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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

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