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Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

<strong>Solutions</strong> <strong>to</strong> <strong>Homework</strong> <strong>Questions</strong> 1<br />

Chapt15, Problem-1: A 4.5 x 10 –9 C charge is located 3.2 m from a –2.8 x 10 –9 C charge. Find<br />

the electrostatic force exerted by one charge on the other.<br />

Solution:<br />

Since the charges have opposite signs, the force is !attractive .<br />

The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get<br />

F = k e q 1 q 2<br />

#<br />

r 2 = 8.99 ! 10 9 N "m2 & ( 4.5 ! 10 )9 C) ( 2.8! 10 )9 C)<br />

%<br />

$<br />

C 2 (<br />

' 3.2 m<br />

( ) 2 = 1.1! 10"8 N<br />

Chapt15, Problem-3: An alpha particle (charge = +2.0e) is sent at high speed <strong>to</strong>ward a gold<br />

nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0 x 10 –14 m<br />

from the gold nucleus?<br />

Solution:<br />

Since the charges have opposite signs, the force is !repulsive<br />

The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get<br />

F = k e( 2 e) ( 79 e)<br />

r 2<br />

#<br />

= 8.99 ! 10 9 N "m2 &<br />

%<br />

$<br />

C 2 (<br />

'<br />

( ) 2<br />

( 2.0!10 )14 m) 2 =<br />

( 158) 1.60 !10 )19 C<br />

91 N ( repulsion)<br />

Chapt15, Problem-5: The nucleus of 8 Be, which consists of 4 pro<strong>to</strong>ns and 4 neutrons, is very<br />

unstable and spontaneously breaks in<strong>to</strong> two alpha particles (helium nuclei, each consisting of 2 pro<strong>to</strong>ns and<br />

2 neutrons). (a) What is the force between the two alpha particles when they are 5.00 x 10 –15 m apart, and (b)<br />

what will be the magnitude of the acceleration of the alpha particles due <strong>to</strong> this force? Note that the mass of<br />

an alpha particle is 4.0026 u.<br />

Solution:<br />

Since the charges have opposite signs, the force is !repulsive<br />

The magnitude so the force is given by Coulomb’s Law, so making the substitutions we get<br />

*<br />

(a) F = k e ( 2e)2<br />

r 2 = 8.99 !10 9 N "m2<br />

4 1.60 !10 )19 2<br />

#<br />

C &<br />

-<br />

#<br />

&<br />

, $<br />

' /<br />

+ ,<br />

./<br />

%<br />

$<br />

C 2<br />

(<br />

' 5.00 ! 10 )15 =<br />

2<br />

36.8 N<br />

#<br />

$<br />

m&<br />

'<br />

(b) The mass of an alpha particle is m = 4.0026 u , where u = 1.66! 10 "27 kg is the unified mass unit.<br />

Applying New<strong>to</strong>n’s 2 nd law, the acceleration of either alpha particle is then<br />

a = F m = 36.8 N<br />

4.0026 1.66 !10 "27 kg<br />

( ) = 5.54 !1027 ms 2<br />

Of course from New<strong>to</strong>n’s 3rd law, both alpha particles experience the same force, and hence undergo the<br />

same acceleration.<br />

1


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Problem-7: Suppose that 1.00 g of hydrogen is separated in<strong>to</strong> electrons and pro<strong>to</strong>ns.<br />

Suppose also that the pro<strong>to</strong>ns are placed at Earth’s North Pole and the electrons are placed at the South<br />

Pole. What is the resulting compressional force on Earth?<br />

Solution:<br />

1.00 g of hydrogen contains Avogadro’s number of a<strong>to</strong>ms, each containing one pro<strong>to</strong>n and one electron. Thus,<br />

each charge has magnitude q = N A e . The distance separating these charges is r = 2 R E , where R E is the<br />

Earth’s radius. Thus applying Coulomb’s Law,<br />

F = k e( N A e ) 2<br />

( 2 R E ) 2<br />

#<br />

= 8.99 ! 10 9 N "m2 & ( 6.02 !10 23<br />

)1.60 ! 10 )19 C<br />

%<br />

$<br />

C 2 (<br />

' 4 6.38 ! 10 6 m<br />

[ ( )] 2<br />

( ) 2 = 5.12 !105 N<br />

Chapt15, Problem-8: An electron is released a short distance above Earth’s surface. A second<br />

electron directly below it exerts an electrostatic force on the first electron just great enough <strong>to</strong> cancel the<br />

gravitational force on it. How far below the first electron is the second?<br />

Solution:<br />

The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus by<br />

using Coulomb;’s Law and New<strong>to</strong>n’s 2 nd Law (applied <strong>to</strong> gravity), we have<br />

k e e 2 r 2 = m e g<br />

so rearranging this expression, and then making the substitutions, we get<br />

r =<br />

( )( 1.60 !10 #19 C ) 2<br />

( )( 9.80 m s 2<br />

)<br />

k e e2<br />

m e<br />

g = 8.99 !10 9 N" m 2 C 2<br />

9.11 !10 #31 kg<br />

Chapt15, Problem-11: Three charges are arranged<br />

as shown in Figure P15.11. Find the magnitude and direction<br />

of the electrostatic force on the charge at the origin.<br />

Solution:<br />

In the sketch <strong>to</strong> the right, F R is the resultant of the<br />

forces F 6 and F 3 that are exerted on the charge at<br />

the origin by the 6.00 nC and the –3.00!nC charges<br />

respectively. Applying Coulomb’s Law <strong>to</strong> each, we get<br />

#<br />

F 6<br />

= 8.99 !10 9 N" m 2 & 6.00 !10 )9 C<br />

%<br />

$<br />

C 2 (<br />

' 0.300 m<br />

= 3.00! 10 )6 N<br />

#<br />

F 3 = 8.99 !10 9 N" m 2 &<br />

%<br />

$<br />

C 2 (<br />

'<br />

( )( 5.00 !10 )9 C )<br />

( ) 2<br />

( 3.00! 10 )9 C) 5.00 ! 10 )9 C<br />

( )<br />

0.100 m<br />

From the Superposition Principle, the resultant is<br />

F R = ( F 6 ) 2 #<br />

+ F 3<br />

%<br />

$<br />

( ) 2 = 1.38 !10 "5 N at ! = tan "1 F 3<br />

or F R = 1.38 !10 "5 N at 77.5° below " x axis<br />

= 5.08 m<br />

( ) 2 = 1.35! 10 )5 N<br />

F 6<br />

&<br />

( = 77.5° ,<br />

'<br />

F 6<br />

5.00 nC<br />

!<br />

F R<br />

F 3<br />

0.300 m<br />

0.100 m<br />

-3.00 nC<br />

6.00 nC<br />

2


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Problem-13: Three point charges are located<br />

at the corners of an equilateral triangle as in Figure P15.13.<br />

Calculate the net electric force on the 7.00 µC charge.<br />

Solution:<br />

The forces on the 7.00 µC charge are shown in the sketch <strong>to</strong><br />

the right. Applying Coulomb’s Law <strong>to</strong> calculate each force,<br />

we get<br />

#<br />

F 1<br />

= 8.99! 10 9 N "m 2 &<br />

%<br />

$<br />

C 2 (<br />

'<br />

= 0.503 N<br />

#<br />

F 2<br />

= 8.99 !10 9 N" m 2 &<br />

%<br />

$<br />

C 2 (<br />

'<br />

= 1.01 N<br />

From the superposition principle, we known<br />

!F x = ( F 1 + F 2 )cos 60.0°=0.755 N ,<br />

and !F y = ( F 1 " F 2 )sin 60.0°="0.436 N<br />

( 7.00! 10 )6 C) 2.00 ! 10 )6 C<br />

( )<br />

0.500 m<br />

( ) 2<br />

( 7.00 !10 )6 C) 4.00 ! 10 )6 C<br />

( )<br />

0.500 m<br />

( ) 2<br />

So the resultant force on the 7.00 mC charge is<br />

F R =<br />

or F R =<br />

60.0°<br />

( ) 2 $ #F<br />

= 0.872 N at ! = tan "1 y '<br />

& ) = "30.0° ,<br />

% #F x (<br />

0.872 N at 30.0° below the + x axis<br />

6.00 nC –3.00 nC<br />

0.600 m<br />

x<br />

( )q<br />

x 2 , or 2 x 2 = ( x + 0.600 m ) 2<br />

2 ! 1 = 1.45 m beyond the - 3.00 nC charge<br />

(!F x ) 2 + !F y<br />

Chapt15, Problem-16: A charge of 6.00x10 –9 C and a charge of<br />

–3.00x10 –9 C are separated by<br />

a distance of 60.0 cm. Find the position at which a third charge of 12.0 x 10 –9 C can be placed so that the<br />

net electrostatic force on it is zero.<br />

Solution:<br />

The required position is shown in the<br />

sketch <strong>to</strong> the right. Note that this places q<br />

closer <strong>to</strong> the smaller charge, which will<br />

allow the two forces <strong>to</strong> cancel. Applying Coulmob’s<br />

Law, and requiring that F 6 = F 3 gives<br />

k e ( 6.00 nC )q<br />

( x + 0.600 m ) 2 = k e 3.00 nC<br />

Solving for x gives the equilibrium position as<br />

x = 0.600 m<br />

y<br />

F 1<br />

+ 7.00 µC<br />

F 2<br />

x<br />

+<br />

2.00 µC 0.500 m<br />

–<br />

-4.00 µC<br />

F 3<br />

q<br />

F 6<br />

3


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Problem-19: An airplane is flying through a thundercloud at a height of 2 000 m.<br />

(This is a very dangerous thing <strong>to</strong> do because of updrafts, turbulence, and the possibility of electric<br />

discharge.) If there are charge concentrations of +40.0 C at height 3 000 m within the cloud and –40.0 C at<br />

height 1 000 m, what is the electric field E at the aircraft?<br />

Solution:<br />

We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions,<br />

one due <strong>to</strong> each concentration.<br />

The contribution due <strong>to</strong> the positive charge at 3000 m altitude is<br />

q<br />

E + = k e<br />

r 2 = #<br />

8.99 N" m 2 & ( 40.0 C)<br />

%<br />

$<br />

!109 C 2 (<br />

' 1000 m<br />

( ) 2 = 3.60 !105 NC ( downward)<br />

The contribution due <strong>to</strong> the negative charge at 1000 m altitude is<br />

q<br />

E ! = k e<br />

r 2 = $<br />

8.99 N# m 2 ' ( 40.0 C)<br />

&<br />

%<br />

"109 C 2 )<br />

( 1000 m<br />

( ) 2 = 3.60 "105 NC ( downward)<br />

From the Superposition Principle, the resultant field is then<br />

E = E + + E ! = 7.20 !10 5<br />

( )<br />

N C downward<br />

Chapt15, Problem-27: In Figure P15.27,<br />

determine the point (other than infinity) at which<br />

the <strong>to</strong>tal electric field is zero.<br />

Solution:<br />

If the resultant field is zero, the<br />

contributions from the two charges must<br />

be in opposite directions and also have<br />

equal magnitudes. Choose the line<br />

connecting the charges as the x-axis, with<br />

the origin at the –2.5!µC charge. Then, the<br />

two contributions will have opposite<br />

r 2 = 1.0 m + d<br />

directions only in the regions x < 0 and<br />

x > 1.0 m . For the magnitudes <strong>to</strong> be equal, the point must be nearer the smaller charge. Thus, the point of<br />

zero resultant field is on the x-axis at x < 0.<br />

Requiring equal magnitudes gives k e q 1<br />

r 1<br />

2<br />

= k e q 2<br />

r 2<br />

2<br />

or<br />

Thus, ( 1.0 m + d)<br />

2.5<br />

6.0 = d<br />

r 1 = d<br />

1.0 m<br />

+x<br />

E2 E 1 q 1 = –2.5 µC q 2 = 6.0 µC<br />

2 .5 µC<br />

d 2 =<br />

6.0 µC<br />

( 1.0 m + d) 2 .<br />

Solving for d yields d = 1.8 m , or 1.8 m <strong>to</strong> the left of the ! 2 .5 µC charge<br />

+y<br />

4


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Problem-28: Figure P15.28 shows<br />

the electric field lines for two point charges separated<br />

by a small distance. (a) Determine the ratio q 1 /q 2 . (b)<br />

What are the signs of q 1 and q 2 ?<br />

Solution:<br />

The magnitude of q 2 is three times the magnitude of q 1<br />

because 3 times as many lines emerge from q 2<br />

as enter q 1 . q 2 = 3 q 1<br />

(a) Then, q 1 q 2 =!1 3<br />

(b)<br />

q 2 > 0 because lines emerge from it,<br />

and q 1 < 0 because lines terminate on it<br />

Chapt15, Problem-35: If the electric field strength in air exceeds 3.0 x 10 6 N/C, the air becomes a<br />

conduc<strong>to</strong>r. Using this fact, determine the maximum amount of charge that can be carried by a metal sphere<br />

2.0 m in radius.<br />

Solution:<br />

For a uniformly charged sphere, the field is strongest at the surface.<br />

Thus, E max = k e q max<br />

R 2 ,<br />

or<br />

q max = R2 E max<br />

k e<br />

( )<br />

( ) 2 3.0 !10 6 NC<br />

= 2.0 m<br />

8.99 !10 9 N" m 2 C 2 = 1.3 !10 "3 C<br />

Chapt15, Problem-41: A 40-cm-diameter loop is rotated in a uniform electric field until the<br />

position of maximum electric flux is found. The flux in this position is measured <strong>to</strong> be 5.2 x 10 5 N · m 2 /C.<br />

Calculate the electric field strength in this region.<br />

Solution:<br />

From the definition of Electric Flux ! E = EAcos" and ! E = ! E, max when ! = 0°<br />

Thus, E = ! E, max<br />

A<br />

( )<br />

= ! E, max<br />

"d 2 4 = 4 5.2 #105 N$ m 2 C<br />

" 0.40 m<br />

( ) 2 = 4.1! 106 NC<br />

5


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Problem-57: Two 2.0-g spheres are suspended by<br />

10.0-cm-long light strings (Fig. P15.57). A uniform electric field is<br />

applied in the x direction. If the spheres have charges of –5.0 x 10 –8 C<br />

and +5.0 x 10 –8 C, determine the electric field intensity that enables<br />

the spheres <strong>to</strong> be in equilibrium at ! = 10°.<br />

Solution:<br />

The sketch <strong>to</strong> the right gives a free-body diagram of the<br />

positively charged sphere. Here, F 1 = k e q 2 r 2 is the attractive<br />

force exerted by the negatively charged sphere and F 2 = qE<br />

is exerted by the electric field.<br />

!F y = 0 " T cos10°=mg or T = mg<br />

cos 10°<br />

!F x = 0 " F 2 = F 1 +T sin10° or qE= k e q 2<br />

r 2<br />

+ mg tan10°<br />

10° y<br />

T<br />

x<br />

At equilibrium, the distance between the two spheres is r = 2( Lsin 10° ).<br />

Thus,<br />

E =<br />

k e<br />

q<br />

4( Lsin 10° )<br />

mg tan10°<br />

2<br />

+<br />

q<br />

F 1<br />

mg<br />

F 2<br />

( )( 5.0 !10 #8 C)<br />

= 8.99 ! 109 N"m 2 C 2<br />

4 ( 0.100 m)sin10°<br />

( ) 9.80 m s 2<br />

( 5.0 !10 #8 C)<br />

[ ] 2 + 2.0 !10 #3 kg<br />

and so the electric field strength required is E = 4.4 !105 NC .<br />

( )tan10°<br />

,<br />

Chapt15, Conceptual-2: Hospital personnel must wear special conducting shoes when they work<br />

around oxygen in an operating room. Why? Contrast with what might happen when personnel wear rubbersoled<br />

shoes.<br />

Solution:<br />

To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could discharge<br />

with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.<br />

Chapt15, Conceptual-12: Is it possible for an electric field <strong>to</strong> exist in empty space? Explain<br />

Solution:<br />

An electric field once established by a positive or negative charge extends in all directions from the<br />

charge. Thus, it can exist in empty space if that is what surrounds the charge.<br />

Chapt15, Conceptual-14: Would life be different if the electron were positively charged and the<br />

pro<strong>to</strong>n were negatively charged? Does the choice of signs have any bearing on physical and chemical<br />

interactions? Explain<br />

Solution:<br />

No. Life would be no different if electrons were positively charged and pro<strong>to</strong>ns were negatively charged.<br />

Opposite charges would still attract, and like charges would still repel. The designation of charges as<br />

positive and negative is merely a definition.<br />

6


Physics 112 <strong>Homework</strong> 1 (solutions) (2004 Fall)<br />

Chapt15, Conceptual-16: Why should a ground wire be connected <strong>to</strong> the metal support rod for<br />

a television antenna?<br />

Solution:<br />

The antenna is similar <strong>to</strong> a lightning rod and can induce a bolt <strong>to</strong> strike it. A wire from the antenna <strong>to</strong> the<br />

ground provides a pathway for the charges <strong>to</strong> move away from the house in case a lightning strike does<br />

occur.<br />

7

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