Chapter 26 by John Smith

THEELECTRIC FIELD

26.1. Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure Ex26.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1is

(9.0 x lo9 N m2 / Cz)(l x

ebelow +x-axis

(0.05 m)’

+ (0.05 m)’

C) (cos& - sin e

Because tan6 = 5 cm/5 cm = 1, the angle 6 = 45”. Hence,

zl = ( 1 8 0 0 N / C )(A;-$) -’

Similarly, the electric field from q2is

,!?,,,,, = El + E, = 2(18OO N / C) - i^ = 25502 N / C Thus, the strength of the electric field is 2550 N/C and its direction is horizontal. Assess: Because the charges are located symmetrically on either side of the x-axis and are of equal value, the y-components of their fields will cancel when added.

26.2. Model: The electric field is that of the two charges located on the y-axis. Visualize: Please refer to Figure Ex26.2. The electric field of the positive charge q, at P is away from ql. On the other hand, the electric field of the negative charge q2 at P is toward q2.These two electric fields are then added vertically to obtain the net electric field at P. Solve: The electric field from q1is

(

“I1,

E - l = 4m0 52

@below +x-axis

(9.0 x lo9 N mz / C’)(l.O x

(0.05 m)’

+ (0.05 m)’

C) cos & - sin

Because tan0 = 5 cm/5 cm, 8 = 45”. So,

El = ( 1 8 0 0 N / C )(+Li-L;) 4 5 Jz Similarly, the electric field from q2is

* E,,,,,

- -

= El

+ E2 = 2(1800 N / C)

Thus, the strength of the electric field is 2550 N/C and its direction is vertically downward. Assess: A quick visualization of the components of the two electric fields shows that the horizontal components cancel.

26-1

26-2

Chapter 26

26.3. Model: The electric field is that of the two 1 nC charges located on the y-axis. Visualize: Please refer to Figure Ex26.3. We denote the top 1 nC charge by q1and the bottom 1 nC charge by q2 The electric fields ( E, and E,) of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field Solve: The electric fields from q1and q2are

Erie, at the point P indicated by the dot. (9.0 x lo9 N m' / C')(1 x lo4 C)

along +x-axis

(0.05 m)' @above +x-axis

Because tan@= 10 cm/5 cm, 8 = tan-'(2) = 63.43'. So,

(9.0 x lo9 N m' / C')(I x

E? =

+ (0.05 m)'

(0.10 m)'

(cos63.43'

i =36001N/C

1+ sin63.43'33) = (3221 + 6443) N / C

The net electric field is thus

E,,

a,P

= E,

+ i2= (39222 + 6443) N / C

To find the angle this net vector makes with the x axis, we calculate 644N/C tan@= @ =9.3' 3922 N / C Thus, the strength of the electric field at P is

E,, = J(3922 N / C)'

+ (644N / C)'

= 3975 N/C

and E,, makes an angle of 9.3'above the +x-axis. Assess: Because of the inverse square dependence on distance, E? < E l . Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.

26.4. Model: The electric field is that of the two charges located on the y-axis. Visualize: Please refer to Figure Ex26.4. We denote the positive charge by q1 and the negative charge by q2.The electric field E, of the positive charge q , is directed away from q l , but the field 2, is toward the negative charge vectorially to find the strength and the direction of the net electric field vector. q2.We will add El and Solve: The electric fields from q, and q2are

- (

I,:'

E , = -1 away from q1along 4 m 0 ti-

+ x axis

(9.0 x lo9 N m2 / C')(1 x IO4 C )

(0.05 m)'

i = 36001N/C

@below -x-axis

From the geometry of the figure, 10 cm tan@= -3 8 = 63.43' 5 cm

E,,

E2 =

(9.0 x lo9 N m' / C')(1 x (0.10 m)'

+ (0.05 m)'

= E, + E? =(3278i' - 6441) N / C

(-cos63.43'1-

E,, = J(3278 N / C)'

sin63.43'1) = -(3221

+ (-644

+ 6443) N / C

N / C)? = 3340 N/C

To find the angle this net vector makes with the horizontal, we calculate

Thus. the strength of the net electric field at P is 3341 N/C and

E,,,

makes an angle of 11.1' below the +x-axis.

The Electric Field

26-3

26.5. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. visualize:

(a)

(b)

The dipole consists of charges kq along the y-axis. The electric field in (a) points down. The field in (b) points up. Solve: (a) The dipole moment is

j = (qs, from - to +) = (1.O x IO+ C)(O.OO~O m)f = 2.0 x IO-’’

f cm

The electric field at (10 cm, 0 cm), which is at distance r = 0.10 m in the plane perpendicular to the electric dipole, is

E‘ = ---

~ K E ,r 3

2.0 x 10-”j c m = -18.03 N / C (0.IO m)3

= -(9.0 x lo9 N m2/ C 2 )

The field strength, which is all we’re asked for, is 18.0 N/C. (b) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is

1 2 j 2(2.0 x IO-’’ c m) = 36.03 N / C E = -- = ( 9 . 0 ~lo9 N m 2 / C 2 ) (0.10 m)’ ~ K E ,r3 The field strength at this point is 36.0 N/C.

26.6. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. Visualize: Y I

+ +4

The dipole consists of charges fq along the y-axis. The electric field in (a) points up. The field in (b) points down. Solve: (a) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is

p+ =

r’l? (0.10 rnl3(3603 N / C ) = (2.0 2 ( 1 / 4 ~ ~ , ) 2(9.0x IO9 N m2 / C 2 )

lo-l,jm)

26-4

Chapter 26

By definition, the dipole moment is

= 2.0 x 10-"3 C m = (qs, from

- to +) = q(O.010 m) 3. Thus

(b) Point (10 cm, 0 cm) is in the plane perpendicular to the dipole. The electric field E' = -(1/4=) i / ?is half the strength of the field at an equal distance r on the axis of the dipole. Hence the field strength at this point is 180 N/C.

26.7. Model: The rods are thin. Assume that the charge lies along a line. Visualize:

Glass

z, P, 2, P, z, P,

Plastic

EgiaSs pp1aSoc

At point P

- - - -- - - - - -- - - -.

Zglass ~ p p l a S o c & s y l a s n c

At point P,

The electric field from the glass rod at r = 1 cm from the glass rod is

10 x 1 0 - ~c

Eglxs= (9.0 x lo9 N m' / C')

(0.01 m),/(O.Ol m)'

+ (0.05 m)'

= 1.765 x lo5 N / C

The electric fields from the glass rod at r = 2 cm and r = 3 cm are 0.835 x lo5 N / C and 0.514 x lo5 N / C . The electric field from the plastic rod at distances 1 cm, 2 cm, and 3 cm from the plastic rod are the same as for the glass rod. Point P, is 1.0 cm from the glass rod is 3.0 cm from the plastic rod, point P, is 2 cm from both rods, and point P, is 3 cm from the glass rod and 1 cm from the plastic rod. Because the direction of the electric fields at PI is the same, the net electric field strength 1 cm from the glass rod is the sum of the fields from the glass rod at 1 cm and the plastic rod at 3 cm. Thus At 1.0cm At2.0cm

+ 0 . 5 1 4 ~ 1 0N~/ C = 2 . 2 8 x 1 O S N / C E = 0 . 8 3 5 x 1 0 5 N / C + 0 . 8 3 5 ~ 1 0N~/ C = 1 . 6 7 x 1 0 5 N / C E = 0 . 5 1 4 x 1 0 5 N / C + 1 . 7 6 5 ~ 1 0N~/ C = 2 . 2 8 x 1 0 5 N / C E=1.765x1OS N / C

At3.0cm Assess: The electric field strength in the space between the two rods goes through a minimum. This point is exactly in the middle of the line connecting the two rods. Also, note that the arrows shown in the figure are not to scale.

26.8. Model: The rods are thin. Assume that the charge lies along a line. Visualize: + +

+ + +

+ + + +

++ +

Gght;Engh;T; -~

+ +

Because both the rods are positively charged, the electric field from each rod points away from the rod. Because the electric fields from the two rods are in opposite directions at P,, Pz,and P,. the net field strength at each point is the difference of the field strengths from the two rods.

The Electric Field

26-5

Solve: Example 26.3 gives the electric field strength in the plane that bisects a charged rod: E rod =- 1 4zg0 r

IQl J

The electric field from the rod on the right at a distance of 1 cm from the rod is 10 x 10-~c

Enght = (9.0 x lo9 N m 2 / C')

(0.01 m),/(O.Ol m)'

+ (0.05 m)'

= 1.765 x 10' N / C

The electric field from the rod on the right at distances 2 cm and 3 cm from the rod are 0 . 8 3 5 ~ 10' N / C and 0.514 x 10' N / C. The electric fields produced by the rod on the left at the same distances are the same. Point P, is 1.0 cm from the rod on the left and is 3.0 cm from the rod on the right. Because the electric fields at P, have opposite directions, the net electric field strengths are At 1.0cm

E = 1 . 7 6 5 x 1 0 5 N / C - 0 . 5 1 4 ~ 1 0 'N / C = 1 . 2 5 x 1 0 5 N / C

At2.0cm

E=0.835x105 N / C

At3.0cm

E = 1 . 7 6 5 ~ 1 0 ' N / C - 0 . 5 1 4 ~ 1 0N~/ C = 1 . 2 5 x 1 O S N / C

- 0 . 8 3 5 ~ 1 0N~/ C = O N / C

26.9.

Model: The rod is thin, so assume the charge lies along a line. visualize: Y

Solve: The force on charge q is F charged rod is

E& =-

4K&0

= qE,

. From Example 26.3, the electric field a distance r from the center o f a

: (9.0 x lo9 N m2 / C')(50 x

(0.04 m),/(0.04 m)'

C)i'

+ (0.05 m)'

= 1.757 x 1051N /

Thus, the force is

F = (-5 More generally,

F = (8.78 x IO4

C)(1.757 x 10' N / C)i' = -8.78 x lO"i' N

N, toward the rod).

26.10. Model: We will assume that the wire is thin and that the charge lies on the wire along a line. Solve: From Equation 26.15, the electric field strength of an infinitely long line of charge having linear charge density A is

1 2111 Ehne(r= 5.0 cm) = 4n&, 5.0 x lo-' m

1 214 El,ne(r= 10.0 cm) = 4K&, 10.0 x IO-' m

26-6

Chapter 26

Dividing the above two equations gives

(

N/C Ehe(r= 10.0 cm) = ~ ~ ~ ~ ~ ~ 2 " , j 4 , , ( r = 5 . 0 =-(2000N/C)=lOOO cm 1) 2

26.11. Model: Assume that the wire is thin and that the charge lies on the wire along a Zine. Solve: From Equation 26.15, the electric field for an infinite uniformly charged line is

where r is the distance from the line in the plane that bisects the line. Solving for the linear charge density,

The charge in 1 cm is

la=L(A(= (1.0 x lo-' m)(5.56 x

C / m) = 5.56 x lo-" C = 0.056 nC

Because the electric field is directed toward the line, Q is negative. Thus Q = -0.056 nC.

26.12. Model: Assume that the rings are thin and that the charge lies along circle of radius R. b-2~ = 20 cm + Visualize:

Q2 = +20 nC

Q, = -20 nC

The rings are centered on the z-axis. Solve: (a) According to Example 26.5, the field of the left (negative) ring at z = 10 cm is

(E,): = That is, the field is

zQi

4mCE,(z2 + R2)"'

(9.0 x lo9 N m2 / C2)(0.10m)(-20 x

[(0.10 m)'

+ (0.05 m)']312

C) = -1.288'~lo4 N / C

E, = (1.288 x lo4 N / C, left). Ring 2 has the same quantity of charge and is at the same distance,

so it will produce a field of the same strength. Because Q2is positive, midpoint is

E, will also point to the left. The net field at the

E = E, +E, = (2.58 x lo4 N / C, left) (b) The force is

F = qE = (+1.0 x

C)(2.576 x lo4 N / C, left) = ( 2 . 5 8 ~10" N, left)

26.13. Model: Assume that the rings are thin and that the charge lies along circle of radius R. 2 i = 20 cm + Visualize:

Q,= +20 nC

Q2 = +20 nC

The Electric Field

26-7

Solve: (a) Let the rings be centered on the z-axis. According to Example 26.5, the field of the left ring at z = 10 cm is

zQi

(E1)z

4nE.(z2

R2Y2

(9.0 x lo9 N m2 / C’)(O.lO m)(20 x lo4 C)

=1.28x104 N / C

l(O.10 m)’ +(0.05 m)’]yz

El = (1.288 x lo4 N / C, right). Ring 2 has the same quantity of charge and is at the same distance, so it will produce a field of the same strength. Because Q2 is positive, i2will point to the left. The net field at the midpoint between the two rings is E = E, + Z2 = 0 N/C. That is,

(b) The field of the left ring at z = 0 cm is ( E , ) . = 0 N/C. The field of the right ring at z = 20 cm to its left is

(9.0 x lo9 N mz / C’)(0.20 m)(20 x lo4 C)

( E d z=

= 4110N/C

+ (O.O5)’]Yz 3 E = E, + Z2 = 0 N/C + (41 10 N/C, left) [(0.20 m)’

So the electric field strength is 41 10 N/C.

26.14. Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric field of the disk points toward the disk. The electric field points away from the disk for a positively charged disk. -b 2: = 20 cm 4-~ Visualize:

Solve: (a) The surface charge density on the disk is

From Equation 26.22, the electric field of the left disk at z = 0.10 m is -6.366 x lod C / m2 2(8.85 x C‘ / N m’)

+ (0.05

1 = -38,000 N I C m / ~ .m)2 l~

Chapter 26

26-8

From Equation 26.22, the electric field of the left disk at z = 0.10 m is

Hence,

El =(38,000N/C, left).

Similarly, the electric field of the right disk at z = 0.10 m (to its left) is

.I?~ = (38,000 N / C, right). The net field at the midpoint between the two disks is E = E, + E2 = 0 N / C . @) The electric field of the left disk at z = 0 m is

[

( E ) =I-2&, .?

1' =L= 2Eo

4L3" lo4 m2 = -3.60 x lo5 N / C 2(8.85x1O-l2 C2 / N m 2 )

El = (3.60 x lo5 N/C, left)

Similarly, the electric field of the right disk at z = 0.20 m (to its left) is E2 = (1.075 x lo4 N/C, right). The net field is thus

E = E, + E, = (3.49 x io5 N/C, left) The field strength is 3.49 x lo5 N/C.

26.16. Model: The distance 2.0 mm is very small in comparison to the size of the electrode, so we can model the electrode as a plane of charge. Solve: From Equation 26.26, the electric field of a plane of charge is

Q -

Eplme= -= -2&, 2&,A 2(8.85 x

80 x 1 0 - ~c = 1.13 x lo5 N / C C z / N m2)(0.20 x 0.20 m)

26.17. Motion: The very wide charged electrode is a plane of charge. Solve: For a plane of charge with surface charge density 77, the electric field is

Q = 2 z R 2 ~ , ( E p 1 , ,=) ~2z(0.005 m)'(8.85 x lo-'' C 2 / N m2)(1000N / C) = 1.39 x 10-12C= 1.39 x

Assess: Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have to use the distance z = 5.0 cm given in the problem.

26.18. Model: A spherical shell of charge Q and radius R has an electric field ourside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize: In the case of a metal ball, the charge resides on its surface. This can then be visualized as a charged spherical shell of radius R. Solve: Equation 26.28 gives the electric field of a charged spherical shell at a distance r > R:

In the present case, Em, = 50,000 N / C at r = + ( l o cm) + 2.0 cm = 7.0 cm = 0.07 m. So,

26.19. Model: An insulating sphere of charge Q and radius R has an electric field outside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize: 2 cm 2 cm It-rit6.0 cm -+I+--+

The Electric Field

26-9

Solve: The electric field of a charged sphere at a distance r > R is given by Equation 26.28:

In the present case,

E = E, + E2,where 2, and E,

are the fields of the individual spheres. The distance from the

center of each sphere to the midpoint between is r, = r, = 4 cm. Thus, ( 9 . 0 ~ 1 0N~m 2 /C2)(10x10-9 C) El =

(0.04 m)* (9.0 x lo9 N m2 / C2)(15x

E2 =

(0.04 m)'

= 5.625 x lo4 N / C

C) = 8.438 x lo4 N / C

The fields point in the same direction, so

E = (5.625 x lo4 N / C + 8.438 x lo4 N / C, right) = (1.41 x lo5 N / C, right). The electric field will point left when Q, and Q2 are interchanged. The electric field strength in both cases is 1.41X 10sN/C.

26.20. Model: The electric field is uniform in a region of space between closely spaced capacitor plates. Solve: The electric field inside a capacitor is E = &/&,A. Thus, the charge needed to produce a field of strength E is Q = &,AE = (8.85 x C2 / N m2)(0.04m x 0.04 m)(l.O x lo6 N / C) = 14.2 nC Thus, one plate has a charge of 14.2 nC and the other has a charge of -14.2 nC. Assess: Note that the capacitor as a whole has no net charge.

26.21. Model: The electric field in a region of space between two charged circular disks is uniform. Solve: The electric field strength inside the capacitor is E = Q / & , A . Thus, the area is A = - =Q &,E

(1.5 x 109)(1.6x C) = 2.71 x lo4 mz = z-D2 4 (8.85 x lo-'' C2 / N m2)(1.0 x lo5 N / C)

Assess: As long as the spacing is much less than the plate dimensions, the electric field is independent of the spacing and depends only on the diameter of the plates.

26.22. Model: The electric field in a region of space between the plates of a parallel-plate capacitor is uniform. Solve: The electric field inside a capacitor is E = Q / & , A . Thus, the charge needed to produce a field of strength E is Q = &,AE = (8.85 x

C2 / N m2)[z(0.02 m)*](3x lo6 N / C) = 33.4 nC

The number of electrons transferred from one plate to the other is

26.23. Model: The charged plastic bead is a point charge.

visualize:

qzt W

The bead hangs suspended in the air when the net force acting on the bead is zero. The two forces that act on the bead are the electric force and the weight. Because the bead is negatively charged, the electric field must be pointed downward to cause an upward force, which will balance the weight.

26-10

Chapter 26

Solve: For the bead to be in static equilibrium, (F,,,)y= qE -mg = 0 N Thus the required field is

(0.10 X lo9 kg)(9.8 N / kg) E =3= =6.13x105 N / C 4 (1.0 x 10’0)(1.60x C)

E = (6.13 x lo5 N / C , down).

26.24. Model: The disks form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the proton will have a constant acceleration. Visualize: Y

Solve:

M R=lcm (a) The two disks form a parallel-plate capacitor with surface charge density

From Equation 26.29, the field strength inside a capacitor is 3.18 x lo-’ C / m2 =3.6Ox1O6 N / C 8.85 x lo-’’ C2 / N m 2

E = - q= E,

(b) The electric field points toward the negative plate, so in the coordinate system of the figure E = -3.60 x 1 0 6 jN / C . The field exerts forces = qFotonE = eE on the proton, causing an acceleration with a y-component that is

(1.6 X

a =-=Y=

’

lo6 N / C) = -3.45 x 1014 m / s2 1.67 x 10-27kg C)(-3.60

After the proton is launched, this acceleration will cause it to lose speed. To just barely reach the positive plate, it should reach v, = 0 m/s at y , = 1 nun. The kinematic equation of motion is

v,’ = 0 m2 / s2 = v,’ + 2uyAy 3

-/,

v, =

= ,/-2(-3.45 x l O I 4 m / s2)(0.001m) = 8 . 3 0 ~lo5 m / s

The acceleration of the proton in the electric field is enormous in comparison to the gravitation acceleration g. That is why we did not explicitly consider g in our calculations.

Assess:

26.25. Model: The electric field is uniform, so the electrons will have a constant acceleration. = 5.0 x 10’ d S

ir-------Ax=1.2cm------.I

Solve: A constant-acceleration kinematic equation of motion is vf = v,z + 2a(Ax)

v? - v: - (5.0 x 10’ m / s)2 - (0 m / s)’ *u = 2Ax ~(1.2x IO-‘ m)

= 1.042 x 10” m / s2

The Electric Field

26-11

The net force on the electron in the electric field is F = qE = mu. Thus, kg)(1.042 x 1017 m / s2)

(9.1 1 x

ma E=-= 4

1.60 x 10-l~c

=5.93x1Os N / C

Hence, the field strength is 5.93 X lo5 N / C .

26.26. Model: The infinite charged plane produces a uniform electric field. Solve: (a) The electric field of a plane of charge with surface charge density 17 is

where in is the mass, q is the charge, and a is the acceleration of the electron. To obtain 17 we must first find a. From the kinematic of motion equation v: = v,’ + 2 a h , m~/ s ) ‘ -(Orn/s)’ - v o2 - ( 1 . 0 ~ 1 0 = 2.5 x 10” m / s2 2A.r ~(2.x 0 IO-’ m)

a=--

a q =

kg)(2.5 x 10” m / s’)

3(8.85 x IO-” C’ / N m2)(9.11 x

i . 6 0 ~ 1 0 -c~ ~ (b) Using the kinematic equation of motion v1 = vo + adt, v -v,, - 1 . 0 ~ 1 0m~/ s - o m / s At=-= 4.0 x 2.50 x 10” m / s2 U

= 2.52 x

C / m’

26.27. Model: The infinite negatively charged plane produces a uniform electric field that is directed toward the plane. Visualize:

v I I

I I

“0

Lh - - *--- - -

I I \I/

v, =om/s

- - - * - - - - a t

Known 7 = -2.0 x C/m2 vo = 2.0 x 106 m / s v, = 0 mJs mp = 1.67 X lo-” kg

Find AX

Solve: From the kinematic equation of motion v: = 0 = vi + 2u& and F

= qE = mu,

--2

-mv; = qE - -vo a&=-

m 2Ax 2qE Furthermore, the electric field of a plane of charge with surface charge density q is E = ~ / 2 &Thus, ~ . &=--

-mi’;&, ’ - -(1.67 x

q17

kg)(2.0 x lo6 m / s)’(8.85 x lo-’’ C’ / N m‘) (1.60 x

C)(-2.0 x lo6 C / m’)

= 0.185 m

26.28. Model: Because r >> s, we can use Equation 26.12 for the electric field in the plane perpendicular to the dipole.

26- 12

Solve:

Chapter 26

(a) From Newton’s third law, the force of Q on the dipole is equal and opposite to the force of the dipole

on Q. You can see from the diagram that F&,,eonP is down and FQondiPle is up, in the direction of magnitude of the dipole field at the position of Q is

3 . The

The magnitude of Fhpole on is

FQon

has the same magnitude, so

Le, direction of 4 m 0 r3 (b) The electric field of charge Q exerts a torque on the dipole. The field 8= 90”. The torque is given by Equation 26.34:

E‘

is perpendicular to the dipole

3 , so

26.29. Model: The size of a molecule is ~ 0 . 1nm. The proton is 2.0 nm away, so r >> s and we can use Equation 26.12 for the electric field in the plane that bisects the dipole. Visualize:

++ -

&pole

Fori proton

Solve: You can see from the diagram that F,,leoapmfon is opposite to the direction of j . The magnitude of the dipole field at the position of the proton is 5.0 x Cm Edpole = 1 P = ( 9 . 0 x 1 O 9 Nm2/C2) = 5.624 x lo5 N/C 4 m 0 1’ (2.0 x 1 0 - ~rn13

The magnitude of

Fdpole on

is C)(5.624 x IO5 N/C) = 9.0 x

F,,leonpmmn = eEapole= ( 1 . 6 0 ~

Including the direction, the force is F,,,,,leonpromn = (9.0 X

N, direction opposite

3).

26.30. Model: The external electric field exerts a torque on the dipole moment of the water molecule. Solve: From Equation 26.34, the torque exerted on a dipole moment by an electric field is z = pEsin0. The maximum torque is exerted when sin0 = 1 or 0 = 90”. Thus,

zmax= pE = (6.2 x

C m)(5.0 x 10’ N I C) = 3.10 x lo-*’ N m

26.31. Model: The electric field is that of three point charges q,, q2and q3. Visualize: Please refer to Figure p26.31. Assume the charges are in the x-y plane. The 5 nC charge is q,, the 10 nC charge is q3,and the -5 nC charge is q2.The net electric field at the dot is E,,, = El + E, + E3.The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: The electric field produced by q, is 1

lqll

El =-= 47r&0 ‘12

( 9 . 0 ~ 1 0Nm’ ~ /C2)(5.0x10” C) (0.02 m)’

= 112,500 N / C

The Electric Field

26-13

El points away from q l , so in component form E, = 112,500~N / C. The electric field produced by q2 is E2 = 28,120 N / C.

E2 points toward q2,so 1

= 28,1203 N / C . Finally, the electric field produced by q3 is

Iq31 (9.0 x lo9 N m2 / C')(10.0 x lo4 C) = 45,000

E3 = -- = 4 m 0 r,'

(0.02 m)'

+ (0.04 m)'

& points away from q3and makes an angle 4 = tan-'(4 / 2) = 63.43" with the x-axis. So, E, = E3cos42 - E3s i n k = (20,1302 -40,2503)

N/C

Adding these three vectors gives

E,,,

= E,

+ E, + E, = (132,6007 - 12,1303) N / C

This is in component form. The magnitude of the field is

-4

= ,/(132,600 N / C)'

E,, =

+ (-12,130

N / C)' = 133,200 N / C

and its angle from the x-axis is 8 = tan-'(lE,/Eyl) = 5.23'. We can also write

= (133,200 N / C, 5.23'

below

the +x-axis).

26.32. Model: The electric field is that of three point charges q,,q2and q3. Visualize: Please refer to Figure â&#x201A;Ź96.32. Assume the charges are in the x-y plane. The -5 nC charge is q,, the bottom 10 nC charge is q3,and the top 10 nC charge is q2.The net electric field at the dot is Em = El + E, + E,. The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: The electric field produced by ql is 1

lq11

E, = -- 47r&0 '12

(9.0 x IO9 N m' / C2)(5.0x

(0.02 m)'

= 112,500 N / C

El points toward q l , so in component form El = -112,500j N / C. The electric field produced by q2 is E2 = 56,250 N / C. E, points away from q2,so E, = -56,2502 N / C . Finally, the electric field produced by q3is 1 1q31- ( 9 . 0 ~IO9 N m' / C')(lO.O x E, = -- 4m0 r3, (0.02 m)* + (0.04 m)'

= 45,000 N / C

E, points away from q3and makes an angle @ = tan-'(2 / 4) = 26.57' with the x-axis. So, E, = -E3 cos& + E3s i n k = (-40,2502 + 20,130j) N / C Adding these three vectors gives

E,,, = E, + E, + E, = (-96,5001 - 92,4005) N / C This is in component form. The magnitude of the field is E,,, =

,/-

= J(96.500 N / C)'

+ (92,400 N / C)'

= 133,600 N / C

and its angle from the x-axis is 8 = tan-'(~Ex/Ey~] = 43.8'. We can also write E,,, = (133,600 N / C, 43.8" below the -x-axis).

26.33. Model: The electric field is that of three point charges q,,q2and q3. Visualize: Please refer to Figure P26.33. Assume the charges are in the x-yplane. The 10 nC charge is ql, the -10 nC charge is ql, and the 5 nC charge is qz. The net electric field at the dot is E,,, = El + I!$ + E,. The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: The electric field produced by ql is

14,1

1- E, = 4 7 7 ~r,'~

(9.0 x IO9 N m' / C2)(10.0x 10" (0.03 m)'

= 100,000 N / C

26-14

Chapter 26

points away from q l , so in component form

is E2 = 28,125 N / C .

& points away from q2,so 2,

E, = -100,ooO~N / C.

The electric field produced by q2

= -28,1251 N / C . Finally, the electric field produced by q3is

1 1q = (9.0 x lo9 N m2 / C2)(10.0x lo-' C) = 36,000 N / C E, = -2 4m0 r: (0.03 m)' + (0.04 m)'

E, points toward q3 and makes an angle

@ = t d ( 3 / 4) = 36.87" with the x-axis. So,

E, = E3 cos@; + E3sin& = (28,8001 + 21,600j) N / C Adding these three vectors gives

Enet= El + E2 + E, = (6751 - 78,400J)

N/C

This is in component form. The magnitude of the field is

E,,, =

,/=

= J(675 N / C)'

+ (78,400 N / C)'

= 78,400 N / C

and its angle from the x-axis is 8 = tar-'(lEX/Ey/)= 89.5". We can also write

= (78,400 N / C, 89.5" below

the +x-axis).

26.34. Model: The electric field is that of three point charges q1= -Q, q2= -Q and q3= +4Q, Visualize: Assume the charges are in the x-y plane. The net electric field at point P is E,,, = El + E, + E,. The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. V

-.,

Solve: (a) The electric field produced by q1points toward q1and is given by

The electric field produced by q2points toward q2 and is given by

=-(--)j. 1

Q -

4m, L2 The electric field produced by q3is

E, points away from q, and makes an angle 4 = tar-'(L/L)

= 45" with the x-axis.

The Electric Field

Adding these three vectors gives

- - -

Enel= El + E2 + E3 =

(b) The force on the charge is F = mu = qE,,

. Therefore,

26.35. Model: The electric field is that of two charges -q and +2q located at x = h. Visualize: v I

E4 net

Solve: (a) At point 1, the electric field from -4 is

E-,

points toward -q and makes an angle $, = tan-'(2u/a) = 63.43' below the -x-axis, hence

The electric field from the +2q is

E +2q

2q =-- 1 2q =- 1 47E0 u2 +(2u)' 4 m 0 5u?

E+?, points away from +2q and makes an angle

= tan-'(2u/u) = 63.43' above the -x-axis. So,

Adding these two vectors,

At point 2, the electric field from -q points toward -q, so

E-,

-(1

2);

4m0 u2

26-15

26- 16

Chapter 26

The electric field from +2q points away from +2q, so

Adding these two vectors,

At point 3, the electric field from -q points toward -q, so

The electric field from +2q points away from +2q, so +2q

=;)”(‘ 4m0 a’

Adding these vectors,

Point 4 is a mirror image of point 1. Since points to the left and down. Thus,

E, net

E4,,, has a reversed y-component and

points to the left and up,

(b) The four electric field vectors at points 1-4 are shown in the figure.

26.36. Model: The electric field is that of two positive charges. Visualize:

E (N/C) I

800,Ooo

200.000

ljp/-qP.. E‘ x _-8 , , .f

-6

The figure shows E, and E, due to the individual charges. The total field is E = E, + E,. Solve: (a) From symmetry, the y-components of the two electric fields cancel out. The x-components are equal and add. Thus, E = 2 ( E , ) x1’ - 2E,cosei

The field strength and angle are E, =&=

141

case = -X =

4m0r2 4m0(x2+ ?/4)

*E=

24X

47CEo(x’ + s?/4)3i2

(b) The electric field at position x is E = (9.0 x lo9 N m’ / C’)(2)(1.0 x [x’

+ (0.003 m)’]’’’

18x

C)x [x’

+ (0.003 m)2]3’2

NIC

The Electric Field

where x has to be in meters. We can now evaluate

I? for different values of x:

2 4

0.002 0.004 0.006 0.010

6 10

768,000 576,000 358,000 158,000

+ -.

26-17

+ 0 NIC as x + 0 m and as

The graph is shown in the figure above.

26.37. Model: The electric field of a dipole is that of two opposite charges +q that make the dipole. Visualize: Please refer to Figure 26.7. The figure shows a dipole aligned on the y-axis, so the x-axis is the - - . bisecting axis. The field at a point on the x-axis is Edpole= E+ + E-. Solve: From the symmetry of the situation we can see that the x-components of the two contributions to the electric field will cancel, that they have equal ycomponents, and that I?hple points in the -y-direction. Thus, (Ehple)x= O N / C where 0 is the angle

(Edple),,=-2E+sine

E+ makes with the x-axis. From the geometry of the figure, SI2

sine = [xz

+ (s I 2)*]1/2

For a point charge +q. the field is 1 E, = -

4ZE0 x2

+(S/2)*

Combining these pieces gives the dipole field at distance x along the bisecting axis:

x >> s, then (xz sz / 4)â&#x20AC;?â&#x20AC;&#x2122; = x3.n u s

1 9s; Ehple - -4Z&0 x3

If we note that then

= qsj^ and if we replace x with a more general variable r to denote the distance from the dipole,

This is Equation 26.12.

26.38. Model: The electric field is that of three point charges. Visualize: Y

The field at points on the x-axis is E,,, = E, + E>+ E3.

26- 1 8

Chapter 26

Solve: (a) We note from the symmetry that the y-component of the net field will have only an x-component. The x-components of

El and E3 cancel. Since i2has no y-component, E, and E, are equal, so

(Ena)x= E2-2E1cos€J =ON/C Note that the signs of q1and q3were used in writing this equation. The electric field strength due to q2 is

The electric field strength due to q1is 1-1911 1 4 El = -4n&,‘1‘ 4m0 x 2 +d‘ From geometry, X

cos@=-= ‘1

Assembling these pieces, the net field is 1 2q -----

+ E,,, -

[

=2q --

4m, x2

2.(

+d?)3’2]’

(b) For x << d, the observation point is very close to q2 = +2q. Furthennore, at x = 0 m the fields and E, are nearly opposite to each other and will nearly cancel. So for x << d we expect the field to be that of a point charge +2q at the origin. To test this prediction, we note that for x << d

This is, indeed, the field on the x-axis of point charge 2q at the origin. For x >> d, the three charges appear as a single charge of value qnec= q, + q2 + q3= 0. So we expect E = 0 when x >> d . In this limit,

so the field does rapidly become zero, as expected. (c) A graph of E, is shown in the figure above.

26.39. Model: The rods are lines of charge with a uniform linear charge density.

Visualize:

Triangle center

-L-

Solve: From Example 26.3, the electric field strength E at distance d from the center of a charged rod is

The Electric Field

26- 19

Because point P is equidistant from the center for each of the three rods, the electric field strength is the same for each rod. We have I O X ~ O - ~c

E = E, = E2 = E3 = ( 9 . 0 IO9 ~ N m' / C') -/d ,

From the geometry in the figure, d = tan30" 3 d = 3L

*E=

E, is along the -y

direction,

(4)-

tan30" = (0.05 m)tan30째 = 0.02887 m

90Nm2/ C (0.02887 m)J(0.02887 m)'

+ (0.05 m)'

= 54,000 N / C

makes an angle of 30" below the -x axis, and

makes an angle of 30" below

the +x axis. In component form,

E, = -(54,000

N / C)j

= (54,000 N / C)(-cos30"~- sin30'yf)

E, = (54,000 N / C)(cos30째~- sin30'y) Adding these three vectors,

Erie, = -2(54,000

N / C)sin30"; - (54,000 N / C ) j = -1.08 x 10'3 N / C

Thus, the electric field strength at the center of the triangle is 1.08 x 10' N/C.

26.40. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P26.40. The line charges lie on the x-axis. Solve: (a) From Equation 26.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is

For the left and right line charges, the electric fields are

Ele, makes an angle $ above the +x-axis and of the figure,

Enghtmakes the same angle 4 above the -x-axis.

We now add these two vectors to find E,,, = E,,,

Thus the field strength is

+ E,,,, . The x-components cancel to give

From the geometry

26-20

Chapter 26

(b)To draw a graph of E,, versus y , we calculated E,, at a few selected values of y. y (in units of d)

0 f 0.5 f l f 2 f 3 f 4

-4

0 f4 f 3.2 f 1.88 f 1.30 f 0.98

-3

-2

-1

Yld

26.41. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P26.41. The line charges lie on the x-axis. Solve: (a) From Equation 26.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is

E = - - 1 21ill 4ncO r For the left and right line charges, the electric fields are 1 2il Elen = E n g h l = 4n&0 =

-/,

dl2

Thus the field strength is

41cEoJ-

4il

The Electric Field

26-21

(b) To draw a graph of Ene,versus y, we calculated E,, at a few selected values of y.

y (in units of d)

~ T I E ,d

X 4 1.6 0.47 0.22 0.12

0 0.5 1 .o 2.0 3.0 4.0

+ +, + +

Y ++ ++ + + + + + + + + + + + + + + + + ?

+ +

+ Solve: The field at distance r from an infinite line of charge is

E = - - r1 2A, 47r&,, r It points straight away from the line. With two perpendicular lines, the field due to the line along the x-axis points in the y-direction and depends inversely on distance y. Similarly, the field due to the line along the y-axis points in the x-direction and depends inversely on distance x. That is

E, =-- 1 2il 4Z&, x

The field strength at this point in space is

E, =-- 1 22. 4% Y

26-22

Chapter 26

26.43. Model: The electric field is that of an infinite line charge. ViSualiZe:

Proton

End view of line of charge

Solve: The wire must be negative to attract the proton. From Equation 26.16, the electric field strength due to an infinite line of charge at a distance r from the line charge is

E=--- 1 214 4nt0 r The force on the proton due to this electric field causes the centripetal acceleration:

Using v = 2m/T = 2@, m ( 4 n Z r 2 f ' )- ( 1 . 6 7 ~ 1 0 - ~kg)4n2(1.0x ' n ~) ~( l . O lo6 x s-')* = 2.29 x 2q (9.0 x lo9 N m2 / C2)2(1.60 x C)

I4 = (4neO>

C/m

Because the wire has to be negative, A = -2.29 nC I m .

26.44. Model: The electric field is that of a Line of charge of length L. Visualize: X

ItX

- i

-x-

The origin of the coordinate system is at the center of the rod. Divide the rod into many small segments of charge Aq and length Ax'.

Solve: (a) Segment i creates a small electric field E, at point P that points to the right. The net field

E will point

to the right and have E, = E: = 0 N/C. The distance to segment i is x', so

E, = ( E j ) =

41cE0(x-x;)

3 E,

= c ( E j ) x=

- x F 1 4m0

Aq (x-x;)

Aq is not a coordinate, so before converting the sum to an integral we must relate the charge Aq to length Ax'. This is done through the linear charge density A = Q/L, from which

Aq = &'=

-AX' Q

L With this charge, the sum becomes

Now we let Ax'+ &'and replace the sum by an integral from x' = - 4 L to x' = + + L . Thus,

E, = (Q/L) dx' - (Q/L)[ 1 4Z&, - L / z ( x - x f ) 2 4Zt0 x - x ' The electric field strength at x is

ILl2

=(p/L) 4nâ&#x201A;Ź,

L Q * E -= - 1 - L2/4 4 7 X ~I -~L'/4

;

The Electric Field

26-23

(b) For x >> L ,

E=-- 1 Q 47r&0x 2 That is, the line charge behaves like a point charge. (e) Substituting into the above formula E = ( 9 . 0 lo9 ~ N m’ / C 2 )

3.0 x lo4 C (3.0 x lo-’ m)‘ - (4x 5.0 x lo-* m)

=9.82x104 N / C

26.45. Model: The electric field is that of a line charge of length L. Visualize: Please refer to Figure P26.45. Let the bottom end of the rod be the origin of the coordinate system. Divide the rod into many small segments of charge Aq and length Ay‘. Segment i creates a small electric field at the point P that makes an angle 8 with the horizontal. The field has both x and y components, but E, = 0 N/C. The distance to segment i from point P is (x’ + Y ’ ~ ) . Solve: The electric field created by segment i at point P is 11‘

The net field is the sum of all the

Ei,which gives E = Ei. Aq is not a coordinate, so before converting the sum

to an integral we must relate charge Aq to length Ay’. This is done through the linear charge density d = Q/L, from which we have the relationship

With this charge, the sum becomes

Now we let Ay‘+ dy’and replace the sum by an integral from y’ = 0 m to y’ = L . Thus,

26.46. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Visualize: Please refer to Figure 26.15. Solve: From Example 26.5, the on-axis field of a ring of charge Q and radius R is

When i<< R, this means we are near the center of the ring. At that point, segments of charge i and j that are 180’ apart create fields E, and E, that cancel each other. When the fields of all segments of charge around the ring are added. the net result is zero. This is indicated by the above expression because when := 0 m, the electric field is zero. When :>> R, then ( - 2 + R’)”’ = ( : 2 ) 3 i z =

26-24

Chapter 26

If this is used in the expression for E,, we get Ez =-- zQ 4iT&,Z3

1 Q 4Z&, z2

This is the field of a point charge Q as seen along the z-axis.

26.47. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Solve: (a) From Example 26.5, the on-axis field of a ring of charge Q and radius R is

E, =- 1 ZQ 4Z&, (zz + Rz)31z For the field to be maximum at a particular value of z, dE/dz = 0. Taking the derivative,

4 3 12)(2z)

1 (2â&#x20AC;&#x2122;

+ R2)li2- (zz + R2)512

(b) The field strength at the point z = R/& is

26.48. Model: Assume that the semicircular rod is thin and that the charge lies along the semicircle of radius R. Visualize:

Segment i

The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge Aq and arc length As. Segment i creates a small electric field E, at the origin. The line from the origin to segment i makes an angle 8 with the x-axis. Solve: Because every segment i at an angle 8 above the axis is matched by segment j at angle 8 below the axis, the y-components of the electric fields will cancel when the field is summed over all segments. This leads to a net field pointing to the right with

E, =c(Ei)x = CEjcosBi

E, = O N / C

Note that angle 8, depends on the location of segment i. Now all segments are at the same distance r, = R from the origin, so

The linear charge density on the rod is the arc length As through

A = QL,where L is the rodâ&#x20AC;&#x2122;s length. This allows us to relate charge Aq to Aq = ilAs = (QL)As

The Electric Field

26-25

Thus, the net field at the origin is

The sum is over all the segments on the rim of a semicircle, so it will be easier to use polar coordinates and integrate over 8 rather than do a two-dimensional integral in x and y. We note that the arc length As is related to the small angle AQ by As = RA8, so

With AQ + de, the sum becomes an integral over all angles forming the rod. 8 varies from A 8 = 4 2 to 8 = +d2. So we finally arrive at

Since we're given the rod's length L and not its radius R , it will be convenient to let R = U1c. So our final expression for E , now including the vector information, is

(b) Substituting into the above expression, E=

(9.0 x lo9 N m z I C 2 ) 2 z ( 3 0x (0.10 m)'

C ) = 1.70 x io5 N I c

26.49.

Model: Assume that the quarter-circle plastic rod is thin and that the charge lies along the quarter-circle of radius R. Visualize: Y

The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge Aq and arc length As. Solve: (a) Segment i creates a small electric field E, at the origin with two components:

(E,)x = E, cos8,

(E,)"= E, sine,

Note that the angle 8, depends on the location of the segment i. Now all segments are at distance r, = R from the origin, so 1- 4 E, = -- = 4ZE, r2 4XE0 R' The linear charge density of the rod is A = Q/L, where L is the rod's length ( L = quarter-circumference = zR12). This allows us to relate charge Aq to the arc length As through

26-26

Chapter 26

Using As = RA8, the components of the electric field at the origin are cose, =-- ' ' (2Q)RA6'cos8i = =1 Aqq ~ 4n0 R2 IrR

(EiIx

(b) The x- and y-components of the electric field for the entire rod are the integrals of the expressions in part (a) from 8 = 0 rad to 8= d 2 . We have

( c ) The integrals are

The electric field is

26.50. Model: Assume that the plastic sheets are planes of charge. Visualize: Please refer to Figure P26.50. Solve: At point 1 the electric field due to the left sheet and the right sheet are

* E",, = E,,,+ E,,,

= --v o

26.51.

Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electron and proton will have constant acceleration. Visualize:

c -

.FX d = 1 cm

x, = point where

particles pass

The negative plate is at x = 0 m and the positive plate is at x = d = 1 cm. Solve: Both particles accelerate from rest (vo = 0 d s ) , so at time t their positions are x, = xe0++a/

= +act'

x, = xpo+ +apt' = d ++apt'

At some instant of time t , the electron and proton have the same position: -re= x, = x , . This is the point where they pass each other. At this instant, x, = +act,' x, = d + +apt,' These are two equations in the two unknowns s, and t , . From the first equation, tt,' = x , / a p. Using this in the second equation gives X,

=d+Lx,

1 + a,/%

The Electric Field

26-27

To finish, we need to find the accelerations of the electron and proton. Both particles are in a parallel-plate capacitor with E,,, = Q / @ . The field points to the left, so E, = -Q/&4. The proton’s acceleration is

Consequently, the acceleration ratio is a p / a s= rn,/mP. Using this, the point where the two charges pass is XI

1 cm kgl1.67 x

d 1+ m , / m P 1+ 9.1 1 x

= 0.9995 cm

26.52. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the proton. Visualize: Y

Known xo =yo = 0 m v&= 1.0 x 1 0 6 d s

= 2.0 cm 1) = 1.0 x

voy=ods

x’,

C/m’

Find

1+++++++++++++1

Solve: From Equation 26.29, the electric field between the parallel plates

Using the kinematic equation Y, = yo + voy(tl - t o ) + +a,(t,

-Yo

?f = ( V / E , ) ~ The . force on the proton

-to)*,

To determine t, - to, we consider the horizontal motion of the proton. The proton travels a distance of 2.0 cm at a constant speed of 1.0 x l o 6 m/s. The velocity is constant because the only force acting on the proton is due to the field between the plate along the y direction. Using the same kinematic equation, 2.0 x IO-’ m A X = ~ . O X ~ O - *m = v o x ( t l - t o ) + O m ~ ( t , - t o ) = = 2.0 x s 1 . 0 ~ 1 0m~/ s *Ay=+

(1.60 x

C)(1.0 x 10- C / m2)(2.0x lo-’ s)’

(1.67 x lo-’’ kg)(8.85 x IO-’’ C2 / Nm‘)

= 2.2 mm

26.53. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the electron. Visualize: Please refer to Figure 26.53. The condition for the electron to not hit the negative plate is that its vertical velocity should just become zero as the electron reaches the plate. Solve: The force on the electron inside the capacitor is

F = mz = qE *av =

=42

-(1.60~10-’~ c)(i.ox104 N / C ) 9 . 1 1 ~ 1 0 - ~kg’

= -1.756 x lOI5 m / s2

26-28

Chapter 26

The initial velocity vo has two components: vox = vo cos45" and voy = vo sin45". Because the electric field inside the capacitor is along the +y direction, the electron has a negative acceleration that reduces the vertical velocity. We require vlY= 0 m/s if it is not to hit the plate. Using kinematics, ,v: = viy + 2a,(y, - y o )

-/,

voy =

= ,/-2(-1.756 3 Yo

(0 m / s)' = viy + 2a,Ay

x IOl5 m / s2)(0.02m) = 8.381 x lo6 m / s

8.381 x lo6 m / s =1.19x107 m / s sin 45"

26.54. Model: Assume that the electric field inside the capacitor is constant, so constant-acceleration kinematic equations apply. Visualize: Please refer to Figure P26.54. Solve: (a) The force on the electron inside the capacitor is

Because E is directed upward (from the positive plate to the negative plate) and q = -1.60 x lO-I9 C , the acceleration of the electron is downward. We can therefore write the above equation as simply a? = qE/m. To determine E, we must first find ay.From kinematics,

x1 = xo + vo,(tl - to)+ +a,(t, -to)' 3

(tl - t o ) =

3 0.04 m

= o m + vo cos450(t, - t o )+ o m

(0.04 m) = 1 . 1 3 1 4 ~lo-* s (5.0 x lo6 m / s)cos45"

Using the kinematic equations for the motion in the y direction,

* o m 1s = vo sin450+ (:)(5+)

v,, = voy + a?(?) *E=-

2(9.1 x 2 m vo sin450 --

kg)(5.0 x lo6 m I s)sin45"

(-1.60 x lO-I9 C)(1.1314 x lo-' s)

s(t1 - to)

= 3550 N /C

(b) To determine the separation between the two plates, we note that yo = 0 m and voy = (5.0 X lo6 m / s)sin45", but at y = y , , the electron's highest point, v,! = 0 m/s. From kinematics, vtv =

viY+ 2a,(y, - y o )3 o m2 / s2 = v,' sin245" + 2ay(y1- y o )

From part (a),

a" = - =

. r n

(-1.60

C)(3550 N / c) = -6.242 x 1014 m / s2 9.1 x kg

(5.0 x lo6 m / s)' *yl-yo

=-4(-6.242x1Ol4 m / s 2 )

=0.010m = 1.0cm

This is the height of the electron's trajectory, so the minimum spacing is 1 .O cm.

26-29

The Electric Field

26.55. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electrons will have a constant acceleration. Visualize:

Solve: (a) The bottom plate should be positive. The electron needs to be repelled by the top plate, so the top plate must be negative and the bottom plate positive. In other words, the electric field needs to point away from the bottom plate so the electron’s acceleration a is toward the bottom plate. (b) Choose an xy-coordinate system with the x-axis parallel to the bottom plate and the origin at the point of entry. Then the electron’s acceleration, which is parallel to the electric field, is 2 = uj^.Consequently, the problem looks just like a Chapter 6 projectile problem. The kinetic energy K = +mv,’ = 3.0 x lo-’’ J gives an initial speed vo = ( 2 ~ / m ) ”=* 8.1 15 x lo6 m / s. n u s the initial components of the velocity are vxo = vo cos45” = 5.74 x lo6 m /

vyo = vosin45” = 5.74 x lo6 m / s

What acceleration ii will cause the electron to pass through the point (x,,y , ) = ( 1 cm, 0 cm)? The kinematic equations of motion are

x1= xo + vxot,+ + a , ( = vxotl= 0.01 m From the x-equation, t, = x I / v x o= 1.742 x

y, = yo + v,,t,

+ + a / , = vyotI+ + a t ; = o m

s . Using this in the y-equation gives 2v t

a = -yo1- -6.59 x 1015m / s2 t:

But the acceleration of an electron in an electric field is

<,,

(9.1 1 X kg)(-6.59 x 1OIs m / s 2 ) - 4 E -eE % E = - - =ma = 37,500 N / C m m m e 1.60 x 10-l9 c (c) The minimum separation d,, must equal the “height” y,, of the electron’s trajectory above the bottom plate. (If d were less than y-, the electron would collide with the upper plate.) Maximum height occurs at t = %t, = 8.71 x lo-’’ s . At this instant, Y , = v,,t + + a t 2 = 0.0025 m = 2.5 mm Thus, d,

= 2.5 IILIII.

26.56. Model: Assume that the plates form a parallel-plate capacitor. The electric field inside the capacitor is uniform, so that the protons will have a constant acceleration. Visualize:

1: a@ .-

v‘ 2.0 cm ____)/

Solve: (a) The acceleration needed to slow the protons to 2.0 x IO’ m/s can be obtained using the kinematic equation of motion v i = vi + 2 a ( x , - xo). We have

”; 7

2 ( x , - xo

- (2.0 x 10’ m / s)’ - (2.0 x lo6 m / s)’ = -9.9 x 10’’ m Is’

2(0.02 m)

26-30

Chapter 26

The force on the proton due to the electric field between the plates is F = mu = qE, . The electric field is

E =-= 4

(1.67 x

kg)(-9.9 x 1OI3 m / s2)

+i.6 x 10-l9 c

= -1.033 x lo6 N / C

The minus sign indicates that the field points to the left. Because E = q / g o for a parallel-plate capacitor, where E is the field strength, C2 / Nm2)(1.033x lo6 N / C) = 9.14 x 10" C / m'

q = E,E = (8.85x

It is clear that the electric field must oppose the motion of the protons. This requires that the first plate be negatively charged and the second plate be positively charged. (b) Let us calculate E that makes protons come to rest completely. Combining the equations for E and u in part (a),

rn(4 4) - (1.67 x 2q(x, - xo

kg)[(O m / s)'

- (2 x lo6 m / s)'] =1.O4x1O6 N / C

C)(0.02 m)

2(1.6 x

q = g o E = (8.85 x lo-'' C 2 / Nm2)(1.04 x lo6 N / C) = 9.24 x 10" C / m2

Because the surface charge density in your device is larger than 9.24 X 10" C / m 2 , the protons will not reach the positive plate. The beam will in fact reverse its direction. The protons will not be stopped, so no, the device does not work.

26.57.

Model: The two electrodes form a parallel plate capacitor. The electric field inside the electrodes is uniform, so the electrons have constant acceleration. Visualize:

Electrodes

-2cm Charged drop

Solve: (a) The ink drops are deflected up or down as they experience an electric field between the two parallel electrodes. The electric field exerts a force on the ink drops which is F=qE=muy

E=mu,lq

We therefore need to determine the mass m, the charge q. and the acceleration uy of the ink drops. We have x 10"m)l 4 = (8 x 105)(1.60x

C) = 1.28 x

= 1.131 x lo-" kg

At maximum deflection, the drop's angle upon exiting the plates must be

)

3mm t a n s = - -vy( - - = H . 1 5 * v y vx 2.0cm

=(H0.15)v,= ( H . l 5 ) ( 2 O m / s ) = 3 i 3 m / s

From the kinematic equation vly = voy+ a,(t, - t o ) ,

ay = VlY - V o y

t, - to

a, = +3.0m/s-O m / s

t, -10

We can obtain r, - ro from the x-motion between the plates as follows:

3 1,

xi - xo = 6 mm = vox(t, - t o )+ 0 m xi = xo + vox(t, - t o )+ +a,(t, - to)' k3.0 m / s 6.0 x m 6.0 x m = 3.0 x lo4 s ja, = = k1.0 x 10' m / s' -to = VOX 20m/s 3.0 x lo4 s

The Electric Field

26-3 1

We are now in a position to obtain the field strength E from the equation

E=-

(1.131 xlO-" kg)(l.Ox lo4 m / s2) 1.28 x 10-l3 c

=8.84x105 N / C

(b) The surface charge density is q = &,E = (8.85 x lo-'' C2 / Nm2)(8.84x lo5 N / C) = 7.82 x lod C / m2 The area of each plate is

Aplare = (6.0 x

m) = 24.0 x 10" m2

m)(4.0 x

Thus the charge on the plates is Q = +T,V$$,, = f1.88 x lo-'' C = M.188 nC. Assess: Because of the large acceleration due to the electric field, the acceleration due to gravity can be ignored in the calculations.

26.58. Model:

An orbiting electron experiences a force that causes centripetal acceleration.

Visualize:

Solve: The electron orbits at a radius r = R + h = 3(2 mm) + 1 mm = 2 mm. The force that causes the circular motion is simply the electric force

F given by Coulomb's law:

Klapkm)qe1,1- K e a p b a e

_--

C)(1.0 x

( 9 10' ~ N m2 / C2)(1.6x

0r'

(2.0 x

m)'

= 3.60 x 10-l3 N

For circular motion,

Visualize:Model: An orbiting proton experiences a force that causes centripetal acceleration. 26.59.

Solve: The proton orbits at a radius r = 5 mm + 1 mm = 6 mm = 6.0 x motion is simply the electric force F given by Coulomb's law:

m. The force that causes the circular

The speed of the proton is

2m T

"=-=

21~(6.0 x

1.ox 10" s

= 3.77 x io4 m / s

(1.67 x lo-" kg)(3.77 x lo4 m / s)?(6.0 x lo-? m) =)

I Q d=

(9.0 x 10' N m2 / C2)(1.60x lo-'' C)

= 9.89 x lo-'' C

The charge on the ball must be negative to hold the proton in its orbit, so Q = -9.89 x lo-'' C.

26-32

Chapter 26

26.60. Model: The electron orbiting the proton experiences a force given by Coulomb’s law. visualize:

Solve: The force that causes the circular motion is

where we used v = 2m/T = 2 m f . The frequency is = 7.16 x l O I 5

26.61. Model: The electron orbiting the proton experiences a force given by Coulomb’s law. Visualize:

Solve: The force that causes the circular motion is F = - - -1 qpIqe1 4 m 0 r’

- me(4n2r2f2)

where we used v = 2 m / T = 2 m f . The radius is

(9.0 x lo9 N m‘ / C’)

(1.60 x 10-l~c)(i.60 x 10-l~c) (9.1x

kg)41c2(1.0x lo1*s”)‘

]

= 1.86 x lo-’ m = 18.6 nm

26.62. Model: The orbital motion of the positron-electron system about their center of mass is due to the electric force between the positron and the electron. Solve: The distance between the charges is twice the radius of their orbits about the center of mass. The force that causes the circular motion is

where we used v = 2@. The frequency is = L U X 1 0 ’ ~HZ

9.1 x

kg)16~’(0.5x 10” m)

26.63. Model: The electric field at the dipole’s location is that of the ion with charge q. Visualize:

Dipole

The Electric Field

26-33

Solve: (a) We have p = aE . The units of a are the units of p / E and are ~m ---C* m - c2s2 -N/C N kg (b) The electric field due to the ion at the location of the dipole is

Because

p = aE ,the induced dipole moment is

From Equation 26.1 1, the electric field produced by the dipole at the location of the ion is

The force the dipole exerts on the ion is

According to Newtonâ&#x20AC;&#x2122;s third law, Fdpleon ,on -

Visualize:

-con on

Therefore,

+ S

+ ++ + + +

It points straight away from the line. The dipole consists of charge +q at distance r + s/2 and charge -q at distance r - SI?. The net force on the dipole due to the field of the line is

(r- s/2)-(r+ s/2) (T+

s/2)(r- s/2)

where we used p = qs for the dipole moment. The minus sign shows that this is an attractive force, toward the line of charge. If r >> s, the (SI?)â&#x20AC;&#x2122; term in the denominator is negligible and we find that the line of charge exerts an attractive force of magnitude

26-34

Chapter 26

26.65. Solve: (a) Charges k2.0 nC form an electric dipole. The electric field strength 2.5 cm from the dipole in the plane perpendicular to the dipole is 1150 N/C. What is the charge separation? (b) Solving for s, S=

(1 150 N / C)(0.025 m)' (9 x io9 N m 2 / c2)(2.0 x 1 0 - ~C)

= 9.98 x lo4 m = 1.0 x

m = 1.0 mm

. 26.66. Solve: (a) A very long charged wire has linear charge density 200 nC/m. At what distance from the wire is the field strength 25,000 N/C? (b) Solving for r, r=

(9 x io9 N m2 / c2)2(2.0x io-'

c / m)

25,000 N / C

= 0.144 m = 14.4 cm

26.67. Solve: (a) At what distance along the axis of a charged disk is the electric field strength half its value at the center of the disk? Give your answer in terms of the disk's radius R. (b) Dividing both sides of the equation by q/2e0 ,

-1

26.68. Solve: (a) A proton is released from the positive plate of a parallel-plate capacitor and accelerates toward the negative plate at 2.0 x lo'* m/s2. If the capacitor plates are 2.0 cm x 2.0 cm squares, what is the magnitude of the charge on each? (b) Solve the first equation for E and substitute into the second equation. The charge is

(1.67 x lo-" kg)(2.Ox 10l2 m / s2)(8.85x

C2 / N m2)(0.020 m)2

1.60 x 10-1~ c

= 7.39 x lo-"

26.69. Model: The long charged wire is an infinite line of charge. The charges on the wire and the plastic rod are uniform. Visualize: Please refer to Figure F'26.69. The plastic stirrer is located on the x-axis. Solve: The electric field of the infinite line of charge at a distance x from its axis is 1 21 47r&0 x

E, =--

Because the electric field is a function of x, the plastic stirrer experiences an electric field that varies along the length of the stirrer. We have handled such problems earlier. Take a small segment of the charge on the stirrer and calculate the electric force due to the line charge on this charge segment Aq. A summation of all such forces on the charge segments Aq will yield the net force on the stirrer. This procedure is equivalent to integrating the electric force on a small segment Aq over the length of the stirrer. The force on charge Aq of length Ax at position x due to the infinite line of charge is

In the above expression, d'= Q/L is the linear charge density of the stirrer and 1 is the linear charge density of the plastic rod. We have also used the relation Aq/& = A'. Changing Ax + du and integrating x from x = 0.02 m to x = 0.08 m, the total force on the stirrer is

(

= (9.0 x lo9 N m' / C')2(1.0 x lo-' C / m)

lox lo-'

0.06 m

]

ln(4) = 4 . 1 6 ~IO" N

‘The Electric Field

26-35

26.70. Model: The rod is thin and is assumed to be a line of charge of length L. Visualize: Y L12 h

-L/2

L/2

Solve: (a) The kversus-y graph over the length of the rod is shown in the figure. (b) Consider a segment of charge Aq of length Ay at a distance y from the center of the rod. The amount of charge in this segment is

as = h

Y = alYlAY

Converting Aq to dq, Ay to dy, and integrating from y = -L/2 to y = +W2, the total charge is LIZ

LIZ

+LIZ

Q=jdq =

aydy = 20ri/

alyldy = 2 -LIZ

aLZ 4

Thus the constant a is a=-4Q L2 (c) With the origin of the coordinate system at the center of the rod, consider two symmetrically located charge

segments Aq with length Ay. The electric field at P from the top charge segment makes an angle 8 below the x-axis and the electric field of the bottom charge segment makes an angle 8 above the x-axis. Because the charge density is symmetric about the origin (i.e., A (at -y) = A (at y ) ) . the y-components of the two contributions cancel out. Thus, we have to calculate only the x-component of the electric field at P. Because the electric field strength of the lower half of the rod is the same as that of the upper half, we only need to obtain the electric field strength of half the rod, then multiply by two. The electric field along the x-direction due to a charge segment Aq is

4 AEx =- 1 47E0 (x’ + y’)

x --- 1 X4YIAY h Y 4?E0 (2+ yZ) JTx’+y? 4?E0 ( x Z + ),’)3’2

1 case = -

Changing AE to dE, Ay = dy, integrating y from y = 0 m to y = L/2, and multiplying by 2 to take into account the entire rod, the electric field is LIZ

The last step used the expression for a from part (b).

2ax

26-36

Chapter 26

26.71. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet. Visualize: E

Solve: (a) Divide the sheet into many long strips parallel to the y-axis and of width Ax. Each strip has a linear charge density A = qAx and acts like a long, charged wire. At the point of interest, strip i contributes a small field

Ei of strength E,=-=-2 A

217Ax

47rEO< 4ZEO< By symmetry, the x-components of all the strips will add to zero and the net field will point straight away from the sheet along the z-axis. The net field’s z-component will be

Ez = C ( E Z ) = : ZE, I

The distance

< = (x,’ + z‘)”’

and cos8, = z/r = z/(x,’ + z ’ )

2qAx = E47rEE,(X’ 2

If we now let Ax

112

. Thus,

z 2)”2

(.;

z’)”2

=-x-

2 qz Ax 4 m 0 , X, +..

+ dx, the sum becomes an integral ranging from x = M I 2 to x = U2. This gives

From trigonometry, tan-’(-@)= -tan-’(@).So finally,

(b) As 2 + 0 m,

This is the electric field due to a plane as you can see from Equation 26.26. We obtain this result because in the limit as :+ 0 m, the dimension L becomes extremely large. As :+ -,

where we have used QL = A as the charge per unit length of the sheet. This is the electric field due to a long, charged wire. We obtain this result because for I >> L, the infinitely long sheet “looks” like an infinite line charge.

The Electric Field

26-31

(c) The following table shows the field strength E, in units of q / q for selected values of 2 in units of L. A graph of E; is shown in the figure above.

Ezl:

Z/L

0.5

0 0.25 0.50 1.o 2.0 3.0 4.0

0.35 0.25 0.15

0.08 0.05 0.04

26.72. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet. Visualize: E

r, = d -1,

Chargestnpi Charge strip i

L 2L 3L 4L Solve: (a) Consider a point on the x-axis at a distance d from the center of the sheet of charge. (We'll call this distance d to begin with, rather than X , to avoid confusion with x as the integration variable.) Once again, let the sheet of charge be divided into small strips of width Ax. Each strip has a linear charge density A = qAx and acts like a long. charged wire. Strip i is at distance r, = d - x, from the point of interest, so it contributes the small field

2il

E, =I 4lT&,< In this situation all fields +.r-di rection :

E, point

2qAx ? 1 4K&,(d - X i )

in the same direction. Their x-components all add to give a net field in the

We'll replace the sum with an integral from x = -L/2 to x = +U2,giving

E , =-

~ K E "-I.,2

(d - s)

2r7 4m0

= -[-ln(d

Ll?

- x)]-,,? = A [ - l n ( d - L/2) + ln(d + L/2)]= -1n 41CEO

2d+L (2d-L)

26-38

Chapter 26

Now that the integration is complete, we can note that d really is the x-coordinate of the point of interest. Substituting x for d and changing to vector form, we end up with

(b) If the point is very distant compared to width of the sheet of charge (x >> L). then the sheet of charge looks like a line of charge with linear density (charge per unit length) A = qL. Write

If x >> L, then W2x << 1 and we can use the approximation ln(1 + u) = u if u << 1. Thus

This is the field of a line of charge with A = qL, as we expected. (c) The following table shows the field strength E, in units of 2q/4m,, for selected values of z in units of L. A graph of E, is shown in the figure above.

z lL

E, /&

0.75 1.o 1.5 2.0 3.O 4.0

1.61 1.10 0.69 0.51 0.34 0.25

26.73. Model: The electric field is that of a positively charged ring. The ring is thin and the charge lies along circle of radius R . Visualize:

Solve: (a) From Example 26.5, the electric field on the axis of a ring of radius R is

The force on negative charge -q is

The meaning of the negative sign is that the force on the electron points left when the displacement is to the right and to the right when the displacement is to the left. That is, the electric force tries to keep the charge at z = 0 m. (b) For small amplitude oscillations, that is, when z << R, the force is

This is simply Hooke’s law

= -k: where the “spring constant” is

k=-

4 m 0R’

The Electric Field

26-39

A particle that obeys Hooke's law undergoes simple harmonic motion with a frequency given by

C)(1.6 x

(1.0 x 21c

C)(9.0 x lo9 N / m2 C')

= 2 . 0 ~ 1 0 HZ '~

m)' (9.11~10-~ kg)(l.Ox1O6 '

foilo on b:a:l ^ 1 I

26.74. Model: The electric field is that of the ball. This electric field polarizes charge in the aluminum foil. vialii:

'ball

\ Polarized /charge

Edipok

Fball on fall

Net +

E,,= 0

&,,D'PO'e

t-----+

Please refer to Figure CP26.74. Solve: (a) The electric field at the location of the piece of foil is

(b) The electric field inside the foil is not just the field due to the +q and -q charges on the surfaces of the foil. It is

E,, E,,, = Eball+ E,,,,.

the superposition of the ball's field

and the polarization field

E,I,

that arises when the foil is polarized by the

charged ball. That is, In electrostatic equilibrium, the internal field must be E,, = 6 N/C. (An internal field that wasn't zero would cause the motion of electrons and thus prevent the foil from being in electrostatic equilibrium.) When the charged ball is brought near, the foil is polarized just to the point where - E,,, = - E M , , causing E,,, = 0 N/C.

= 6N/C. We know the ball's field, from part (a). The polarization

field is that of a parallel-plate capacitor since the polarized foil, with +q and -q surfaces, is just like a capacitor. The polarization field points upward, E,,, = ( q / & , ) j ,where q is the surface charge density. Adding the two fields and setting the sum to zero gives

Once we know 17, the actual charge on the foil surface of area A = IrR' is

q = qA = q&

R2 = -Q 4h2

(d) The attractive and repulsive forces can each be separately calculated and combined, but it is easier to consider the polarized foil to be a dipole since it has two opposite charges f q separated by a distance t. Because t << h, we can write the on-axis electric field of the foil as the on-axis field of a dipole. Using Equation 26.12, the electric field produced by the dipole is 3qt 2P E,pOle= --4 a ~ , r ~41c&,h3 The dipole exerts a force on the charged ball

eo,, ,, on

= QE,,,, . From Newton's third law,

26-40

Chapter 26

The foil is lifted when FMlonfO,, 1 mg = pVg , where p is the mass density of the foil and V = nR2t is the foilâ&#x20AC;&#x2122;s volume. That is, when

since the foil thickness t cancels. Using the expression for q we found in part (c) ,

The minimum height for which the ball lifts the foil is (9.0 x IO9 N m2 / C2)(100x

2n(2700 kg / m3)(9.8 m / sâ&#x20AC;&#x2122;)

C)2

= 0.0140m = 1.40cm