Solutions to Homework Questions 2
Solutions to Homework Questions 2
Solutions to Homework Questions 2
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Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> <strong>Questions</strong> 2<br />
Chapt16, Problem-1: A pro<strong>to</strong>n moves 2.00 cm parallel <strong>to</strong> a uniform electric field with E = 200<br />
N/C. (a) How much work is done by the field on the pro<strong>to</strong>n? (b) What change occurs in the potential energy<br />
of the pro<strong>to</strong>n? (c) Through what potential difference did the pro<strong>to</strong>n move?<br />
Solution:<br />
(a) The work done is W = F !s cos" = qE<br />
( )!s cos" , or<br />
W = ( 1.60 !10 "19 C) ( 200 N C) ( 2.00 !10 "2 m )cos 0°= 6.40 !10 "19 J<br />
(b) The change in the electrical potential energy is<br />
!PE e = "W = ! 6.40" 10!19 J<br />
(c) The change in the electrical potential is<br />
!V = !PE e<br />
= "6.40 #10 "19 J<br />
q 1.60# 10 -19 C = ! 4.00 V<br />
Chapt16, Problem-3: A potential difference of 90 mV exists between the inner and outer surfaces<br />
of a cell membrane. The inner surface is negative relative <strong>to</strong> the outer surface. How much work is required<br />
<strong>to</strong> eject a positive sodium ion (Na + ) from the interior of the cell?<br />
Solution:<br />
The work done by the agent moving the charge out of the cell is<br />
( )<br />
W input = !W field = !!"PE ( e )=+q "V<br />
= 1.60 !10 "19 #<br />
( C) + 90! 10 "3 J &<br />
%<br />
( =<br />
$ C'<br />
1.4 !10 "20 J<br />
Chapt16, Problem-7: A pair of oppositely charged parallel plates are separated by 5.33 mm. A<br />
potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field<br />
between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much<br />
work must be done on the electron <strong>to</strong> move it <strong>to</strong> the negative plate if it is initially positioned 2.90 mm from<br />
the positive plate?<br />
Solution:<br />
E = !V<br />
(a) d = 600 J C<br />
5.33 "10 #3 m = 1.13! 105 NC<br />
(b) F = q E= ( 1.60 !10 "19 C) ( 1.13 !10 5 NC)= 1.80! 10"14 N<br />
(c)<br />
W = F !s cos"<br />
= ( 1.80 !10 "14 N) [( 5.33 " 2.90)!10 "3 m]cos 0°= 4.38 !10 "17 J<br />
1
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
Chapt16, Problem-8: Suppose an electron is released from rest in a uniform electric field whose<br />
strength is 5.90x10 3 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm?<br />
(b) How fast will the electron be moving after it has traveled 1.00 cm?<br />
Solution:<br />
(a) !V = Ed = ( 5.90 "10 3 Vm)1.00 ( " 10 #2 m )= 59.0 V<br />
(b)<br />
1<br />
2 m ev 2 ! 0 = "KE = W = !"PE e = q ("V<br />
) ,<br />
( )<br />
( )( 59.0 J C)<br />
so v = 2 q !V<br />
= 2 1.60 "10 #19 C<br />
m e<br />
9.11 "10 #31 kg<br />
= 4.55! 106 ms<br />
Chapt16, Problem-15: Two point charges Q 1 = +5.00 nC and Q 2 = –3.00 nC are separated by<br />
35.0 cm. (a) What is the electric potential at a point midway between the charges? (b) What is the potential<br />
energy of the pair of charges? What is the significance of the algebraic sign of your answer?<br />
Solution:<br />
k e q<br />
V = !<br />
i<br />
(a) i<br />
r i<br />
#<br />
= 8.99 ! 10 9 N "m2 & # 5.00 !10 )9 C<br />
%<br />
$<br />
C 2 (<br />
' 0.175 m ) 3.00 !10 )9 C&<br />
%<br />
( =<br />
$<br />
0.175 m '<br />
103 V<br />
(b) PE = k e q i q 2<br />
r 12<br />
#<br />
= 8.99 ! 10 9 N "m2 &<br />
%<br />
$<br />
C 2 (<br />
'<br />
( 5.00! 10 )9 C) ) 3.00 ! 10 )9 C<br />
0.350 m<br />
( )<br />
= ! 3.85 "10 !7 J<br />
The negative sign means that positive work must be done <strong>to</strong> separate the charges (i.e., bring them up <strong>to</strong><br />
a state of zero potential energy).<br />
Chapt16, Problem-19: In Rutherford’s famous scattering experiments (which led <strong>to</strong> the<br />
planetary model of the a<strong>to</strong>m), alpha particles (having charges of +2e and masses of 6.64x10 –27 kg) were<br />
fired <strong>to</strong>ward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is<br />
fired at 2.00x10 7 m/s directly <strong>to</strong>ward the gold nucleus as in Figure P16.19. How close does the alpha particle<br />
get <strong>to</strong> the gold nucleus before turning around? Assume the gold nucleus remains stationary.<br />
Solution:<br />
From conservation of energy, ( KE + PE e ) f<br />
= ( KE+ PE e ) i<br />
, which gives<br />
( )( 2e)<br />
0 + k e Qq = 1 r f<br />
2 m ! v 2 i + 0 or r f = 2 k eQq<br />
2<br />
m !<br />
v = 2 k e 79e<br />
2<br />
i<br />
m !<br />
v i<br />
#<br />
2 8.99! 10 9 N" m 2 &<br />
%<br />
C 2 ( ( $<br />
'<br />
158 ) ( 1.60! 10)19 C ) 2<br />
r f =<br />
6.64 !10 )27 kg<br />
( )( 2.00!10 7 ms) 2 = 2.74!10 "14 m<br />
2
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
Chapt16, Problem-20: Starting with the definition of work, prove that at every point on an<br />
equipotential surface the surface must be perpendicular <strong>to</strong> the local electric field<br />
Solution:<br />
By definition, the work required <strong>to</strong> move a charge from one point <strong>to</strong> any other point on an equipotential<br />
surface is zero. From the definition of work, W = ( Fcos !)" s , the work is zero only if s = 0 or Fcos! = 0 .<br />
The displacement s cannot be assumed <strong>to</strong> be zero in all cases. Thus, one must require that Fcos! = 0 . The<br />
force F is given by F = qE and neither the charge q nor the field strength E can be assumed <strong>to</strong> be zero in<br />
all cases. Therefore, the only way the work can be zero in all cases is if cos! = 0 . But if cos! = 0 , then<br />
! = 90° or the force (and hence the electric field) must be perpendicular <strong>to</strong> the displacement s (which is<br />
tangent <strong>to</strong> the surface). That is, the field must be perpendicular <strong>to</strong> the equipotential surface at all points<br />
on that surface.<br />
Chapt16, Problem-25: An air-filled capaci<strong>to</strong>r consists of two parallel plates, each with an area of<br />
7.60 cm 2 , separated by a distance of 1.80 mm. If a 20.0-V potential difference is applied <strong>to</strong> these plates,<br />
calculate (a) the electric field between the plates, (b) the capacitance, and (c) the charge on each plate.<br />
Solution:<br />
E = !V<br />
(a) d = 20.0 V<br />
1.80" 10 -3 m = 1.11" 104 Vm= 11.1 kV m directed <strong>to</strong>ward the negative plate<br />
( )( 7.60" 10 #4 m 2<br />
)<br />
(b) C = ! 0 A = 8.85 " 10#12 C 2 N$ m 2<br />
d<br />
1.80 "10 -3 m<br />
= 3.74 !10 "12 F = 3.74 pF<br />
(c) Q = C (!V<br />
)= 3.74" 10 #12 F ( )= 7.47 "10 #11 C = 74.7 pC on one plate and ! 74.7 pC on the<br />
other plate.<br />
( ) 20.0 V<br />
Chapt16, Problem-28: A small object with a mass of 350 mg carries a charge of 30.0 nC and is<br />
suspended by a thread between the vertical plates of a parallel-plate capaci<strong>to</strong>r. The plates are separated by<br />
4.00 cm. If the thread makes an angle of 15.0° with the vertical, what is the potential difference between the<br />
plates?<br />
Solution:<br />
mg<br />
!F y = 0 " T cos15.0°=mg or T =<br />
cos15.0°<br />
or<br />
!F x = 0 " qE = T sin15.0°=mg tan15.0°<br />
mg tan15.0°<br />
E =<br />
q<br />
mgd tan15.0°<br />
!V = Ed =<br />
q<br />
( )( 9.80 m s 2<br />
) 0.0400 m<br />
!V = 350 "10 #6 kg<br />
30.0 " 10 #9 C<br />
( )tan15.0°<br />
15.0°<br />
T<br />
= 1.23 "10 3 V = 1.23 kV<br />
mg<br />
F = qE<br />
3
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
Chapt16, Problem-31: (a) Find the equivalent capacitance of the<br />
group of capaci<strong>to</strong>rs in Figure P16.31. (b) Find the charge on and the<br />
potential difference across each.<br />
Solution:<br />
Using the rules for combining capaci<strong>to</strong>rs in series and in parallel,<br />
the circuit is reduced in steps as shown below. The equivalent capaci<strong>to</strong>r is<br />
shown <strong>to</strong> be a 2 .00 µF capaci<strong>to</strong>r.<br />
4.00 µF<br />
3.00 µF<br />
6.00 µF<br />
3.00 µF<br />
2.00 µF<br />
a b c<br />
a b c<br />
a<br />
c<br />
2.00 µF<br />
12.0 V<br />
Figure 1<br />
(b) From Figure 3: Q ac = C ac !V<br />
12.0 V<br />
Figure 2<br />
( ) 12.0 V<br />
( ) ac<br />
= 2 .00 µF<br />
12.0 V<br />
Figure 3<br />
( )= 24.0 µC<br />
From Figure 2: Q ab =Q bc = Q ac = 24.0 µC<br />
Thus, the charge on the 3.00 µF capaci<strong>to</strong>r is Q 3 = 24.0 µC<br />
Continuing <strong>to</strong> use Figure 2, (!V) ab<br />
= Q ab<br />
24.0 µC<br />
=<br />
C ab<br />
6.00 µF = 4.00 V ,<br />
and (!V) 3<br />
= (!V) bc<br />
= Q bc<br />
24.0 µC<br />
=<br />
C bc<br />
3.00 µF = 8.00 V<br />
From Figure 1, (!V) 4<br />
= (!V) 2<br />
= (!V) ab<br />
= 4.00 V<br />
and Q 4 = C 4 !V<br />
Q 2 = C 2 !V<br />
( ) 4<br />
= ( 4.00 µF) 4.00 V<br />
( ) 2<br />
= 2.00 µF<br />
( ) 4.00 V<br />
( )= 16.0 µC<br />
( )= 8.00 µC<br />
Chapt16, Problem-44: Two capaci<strong>to</strong>rs C 1 = 25.0 µF and C 2 = 5.00 µF are connected in parallel<br />
and charged with a 100-V power supply. (a) Calculate the <strong>to</strong>tal energy s<strong>to</strong>red in the two capaci<strong>to</strong>rs. (b) What<br />
potential difference would be required across the same two capaci<strong>to</strong>rs connected in series in order that the<br />
combination s<strong>to</strong>re the same energy as in (a)?<br />
Solution:<br />
(a) When connected in parallel, the energy s<strong>to</strong>red is<br />
W = 1 2 C 1 (!V<br />
) 2 + 1 2 C 2 (!V<br />
) 2 = 1 (<br />
2 C 1 + C 2 )(!V) 2<br />
[ ] 100 V<br />
= 1 25.0 + 5.00<br />
2 ( )!10 "6 F<br />
(b) When connected in series, the equivalent capacitance is<br />
1<br />
C eq =<br />
25.0 + 1<br />
'1<br />
! $<br />
# & µF = 4.17 µF<br />
" 5.00%<br />
From W = 1 2 C eq !V<br />
!V = 2W<br />
C eq<br />
=<br />
( ) 2 = 0.150 J<br />
( ) 2 , the potential difference required <strong>to</strong> s<strong>to</strong>re the same energy as in part (a) above is<br />
2( 0.150 J)<br />
4.17 "10 #6 F = 268 V<br />
4
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
Chapt16, Problem-51: A model of a red blood cell portrays the cell as a spherical capaci<strong>to</strong>r—a<br />
positively charged liquid sphere of surface area A, separated by a membrane of thickness t from the<br />
surrounding negatively charged fluid. Tiny electrodes introduced in<strong>to</strong> the interior of the cell show a<br />
potential difference of 100 mV across the membrane. The membrane’s thickness is estimated <strong>to</strong> be 100 nm<br />
and its dielectric constant <strong>to</strong> be 5.00. (a) If an average red blood cell has a mass of 1.00x10 –12 kg, estimate<br />
the volume of the cell and thus find its surface area. The density of blood is 1100 kg/m 3 . (b) Estimate the<br />
capacitance of the cell. (c) Calculate the charge on the surface of the membrane. How many electronic<br />
charges does this represent?<br />
Solution:<br />
(a)<br />
V = m ! = 1.00 "10 #12 kg<br />
1100 kg m 3 = 9.09! 10"16 m 3<br />
Since V = 4!r3<br />
3<br />
(b)<br />
" 3V %<br />
, the radius is r = $<br />
# 4!<br />
'<br />
&<br />
A = 4!r 2 "<br />
= 4!<br />
3V %<br />
$<br />
# 4!<br />
'<br />
&<br />
C = ! " 0 A<br />
d<br />
= 5.00<br />
23<br />
( ) 8.85 !10 "12 C 2 N #m 2<br />
13<br />
, and the surface area is<br />
( )<br />
"<br />
= 4! 3 9.09 ( 10)16 m 3 %<br />
$<br />
'<br />
$ 4! '<br />
#<br />
&<br />
( )( 4.54! 10 "10 m 2<br />
)<br />
23<br />
= 4.54! 10"10 m 2<br />
100 ! 10 "9 =<br />
m<br />
2.01! 10"13 F<br />
(c) Q = C (!V)= ( 2 .01"10 #13 F) ( 100 "10 -3 V)= 2.01! 10 "14 C ,<br />
and the number of electronic charges is<br />
n = Q e = 2.01 !10 "14 C<br />
1.60 ! 10 -19 = 1.26! 105<br />
C<br />
Chapt16, Problem-54: Charges of equal magnitude 1.00x10 –15 C and opposite sign are<br />
distributed over the inner and outer surfaces of the cell wall in Figure P16.54. Find the force on the<br />
potassium on (K + ) if the ion is (a) 2.70 µm from the center of the cell, (b) 2.92 µm from the center, and (c)<br />
4.00 µm from the center.<br />
Solution:<br />
(a) At r = 2.70 µm from the cell center, the ion is located inside the spherically symmetric charge<br />
distributions on the surfaces of the cell wall. From Gauss’s law, the field is zero at such a location, so<br />
F = qE = q 0 ( )= 0 .<br />
5
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
(b) With spherically symmetric charge distributions, consider a spherical gaussian surface of radius<br />
r = 2.92 µm centered on the cell center. From Gauss’s law, the electric field at points on the gaussian<br />
surface is the same as if the <strong>to</strong>tal charge inside the surface, Q = !1.00" 10 !15 C , was located at the<br />
center. Thus,<br />
!<br />
F = qE = q k eQ $<br />
#<br />
" r 2 &<br />
%<br />
*<br />
= ( +1.60 ' 10 (19 !<br />
C) 8.99 '10 9 N )m 2 $ ((1.00 ' 10 (15 C)<br />
-<br />
,<br />
/<br />
#<br />
"<br />
C 2 &<br />
% ( 2.92 '10 -6 m) 2<br />
+ ,<br />
./<br />
F = !1.69" 10 !13 N<br />
or F = 1.69! 10"13 N directed radially inward<br />
(c)<br />
At r = 4.00 µm from the cell center, the ion is located outside a spherically symmetric charge<br />
distribution having zero net charge. From Gauss’s law, the field is zero at such a location, so<br />
F = qE = q 0 ( )= 0 .<br />
Chapt16, Problem-55: A virus rests on the bot<strong>to</strong>m plate of oppositely charged parallel plates in<br />
the vacuum chamber of an electron microscope. The electric field strength between the plates is 2.00x10 5<br />
N/C, and the bot<strong>to</strong>m plate is negative. If the virus has a mass of 1.00x10 –15 kg and suddenly acquires a<br />
charge of –1.60x10 –19 C, what are its velocity and position 75.0 ms later? Do not disregard gravity.<br />
Solution:<br />
The electric field is directed vertically downward with E y = !2.00 " 10 5 NC. Thus,<br />
a y = !F y<br />
m = qE y " mg<br />
m<br />
or a y =+22.2 m s 2<br />
At t = 75.0 ms = 0.0750 s ,<br />
v y = v yi + a y t = 0 + 22.2 m s 2<br />
( )("2.00#10 5 NC)<br />
= "1.60 #10 "19 C<br />
1.00 # 10 -15 kg<br />
( ) 0.0750 s<br />
( 22.2 m s2) 0.0750 s<br />
" 9.80 m s 2 ,<br />
( )= 1.67 m s upward , and<br />
( ) 2<br />
y = v yi t+ 1 2 a y t 2 = 0 + 1 2<br />
= 0.0624 m = 6.24 cm above bot<strong>to</strong>m plate<br />
Chapt16, Conceptual-5: Suppose you are sitting in a car and a 20 kV power line drops across<br />
the car. Should you stay in the car or get out? The power line potential is 20 kV compared <strong>to</strong> the potential<br />
of the ground.<br />
Solution:<br />
[Adapted from the solution in the text] The rubber (a good insula<strong>to</strong>r) tires on the car means that there is<br />
not good ‘electrical contact’ between the rest of the car (including occupants) and the Earth (ground). Thus<br />
if the power line makes electrical contact with the metal of the car, it will raise the potential of the car <strong>to</strong><br />
20kV. It will also most-likely raise the potential of your body <strong>to</strong> 20kV, because you are undoubtedly in good<br />
electrical contact with the car. In itself, this is not a problem – electrostatic equilibrium will be<br />
established, and charge will quickly cease <strong>to</strong> flow. However if you step out of the car, your body (with a<br />
potential of 20 kV) will make contact with the ground, which is at a potential of zero Volts (by definition)<br />
As a result, a current will pass through your body, and you are likely <strong>to</strong> be injured. Thus it is best <strong>to</strong> stay in<br />
the car until help arrives.<br />
6
Physics 112 <strong>Homework</strong> 2 (solutions) (2004 Fall)<br />
Chapt16, Conceptual-6: Why is it important <strong>to</strong> avoid sharp edges or points on conduc<strong>to</strong>rs used<br />
in high-voltage equipment?<br />
Solution:<br />
A sharp point on a charged conduc<strong>to</strong>r would produce a large electric field in the region near the point (see<br />
Section 15.6). Therefore an unintended electric discharge could take place at such a location.<br />
Chapt16, Conceptual-9: [Sorry, I did not spot the typo in the textbook –<br />
here is a rewording of the question] Why is it dangerous <strong>to</strong> <strong>to</strong>uch the terminals of a highvoltage<br />
capaci<strong>to</strong>r even after the voltage source that charged the battery capaci<strong>to</strong>r is disconnected? from the<br />
capaci<strong>to</strong>r? What can be done <strong>to</strong> make the capaci<strong>to</strong>r safe <strong>to</strong> handle after the voltage source has been<br />
removed?<br />
Solution:<br />
The capaci<strong>to</strong>r often remains charged long after the voltage source is disconnected. This residual charge<br />
can be lethal. The capaci<strong>to</strong>r can be safely handled after discharging the plates by short-circuiting the<br />
device with a conduc<strong>to</strong>r, such as a screwdriver with an insulating handle.<br />
Chapt16, Conceptual-15: If you were asked <strong>to</strong> design a capaci<strong>to</strong>r where small size and large<br />
capacitance were required. What fac<strong>to</strong>rs would be important in your design?<br />
Solution:<br />
You should use a dielectric-filled capaci<strong>to</strong>r whose dielectric constant is very large. Furthermore, you should<br />
make the dielectric as thin as possible, keeping in mind that it cannot be <strong>to</strong>o thin or else dielectric<br />
breakdown will occur.<br />
7