PHYS 1420: College Physics II
Fall 2009
Quiz 06: Chapters 15—17
1.
r
" q%
E1 = +$k 12 'iˆ
# r1 &
" 10 (10*6 C) %
r
9 N)m 2 (
'iˆ
E1 = + 9 (10 C2 $
2
$# (0.16m) '&
r
E1 = + 3.5 (10 6 N iˆ
Three charges are positioned as shown:
q1 = +10!C, q2 = –8.0!C, and q3 = +12!C.
(
0.12m
+10!C
!
0.16m
C
(
)
(
A)
)
r
# q &
E 2 = "%k 22 ( ˆj
$ r2 '
# 8 )10"6 C) &
r
9 N*m 2 (
( ˆj
E 2 = " 9 )10 C2 %
2
%$ (0.12m) ('
r
E 2 = " 5 )10 6 N ˆj
F1
+12!C
(
–8.0!C
F2
)
C
)
(6 points) Calculate the electrostatic force vector on q3, placed at the origin.
r
# qq &
F1 = "%k 1 2 3 ( ˆj
$ r13 '
# 10 )10"6 C)(12 )10"6 C) &
r
9 N*m 2 (
( ˆj
F1 = " 9 )10 C2 %
2
0.12m
%$
('
(
)
r
F1 = "( 75N) ˆj
(
!
)
r
# qq &
F2 = +%k 2 2 3 (iˆ
$ r23 '
# 8 )10"6 C)(12 )10"6 C) &
r
9 N*m 2 (
(iˆ
F2 = + 9 )10 C2 %
2
%$
('
(0.16m)
r
F2 = +( 33.8N)iˆ
r r r
F = F1 + F2 = 33.8iˆ " 75 ˆj N
!
B) (6 points) Determine the electric field vector at
(
)
(
r " q %"( 4 + ( 3 + %
E 3 = $k 32 '$* -iˆ + * - ˆj'
# r3 &#) 5 , ) 5 , &
" 12 .1006 C) %
(4+
9 N/m 2 (
' = +2.2 .10 6 NC
E 3x = +* - 9 .10 C2 $
2
)5,
0.20m
$# (
) '&
" 12 .1006 C) %
( 3+
9 N/m 2 (
' = +1.6 .10 6 NC
E 3y = +* - 9 .10 C2 $
2
)5,
$# (0.20m) '&
(
)
(
)
E x = E1 + E 3x = +5.7 "10 6 NC
)
E y = E 2 + E 3y = #3.4 "10 6 NC
r
E = +5.7iˆ # 3.4 ˆj "10 6 NC
(
the location (0.16m, 0.12m).
)
E3
!
P
E2
0.12m
+10!C
+12!C
E1
!
2.
–8.0!C
(3 points) How much work is required to completely
separate two charges q1 = —1.5!C and q2 = +3.0!C and
leave them at rest if they were initially 10mm apart?
q1q2
r
($1.5 "10$6 C)(+3.0 "10$6 C)
U=k
0.16m
(
U = 9 "10 9 N#m
C2
2
)
(0.01m)
U = 4.05J
Section 12989
Page 01
!
Quiz 06
3.
Fall 2009
(2 points) A 12-V battery is connected to a parallel
plate capacitor having an area A = 0.15m2 and a plate
separation of d = 3mm. How much charge can this
capacitor store?
C)
(3 points) Find the amount of charge stored on
each capacitor.
Q = CV
Q1 = (0.75µF)(12V) = 9µC
#" A&
Q = CV = % o (V
$ d '
(8.85 )10
Q=
C2
*12
N+m 2
Q2 = (0.20µF)(12V) = 2.4µC
)(0.15m )(12V)
Q3 = (1.05µF)(12V) = 12.6µC
2
(3 )10 m)
*3
Q = CV
Q = 5.31)10*9 C
4.
!
Q = (2.0µF)(12V) = 24µC
!
6.
(2 points) The same 12-V battery is now connected to
a capacitor having A = 0.15m2 and d = 3mm, but this
capacitor has a teflon dielectric (" =2.1) completely
filling the space between the plates. How much energy can this capacitor store?
(3 points) Copper wire has a room temperature resistivity #o = 1.7x10–8$%m. A particular wire is 10m long
and has resistance Ro = 25m$ at room temperature
! (20°C). What is the diameter of this particular wire?
"L "L
=
A #r 2
"L
r2 =
#R
"L
r=
#R
R=
$ # A'
U = CV = &" o )V 2
% d (
1
2
2
1
2
C
(2.1)(8.85 *10+12 N,m
1
2
U=
(3 *10 m)
2
+3
)(0.15m ) (12V)
2
2
2
U = 6.69 *10+8 J
5.
d = 2r = 2
The three capacitors shown are wired with a 12V battery. The capacitances are C1 = 0.75!F, C2 = 0.20!F,
and C3 = 1.05!F.
(1.7 $10
%8
!
d=2
7.
!
"L
#R
& ' m)(10m)
#(0.025&)
= 2.94 $10%3 m
(3 points) A 120-V air conditioner draws 15A of current. The unit runs for 15 minutes. How much energy, in kW!h, is consumed?
E = Pt
E = ( IV ) t
A)
# 1kW &
E = (15A)(120V)%
((0.25hr )
$1000V " A '
E = 0.45kWh
(2 points) Find the equivalent capacitance.
C = C1 + C2 + C3
C = (0.75µF) + (0.20µF) + (1.05µF)
!
C = 2.0µF
B)
(2 points) How much energy is stored?
U = 12 CV 2
!
U=
1
2
(2.0 "10 F)(12V)
#6
2
U = 1.44 "10#4 J
!
Section 12989
Page 02
Name:
Section:
12990
Quiz 06: Chapters 15—17
1.
Three charges are positioned as shown:
q1 = +8!C, q2 = —10.0!C, and q3 = +12!C.
P
0.12m
+8.0!C
+12!C
F2
F1
r
" q %)
E1 = +$k 12 'i
# r1 &
" 8 (10*6 C) %)
r
9 N)m 2 (
'
E1 = + 9 (10 C2 $
2 i
$# (0.16m) '&
r
)
E1 = + 2.8 (10 6 NC i
(
(
–10!C
0.16m
(6 points) Calculate the electrostatic force vector
on q3, shown at the origin.
r
# q q &)
F1 = "%k 1 2 3 ( j
$ r13 '
# 8 )10"6 C)(12 )10"6 C) & )
r
9 N*m 2 (
(j
F1 = " 9 )10 C2 %
2
%$
('
(0.12m)
r
)
F1 = "(60N) j
(
r
# q q &)
F2 = +%k 2 2 3 (i
$ r23 '
# 10 )10"6 C)(12 )10"6 C) &)
r
9 N*m 2 (
(i
F2 = + 9 )10 C2 %
2
%$
('
(0.16m)
r
)
F2 = +( 42.2N) i
!
r r r
)
)
F = F1 + F2 = ( 42.2 i " 60 j )N
B)
(
)
(
)
(
)
r
# q &)
E 2 = "%k 22 ( j
$ r2 '
# 10 )10"6 C) & )
r
9 N*m 2 (
(j
E 2 = " 9 )10 C2 %
2
%$ (0.12m) ('
r
)
E 2 = " 6.25 )10 6 NC j
!
A)
)
)
)
r " q %"( 4 + ( 3 + %
E 3 = $k 32 '$* -iˆ + * - ˆj'
# r3 &#) 5 , ) 5 , &
!
" 12 .1006 C) %
(4+
9 N/m 2 (
' = +2.2 .10 6 NC
E 3x = +* - 9 .10 C2 $
2
)5,
0.20m
$# (
) '&
" 12 .1006 C) %
( 3+
9 N/m 2 (
' = +1.6 .10 6 NC
E 3y = +* - 9 .10 C2 $
2
)5,
$# (0.20m) '&
(
)
(
)
E x = E1 + E 3x = +5 "10 6 NC
E y = E 2 + E 3y = #4.65 "10 6 NC
r
)
)
E = (+5 i # 4.65 j ) "10 6 NC
(6 points) Determine the electric field vector at
the location point P = (0.16m, 0.12m).
!
E3
!
0.12m
+8.0!C
+12!C
P
E1
2.
E2
(3 points) How much work is required to completely
separate two charges q1 = —1.5!C and q2 = +3.0!C and
leave them at rest if they were initially 15mm apart?
–10!C
q1q2
r
($1.5 "10$6 C)(+3.0 "10$6 C)
U=k
0.16m
(
U = 9 "10 9 N#m
C2
2
)
(0.015m)
U = 2.7J
!
Page 01
Quiz 06
3.
Fall 2009
(2 points) A 9-V battery is connected to a parallel
plate capacitor having an area A = 0.15m2 and a plate
separation of d = 4mm. How much charge can this
capacitor store?
C)
(3 points) Find the amount of charge stored on
each capacitor.
Q = CV
Q1 = (0.75µF)(9V) = 6.75µC
#" A&
Q = CV = % o (V
$ d '
(8.85 )10
Q=
C2
*12
N+m 2
Q2 = (0.20µF)(9V) = 1.8µC
)(0.15m )(9V)
Q3 = (1.05µF)(9V) = 9.45µC
2
Q = CV
(4 )10 m)
*3
Q = (2.0µF)(9V) = 18µC
Q = 2.99 )10*9 C
4.
!
6.
(2 points) The same 9-V battery is now connected to
a capacitor having A = 0.15m2 and d = 4mm, but this
capacitor has a teflon dielectric (! =2.1) completely
filling the space between the plates. How much
energy can this capacitor store?
!
(3 points) Copper wire has a room temperature
resistivity !o = 1.7x10–8#$m. A particular wire is 10m
long and has resistance Ro = 25m# at room
temperature (20°C). What is the diameter of this
particular wire?
"L "L
=
A #r 2
"L
r2 =
#R
"L
r=
#R
$ # A'
U = 12 CV 2 = 12 &" o )V 2
% d (
C
(2.1)(8.85 *10+12 N,m
1
2
U=
)(0.15m ) (9V)
2
2
(4 *10 m)
2
R=
+3
2
U = 2.82 *10+8 J
5.
The three capacitors shown are wired with a 9V
battery. The capacitances are C1 = 0.75"F, C2 = 0.20"F,
and C3 = 1.05"F.
d = 2r = 2
!
(1.7 $10
%8
d=2
7.
!
"L
#R
& ' m)(10m)
#(0.025&)
= 2.94 $10%3 m
(3 points) A 120-V air conditioner draws 15A of
current. The unit runs for 20 minutes. How much
energy, in kW!h, is consumed?
E = Pt
E = ( IV ) t
# 1kW & 1
E = (15A)(120V)%
( hr
$1000V " A ' 3
E = 0.60kWh
( )
A)
(2 points) Find the equivalent capacitance.
C = C1 + C2 + C3
C = (0.75µF) + (0.20µF) + (1.05µF)
C = 2.0µF
B)
!
(2 points) How much energy is stored?
U = 12 CV 2
!
U=
1
2
(2.0 "10 F)(9V)
#6
2
U = 8.1"10#5 J
!
Section 12990
Page 02