Física para ciencias e ingenieria serway 7ed vol 2 (solucionario)

The electric field at point P is due to charges other than the test charge.. The net force on the test charge, and hence the electric field at this location, must then be zero.. The forc

LIBROS UNIVERISTARIOS

Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS

DE FORMA CLARA VISITANOS PARA

Electric Forces and

Electric Fields

Quick Quizzes

1. (b) Object A must have a net charge because two neutral objects do not attract each other Since object A is attracted to positively-charged object B, the net charge on A must be negative

2. (b) By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other

3. (c) The electric field at point P is due to charges other than the test charge Thus, it is

unchanged when the test charge is altered However, the direction of the force this field exerts on the test change is reversed when the sign of the test charge is changed

4. (a) If a test charge is at the center of the ring, the force exerted on the test charge by charge

on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring The net force on the test charge, and hence the electric field at this location, must then be zero

5. (c) and (d) The electron and the proton have equal magnitude charges of opposite signs The forces exerted on these particles by the electric field have equal magnitude and

opposite directions The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton

6. (a) The field is greatest at point A because this is where the field lines are closest together The absence of lines at point C indicates that the electric field there is zero

7. (c) When a plane area A is in a uniform electric field E, the flux through that area is

cos

E EA θ

Φ = where θ is the angle the electric field makes with the line normal to the

plane of A If A lies in the xy-plane and E is in the z-direction, then θ = ° and 0

10. (b) and (d) Since the net flux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b) Statement (d) must be true because there would be a net flux through the surface if more lines entered the surface than left it (or vise-versa)

Answers to Even Numbered Conceptual Questions

2. Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is

enhanced by the oxygen

4 Electrons are more mobile than protons and are more easily freed from atoms than are protons

6. No Object A might have a charge opposite in sign to that of B,

but it also might be neutral In this latter case, object B causes

object A to be polarized, pulling charge of one sign to the near

face of A and pushing an equal amount of charge of the

opposite sign to the far face Then the force of attraction

exerted by B on the induced charge on the near side of A is

slightly larger than the force of repulsion exerted by B on the

induced charge on the far side of A Therefore, the net force

on A is toward B

++

++++ +++

16. The antenna is similar to a lightning rod and can induce a bolt to strike it A wire from the antenna to the ground provides a pathway for the charges to move away from the house

in case a lightning strike does occur

18 (a) If the charge is tripled, the flux through the surface is also tripled, because the net flux

is proportional to the charge inside the surface (b) The flux remains constant when the

volume changes because the surface surrounds the same amount of charge, regardless of

its volume (c) The flux does not change when the shape of the closed surface changes (d)

The flux through the closed surface remains unchanged as the charge inside the surface is

moved to another location inside that surface (e) The flux is zero because the charge

inside the surface is zero All of these conclusions are arrived at through an understanding

of Gauss’s law

20. (a) –Q (b) +Q (c) 0 (d) 0 (e) +Q (See the discussion of Faraday’s ice-pail experiment in the

textbook.)

22 The magnitude of the electric force on the electron of charge e due to a uniform electric

field G

is Thus, the force is constant Compare this to the force on a projectile of

mass m moving in the gravitational field of the Earth The magnitude of the gravitational force is mg In both cases, the particle is subject to a constant force in the vertical direction

and has an initial velocity in the horizontal direction Thus, the path will be the same in each case—the electron will move as a projectile with an acceleration in the vertical

direction and constant velocity in the horizontal direction Once the electron leaves the region between the plates, the electric field disappears, and the electron continues moving

in a straight line according to Newton’s first law

E F = E e

Answers to Even Numbered Problems

14. x =0.634d, stable if third bead has positive charge

16. 1.45 m beyond the –3.00 nC charge

18 (a) 2.00 10 N C to the right× 7 (b) 40.0 N to the left

15.4 The attractive forces exerted on the positive charge by

the negative charges are shown in the sketch and have

1.91 e along the diagonal toward the nega

R

k q a

F a

= =

×

15.6 The attractive force between the charged ends tends to compress the molecule Its

magnitude is

2 19

4.89 10 N

2.25 10 N m2.17 10 m

F x

15.7 1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton

and one electron Thus, each charge has magnitude q =N e A The distance separating

these charges is r=2R E, where is Earth’s radius Thus,

( )

E R

2 2

9

10

e A E e

e e

k e r

15.9 (a) The spherically symmetric charge distributions behave as if all charge was located

at the centers of the spheres Therefore, the magnitude of the attractive force is

(b) When the spheres are connected by a conducting wire, the net charge

will divide equally between the two identical spheres

Thus, the force is now

e

k q q F

e

k q q F

e

k q q F

15.11 In the sketch at the right, is the resultant of the

forces that are exerted on the charge at

the origin by the 6.00 nC and the –3.00 nC charges

respectively

R F

2 6

N mC

15.12 Consider the arrangement of charges shown in the sketch at

the right The distance r is

7

6.00 10 C 2.00 10 C

N m8.99 10

6.00 10 C 3.00 10 C

N m8.99 10 3.24

2 9

7.00 10 C 2.00 10 C

N m8.99 10

0.503 N

7.00 10 C 4.00 10 C

N m8.99 10

15.14 Assume that the third bead has charge Q and is located at 0 x d< < Then the forces

exerted on it by the +3q charge and by the +1q charge have magnitudes

This is a position of stable equilibrium if Q >0 In that case, a small displacement from

the equilibrium position produces a net force directed so as to move Q back toward the

equilibrium position

15.15 Consider the free-body diagram of one of the spheres given

at the right Here, T is the tension in the string and F is the e

repulsive electrical force exerted by the other sphere

15.16 The required position is shown in the

sketch at the right Note that this places q

closer to the smaller charge, which will

allow the two forces to cancel.Requiring

0.6

x +

2x = x+0.600 m, or

Solving for x gives the equilibrium position as

15.18 (a) Taking to the right as

positive, the resultant

electric field at point P is

15.19 We shall treat the concentrations as point charges Then, the resultant field consists of

two contributions, one due to each concentration

The contribution due to the positive charge at 3 000 m altitude is

15.22 The force an electric field exerts on a positive change is in the direction of the field Since

this force must serve as a retarding force and bring the proton to rest, the force and hence the field must be in the direction opposite to the proton’s velocity

net

The work-energy theorem, W =KE f −KE i

q x

=

∆

qE F a

v a

and the magnitudes of the fields due to

each of the charges are

2

8.99 10 N m C 8.00 10 C

1.15 10 N C0.250 m

e

k q E r

−

)9

h

−

×

=

15.25 From the symmetry of the charge distribution, students

should recognize that the resultant electric field at the

q

Eur3

Eur2 Eur1

15.26 If the resultant field is to be zero, the contributions

of the two charges must be equal in magnitude

and must have opposite directions This is only

possible at a point on the line between the two

negative charges

Assume the point of interest is located on the

15.27 If the resultant field is zero, the

contributions from the two charges must

be in opposite directions and also have

equal magnitudes Choose the line

connecting the charges as the x-axis, with

the origin at the –2.5 µC charge Then, the

two contributions will have opposite

directions only in the regions x < and 0

For the magnitudes to be equal, the point must be nearer the smaller charge

Thus, the point of zero resultant field is on the x-axis at

Solving for d yields

, or 1.8 m to the left of the − 2.5 Cµ charge

15.28 The magnitude of is three times the magnitude of because 3 times as many lines

emerge from as enter q

(b) q >2 0 because lines emerge from it,

and q1<0 because lines terminate on it

15.29 Note in the sketches at the right that

electric field lines originate on

positive charges and terminate on

negative charges The density of

lines is twice as great for the 2 q−

charge in (b) as it is for the

15.31 (a) The sketch for (a) is shown at

the right Note that four times

as many lines should leave

as emerge from although, for

clarity, this is not shown in this

(b) The field pattern looks the same

here as that shown for (a) with

the exception that the arrows

are reversed on the field lines

15.32 (a) In the sketch for (a) at

the right, note that there

are no lines inside the

sphere On the outside

of the sphere, the field

lines are uniformly

spaced and radially

outward

(b) In the sketch for (b) above,

note that the lines are

perpendicular to the surface

at the points where they emerge They should also be symmetrical about the

symmetry axes of the cube The field is zero inside the cube

15.33 (a) Zero net charge on each surface of the sphere

(b) The negative charge lowered into the sphere repels −5 C to the outsideµ surface,

and leaves +5 C on the insideµ surface of the sphere

(c) The negative charge lowered inside the sphere neutralizes the inner surface, leaving zero charge on the inside This leaves −5 C on the outsideµ surface of the

sphere

(d) When the object is removed, the sphere is left with −5.00 C on the outsideµ

surface and zero charge on the inside

15.34 (a) The dome is a closed conducting surface Therefore, the electric field is

everywhere inside it

At the surface and outside of this spherically symmetric charge distribution, the field is as if all the charge were concentrated at the center of the sphere

2

8.99 10 N m C 2.0 10 C

1.8 10 N C1.0 m

e

k q E R

−

(c) Outside the spherical dome, k q e2

E r

R

3 max

2.0 m 3.0 10 N C

1.3 10 C8.99 10 N m C

e

R E q

15.36 If the weight of the drop is balanced by the electric force, then mg or the

mass of the drop must be

q E eE

16 2

1.6 10 C 3 10 N C

5 10 kg9.8 m s

eE m g

7 3

p

F m

15.38 The flux through an area is Φ =E EAcosθ, where θ is the angle between the direction of

the field E and the line perpendicular to the area A

(a) Φ =E EAcosθ=(6.2 10 N C 3.2 m cos 0× 5 )( 2) ° = 2.0 10 N m C× 6 ⋅ 2

(b) In this case, θ = ° and 90 Φ = 0

15.39 The area of the rectangular plane is A =(0.350 m 0.700 m)( )=0.245 m2

(a) When the plane is parallel to the yz plane, θ= ° , and the flux is 0

15.40 In this problem, we consider part (b) first

(b) Since the field is radial everywhere, the charge distribution generating it must be spherically symmetric Also, since the field is radially inward, the net charge inside the sphere is negative charge

(a) Outside a spherically symmetric charge distribution, the field is k Q e2

E r

Since we have determined that < , we now have Q= −55.7 nC

15.41 cosΦ =E EA θ and Φ =E ΦE, max when θ = ° 0

15.43 We choose a spherical gaussian surface, concentric with the charged spherical shell and

of radius r Then, ΣEAcosθ=E(4πr2)cos 0° =4πr E2

(a) For r (that is, outside the shell), the total charge enclosed by the gaussian

surface is Q q Thus, Gauss’s law gives

(b) Inside the shell, r<a , and the enclosed charge is Q= + q

Therefore, from Gauss’s law, 2

The field for r < a is k q e2 directed radially ou

r

=

EG

tward

15.44 Construct a gaussian surface just barely inside the surface of the conductor, where E 0=

Since E=0 inside, Gauss’ law says

15.45 E =0 at all points inside the conductor, and cosθ =cos 90° = on the cylindrical surface 0

Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s law reduces to cos cap

The charge enclosed by the gaussian surface is Q=σA , where A is the cross-sectional

area of the cylinder and also the area of the end cap, so Gauss’s law becomes

15.46 Choose a very small cylindrical gaussian surface with one end inside the conductor

Position the other end parallel to and just outside the surface of the conductor

Since, in static conditions, E 0= at all points inside a conductor, there is no flux through the inside end cap of the gaussian surface At all points outside, but very close to, a conductor the electric field is perpendicular to the conducting surface Thus, it is parallel

to the cylindrical side of the gaussian surface and no flux passes through this cylindrical side The total flux through the gaussian surface is then Φ =EA , where A is the cross-

sectional area of the cylinder as well as the area of the end cap

The total charge enclosed by the cylindrical gaussian surface is Q=σA, where σ is the charge density on the conducting surface Hence, Gauss’s law gives

10

8.99 10 N m C 1.60 10 C

8.2 10 N0.53 10 m

0.53 10 m 8.2 10 N

2.2 10 m s9.11 10 kg

e

r F v

15.49 The three contributions to the resultant

electric field at the point of interest are

shown in the sketch at the right

The magnitude of the resultant field is

15.50 Consider the free-body diagram shown at the right

15.51 (a) At a point on the x-axis, the contributions by the two

charges to the resultant field have equal magnitudes

given by 1 2 k q e2

r

= = The components of the resultant field are

k q b

x

=+

EG

q q q

q y

x

r a

q

r

Eur1

Eur2

(b) Note that the result of part (a) may be written as ( )

( 2 2)3 2

e R

k Q b E

=+ where Q 2q= is the total charge in the charge distribution generating the field

In the case of a uniformly charged circular ring, consider the ring to consist of a very large number of pairs of charges uniformly spaced around the ring Each pair consists of two identical charges located diametrically opposite each other on the

ring The total charge of pair number i is Q At a point on the axis of the ring, this

pair of charges generates an electric field contribution that is parallel to the axis and has magnitude

i

( 2 2)3 2

e i i

k bQ

=+

Since , the electrical force must be directed downward, aiding the

gravitational force in accelerating the bead Because the bead is positively charged, the electrical force acting on it is in the direction of the electric field Thus, the field

1.00 10 kg 44.1 9

C 3.43 C1.00 10 N C

15.53 Because of the spherical symmetry of the

charge distribution, any electric field

present will be radial in direction If a field

does exist at distance R from the center, it is

the same as if the net charge located within

were concentrated as a point charge

at the center of the inner sphere Charge

located at does not contribute to the

+8.00 mC

–4.00 mC

(a) At r =1.00cm, E =0 since static

electric fields cannot exist within

e E r

=

×

×

2 2 2

(c) At r =4.50cm, E =0 since this is located within conducting materials

(d) The net charge located at r ≤7.00 cm is Q= +4.00 Cµ

15.54 The charges on the spheres will be equal in magnitude and opposite in sign From

2 2

e

F=k q r , this charge must be

( 4 ) ( )2 2

3

1.00 10 N 1.00 m

1.05 10 C8.99 10 N m C

e

F r q

1.05 10 C

6.59 101.60 10 C

q n e

6.59 10

2.51 102.62 10

n N

0.210 N

1.00 10 kg

F g m

×

The period of the pendulum will be

g

(b) Y The force of gravity is a significant portion of the total downward force

acting on the ball Without gravity, the effective acceleration would be

es

2 -3

0.200 N

1.00 10 kg

e F g m

a 2.28% difference from the correct value with gravity included

15.57 The sketch at the right gives a free-body diagram of the

positively charged sphere Here, 2 2

F =k q r is the attractive force exerted by the negatively charged sphere and F2=qE

is exerted by the electric field

15.58 As shown in the sketch, the electric field

at any point on the x-axis consists of two

parts, one due to each of the charges in the

1.00 10 kg 9.80 m scot 37.0 5.00 3.00 cot 37.0 10 N C1.09 10 C 10.9 nC

y x

mg q

15.60 (a) At any point on the x-axis in the range 0< <x 1.00 m, the contributions made to the

resultant electric field by the two charges are both in the positive x direction Thus,

it is not possible for these contributions to cancel each other and yield a zero field

(b) Any point on the x-axis in the range x < is located closer to the larger magnitude 0charge (q=5.00 Cµ ) than the smaller magnitude charge (q =4.00 Cµ ) Thus, the contribution to the resultant electric field by the larger charge will always have a greater magnitude than the contribution made by the smaller charge It is not

possible for these contributions to cancel to give a zero resultant field

(c) If a point is on the x-axis in the region , the contributions made by the two charges are in opposite directions Also, a point in this region is closer to the smaller magnitude charge than it is to the larger charge Thus, there is a location in this region where the contributions of these charges to the total field will have equal magnitudes and cancel each other

15.61 We assume that the two spheres have equal charges, so the

repulsive force that one exerts on the other has magnitude

From Figure P15.61 in the textbook, observe that the distance

separating the two spheres is

From the free-body diagram of one sphere given above, observe

Thus, k q e 2 r2=mgtan10°

0.015 kg 9.8 m s 0.047 m tan10tan10

8.99 10 N m C

e

mgr q

15.63 (a) When an electron (negative charge) moves distance x∆ in the direction of an

electric field, the work done on it is

1.60 10 J1.60 10 C 0.100

e x

(b) The magnitude of the retarding force acting on the electron is , and Newton’s second law gives the acceleration as

e

F = E e e

a= −F m= −eE m Thus, the time required to

bring the electron to rest is

( ) ( )

v v t

eE a m

e y

v0

Thus, if R =1.27 10 m× −3 , we must have

2 2

y

v t a

y

v t a

×

Electrical Energy and

Capacitance

Quick Quizzes

1. (b) The field exerts a force on the electron, causing it to accelerate in the direction opposite

to that of the field In this process, electrical potential energy is converted into kinetic energy of the electron Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron

2 (b), (d) Charged particles always tend to move toward positions of lower potential

energy The electrical potential energy of a charged particle is PE qV= and, for

positively-charged particles, this increases as V increases For a negatively-positively-charged particle, the

potential energy decreases as V increases Thus, a positively-charged particle located at

would move toward the left A negatively-charged particle would oscillate around which is a position of minimum potential energy for negative charges

x=A

x B=

3. (d) If the potential is zero at a point located a finite distance from charges, negative

charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region

4. (c) Both the electric potential and the magnitude of the electric field decrease as the

distance from the charged particle increases However, the electric flux through the

balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size

5. (a) From the conservation of energy, the final kinetic energy of either particle will be given by

KE =KE + PE −PE = +qV −qV = −q V −V = − ∆q V

For the electron, q= −e and ∆ = +V 1 V giving KE f = − −( )(e +1 V)= +1 eV

For the proton, q= +e and ∆ = −V 1 V, so KE = − f ( )(e −1 V)= +1 eV, the same as that of the electron

6. (c) The battery moves negative charge from one plate and puts it on the other The first plate is left with excess positive charge whose magnitude equals that of the negative

charge moved to the other plate

35

7 (a) C decreases (b) Q stays the same (c) E stays the same

(d) ∆V increases (e) The energy stored increases

Because the capacitor is removed from the battery, charges on the plates have nowhere to

go Thus, the charge on the capacitor plates remains the same as the plates are pulled

apart Because

Q A

E= σ =

∈ ∈ , the electric field is constant as the plates are separated

Because ∆V = Ed and E does not change, ∆V increases as d increases Because the same

charge is stored at a higher potential difference, the capacitance has decreased Because

2 2

=Energy stored and Q stays the same while C decreases, the energy stored

increases The extra energy must have been transferred from somewhere, so work was done This is consistent with the fact that the plates attract one another, and work must be done to pull them apart

8 (a) C increases (b) Q increases (c) E stays the same

(d) ∆V remains the same (e) The energy stored increases

The presence of a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant Since the battery holds the potential difference constant while the capacitance increases, the charge stored (Q C V= ∆ ) will increase Because the potential difference and the distance between the plates are both constant, the electric field

capacitance of the capacitor

Answers to Even Numbered Conceptual Questions

2 Changing the area will change the capacitance and maximum charge but not the

maximum voltage The question does not allow you to increase the plate separation You can increase the maximum operating voltage by inserting a material with higher dielectric strength between the plates

4 Electric potential V is a measure of the potential energy per unit charge Electrical

potential energy, PE = QV, gives the energy of the total charge Q

6 A sharp point on a charged conductor would produce a large electric field in the region

near the point An electric discharge could most easily take place at the point

8 There are eight different combinations that use all three capacitors in the circuit These

combinations and their equivalent capacitances are:

All three capacitors in series -

All three capacitors in parallel - = + +

One capacitor in series with a parallel combination of the other two:

12. All connections of capacitors are not simple combinations of series and parallel circuits As

an example of such a complex circuit, consider the network of five capacitors C1, C2, C3, C4, and C5 shown below

C5

This combination cannot be reduced to a simple equivalent by the techniques of

combining series and parallel capacitors

14. The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates

16 (a) i (b) ii

18 (a) The equation is only valid when the points A and B are located in a region where the

electric field is uniform (that is, constant in both magnitude and direction) (b) No The field due to a point charge is not a uniform field (c) Yes The field in the region between a

pair of parallel plates is uniform

Answers to Even Numbered Problems