speed of the particle at a distance 10c from the center of the
spheres.*
\N::
"-
u
\<.
\< Qe.
D
-= ~2 YY\
*= I n th e or'gma
.. I prmte
. d vers,on,
. t h e d',stance1wasc.given as "1 0"
r .
V'L.
e
- )
V
-=0~~
'OCr<'-~
:O::~g,
,,,I,,, ,h,OJ'dmmty ,
,hi" 'ph,,;~1,h,IIh", "If".
cha~ge/anywhere else inside. ~
00
I" ""Id,
\
,",I", ood .,
)
I
(~ Jse Gauss's lawj9.-p o~e that the~le tric field inside the shell is Zero ever where. t5eseribe-the~
~ace
that you use./
/
(
/
/
N
\
/
/
'
j
(b) The charges are now redistributed so that the surface chaJ'g density is no Ion er uniform. Is the
electric field still zero/ever/where inside the shell?
Yes
No /
Justify YOU~7r'
V
It cannot be determined fro
G
\
Now consider a small conducting sphere with charge •. wh Se center is at corner
A of a cubical surface, as
shown below.
/,
~
Bf'----
/
!
I
A
",C
I
'L_
D
(
-=
: >,F
! r-=Tx'lT
"~-r----_I" : I
I
/
"G
'H
I
L
---~u.:l
~
E
)
.
(c) For which faces of the surfacerif arl(~is.the electric flux through that face equal to zero?
_ABCD
__ CDEF
Explain your reasoning.
\
EFGH
\
1_ BCFG
~BGj:(
_ADEH
/
(d) At which corner(s) of the surface doeS th electric field have the least magnitude?
/
/
(e) Determine the ectric field strength at he position(s) ou have indicated in part Cd)in terms of Q, L,
and fundame al constants, as appropriate.
(f) Given that one-ei hth of the sphere at point A is inside the surface, calculate the electric flux
through face CDE .
/
'1\
- p
5. The solid nonconducting cylinder of radius R shown above is very long. It contains a negative charge evenly
distributed throughout the cylinder, with volume charge density p. Point PI is outside the cylinder at a distance
1', from its center C and point Pz is inside the cylinder at a distance rz from its center C. Both points are in the
same plane, which is perpendicular to the axis of the cylinder.
a.
On the following cross-sectional
diagram, draw vectors to indicate the directions ofthe electric field at points P,
and Pz.
b.
Using Gauss's law, derive expressions for the magnitude of the electric field E in terms of
1',
constants for the following two cases.
i. I' > R (outside the cylinder)
(;.cJ.A ~
1.€rv
'C,o
ii. I' < R (inside the cylinder)
]
R. p, and fundamental
"'
,,
•..•..------"
,,
,,
,
/
/
,
,,
I
I
:
RIP
" ' ... _--_ ..••..
,
,
,,
,,
/
/
6. A nonconducting sphere with center C and radius a, has a spherically symmetric electric charge density. The total
charge of the object is Q > O.
a. Determine the magnitude and direction of the electric field at point P, which is a distance R> a to the right of the
2)
sPhere's(ceLn~~
5
b.
711' ("
P
J:!-yro-3
::-;;>
17
~)
'!. 0
=
Po ..}
=-0..----
~
'3 ~or'2.
V
q
-
~
t-l1r 'Go
f"2.
.
)"r
Determine the flux of the electric field through the spherical surface centered at C and passing through P.
,,
/
----
•....•
.•.• ,
,
I
,,
,
\
:
.c'n'ti~
I
~~~~
\
'l;i~'.
,'
Q
,,
/
"
Ip
d';:.7;.'
.•.•
,-r
R
/
_--_... •..
'"
/
~
, .-l
-Q
A point pal1icle of charge -Q is now placed a distance R below point P. as shown above.
c. Determine the magnitude and direction of the electric field at point P.
Of
"f -
E,;\'\..0t'
r-
""f;::"'
=-
V +Cf... -t v - 't
-z.
(Q \7.
+ ~~e2)
.
-= _,_
(" 4rr'Co
\
,/ C!k~,j
.
.,
•••
d.
Now consider four point charges, q" q" q), and q4, that lie in the plane of the page as shown in the diagram above.
Imagine a three-dimensional closed surface whose cross section in the plane of the page is indicated.
i. Which of these charges contribute to the net electric flux through the surface?
Z
1... ~
cz. c:
ii. Which of these charges contribute to the electric field at point PJ?
e.
iii. Are yov:;;swer;;11nd
ii~e
If the ~ge
your answer.
a ~~~~es
e~s~y
Y\O
r
~
thi~t~
'oL~
evJ 81oLe. ~
f.
0c:~ent?
~
~
c:x.JLi
~~0-e
fiel~~~
~
hl ~-clc{
~~
bJL
the:acetl~~'
C-~
J'c«-e .
lfthe field is zero at all points on a surface, does this mean there is no net charge enclosed by the surface? JustiIY your
answer.
S
~
E:: 0 ~
5te
=()