chapter 26

THE ELECTRIC FIELD

26.1. Model: Visualize: both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1 is

The electric field is that of the two charges placed on the y-axis. Please refer to Figure Ex26.1. We denote the upper charge by q1 and the lower charge by q2. Because

(9.0 x lo9 N m2 / Cz)(l x C) ebelow +x-axis (cos & - sin e 3) (0.05 m)’ + (0.05 m)’

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

zl =(1800N/C) (A;-$) -’

Similarly, the electric field from q2 is

,!?,,,,, = El + E, = 2(18OO N / C) - i = 25502 N / C (4 Thus, the strength of the electric field is 2550 N/C and its direction is horizontal. Assess: y-components of their fields will cancel when added.

Because the charges are located symmetrically on either side of the x-axis and are of equal value, the

26.2. Model: 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 q1 is

The electric field is that of the two charges located on the y-axis.

cos & - sin 6) (9.0 x lo9 N mz / C’)(l.O x C) E l = 1 = [ (0.05 m)’ + (0.05 m)’ - ( -- “ I 1 , @below +x-axis

4m0 5 2

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

El =(1800N/C) ( 4 5 +Li-L;) Jz Similarly, the electric field from q2 is

- - * 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: cancel.

A quick visualization of the components of the two electric fields shows that the horizontal components

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 q1 and 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 Erie, at the point P indicated by the dot. Solve: The electric fields from q1 and q2 are

(9.0 x lo9 N m' / C')(1 x lo4 C)

(0.05 m)' along +x-axis i = 3 6 0 0 1 N / C

1 @above +x-axis

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

- (9.0 x lo9 N m' / C')(I x C) E? = (cos63.43' 1 + sin63.43'33) = (3221 + 6443) N / C

(0.10 m)' + (0.05 m)'

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

* @ = 9.3' 6 4 4 N / C

3922 N / C tan@ =

Thus, the strength of the electric field at P is

E,, = J(3922 N / C)' + (644 N / C)' = 3975 N/C

and E,, makes an angle of 9.3'above the +x-axis. Assess: special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.

26.4. Model: 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

q2. We will add El and Solve:

Because of the inverse square dependence on distance, E? < E l . Additionally, because the point P has no

The electric field is that of the two charges located on the y-axis.

vectorially to find the strength and the direction of the net electric field vector. The electric fields from q, and q2 are

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

1 '':I, away from q1 along + x axis i = 36001N/C (0.05 m)'

E , = -- - ( 4 m 0 ti- '"' @below -x-axis

From the geometry of the figure, 10 cm 5 cm

tan@ = - 3 8 = 63.43'

- (9.0 x lo9 N m' / C')(1 x C) 3 E2 = (-cos63.43'1- sin63.43'1) = -(3221 + 6443) N / C

(0.10 m)' + (0.05 m)'

E,, = E, + E? =(3278i' - 6441) N / C E,, = J(3278 N / C)' + (-644 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: field as that of a dipole moment. visualize: Y Y

The distances to the observation points are large compared to the size of the dipole, so model the

(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:

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

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

(a) The dipole moment is

= -18.03 N / C 2.0 x 10-”j c m E‘ = --- = -(9.0 x lo9 N m2 / C 2 ) ~ K E , r3 (0. IO m)3

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-’’ 3 c m) E = -- = ( 9 . 0 ~ lo9 N m2/C2) ~ K E , r3 (0.10 m)’

= 36.03 N / C

The field strength at this point is 36.0 N/C.

26.6. Model: field as that of a dipole moment. Visualize: Y

The distances to the observation points are large compared to the size of the dipole, so model the

I

f it + +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

+ r’l? - (0.10 rnl3(3603 N/C) = (2.0 lo-l,j m) * p = - 2 ( 1 / 4 ~ ~ , ) 2(9.0x IO9 N m2 /C2)

26-4 Chapter 26

Plastic

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

EgiaSs pp1aSoc Zglass ~ p p l a S o c & s y l a s n c - I' z, P, At point P At point P, At point P,

z, P,

2, P,

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

= 1.765 x lo5 N / C 10 x 10-~ c Eglxs = (9.0 x lo9 N m' / C')

(0.01 m),/(O.Ol m)' + (0.05 m)'

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 E=1.765x1OS N / C + 0 . 5 1 4 ~ 1 0 ~ N/C=2.28x1OS N / C

At2.0cm E=0.835x105 N / C + 0 . 8 3 5 ~ 1 0 ~ N/C=1.67x105 N / C

At3.0cm E=0.514x105 N / C + 1 . 7 6 5 ~ 1 0 ~ N/C=2 .28x105 N / C

Assess: in the middle of the line connecting the two rods. Also, note that the arrows shown in the figure are not to scale.

The electric field strength in the space between the two rods goes through a minimum. This point is exactly

26.8. Model: The rods are thin. Assume that the charge lies along a line. Visualize: + + + + + + + + i Gght;Engh;T; -- ~

+ 4- + + + + + + 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.


Solve: Example 26.3 gives the electric field strength in the plane that bisects a charged rod:

1 IQl E =- 4zg0 r J M

rod

The electric field from the rod on the right at a distance of 1 cm from the rod is

(0.01 m),/(O.Ol m)' + (0.05 m)' = 1.765 x 10' N / C 10 x 10-~ c

Enght = (9.0 x lo9 N m2 / 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=1.765x105 N / C - 0.514~10 ' N/C=1.25x105 N / C

At2.0cm E=0.835x105 N / C - 0 . 8 3 5 ~ 1 0 ~ N / C = O N / C

At3.0cm E = 1 . 7 6 5 ~ 1 0 ' N / C - 0 . 5 1 4 ~ 1 0 ~ N/C=1.25x1OS N / C

26.9. Model: The rod is thin, so assume the charge visualize: Y

~ z 5 X

II

i

lies along a line.

- - Solve: charged rod is

The force on charge q is F = qE, . From Example 26.3, the electric field a distance r from the center o f a

- 1 Q : (9.0 x lo9 N m2 / C')(50 x C)i' = 1.757 x 1051 N / c E& =-

4K&0 r , / w ' = (0.04 m),/(0.04 m)' + (0.05 m)'

Thus, the force is

F = (-5 x C)(1.757 x 10' N / C)i' = -8.78 x lO"i' N

More generally, F = (8.78 x IO4 N, toward the rod).

26.10. Model: Solve: density A is

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

1 2111 1 214 El,ne(r = 10.0 cm) = - Ehne(r = 5.0 cm) = - 4n&, 5.0 x lo-' m 4K&, 10.0 x IO-' m

26-6 Chapter 26

Dividing the above two equations gives

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

26.11. Solve:

Model: From Equation 26.15, the electric field for an infinite uniformly charged line is

Assume that the wire is thin and that the charge lies on the wire along a Zine.

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. Visualize:

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

Q, = -20 nC Q2 = +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

(9.0 x lo9 N m2 / C2)(0.10 m)(-20 x C) - 4mCE,(z2 + R2)"' -

= -1.288'~ lo4 N / C zQi

[(0.10 m)' + (0.05 m)']312 (E,): =

That is, the field is 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 Q2 is positive, E, will also point to the left. The net field at the midpoint is

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. Visualize:

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

Q, = +20 nC Q2 = +20 nC


-

Solve: 10 cm is

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

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

l(O.10 m)’ +(0.05 m)’]yz zQi - =1.28x104 N / C

(E1)z = 4nE.(z2 + R 2 Y 2 -

That is, 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, i2 will point to the left. The net field at the

midpoint between the two rings is E = E, + Z2 = 0 N/C. (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)

[(0.20 m)’ + (O.O5)’]Yz = 4 1 1 0 N / C ( E d z =

3 E = E, + Z2 = 0 N/C + (41 10 N/C, left) So the electric field strength is 41 10 N/C.

26.14. field of the disk points toward the disk. The electric field points away from the disk for a positively charged disk. Visualize:

Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric

b- 2: = 20 cm ~-4 g T A A vi

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

= -38,000 N I C 1 -6.366 x lod C / m2 1 2(8.85 x C‘ / N m’) + (0.05 m / ~ . l ~ m)2

26-8 Chapter 26

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-- =L= 4L3" lo4 ' m2 = -3.60 x lo5 N / C - El = (3.60 x lo5 N/C, left) ?. 2&, [ '1 2Eo 2(8.85x1O-l2 C2 / N m 2 )

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 electrode as a plane of charge. Solve: From Equation 26.26, the electric field of a plane of charge is

The distance 2.0 mm is very small in comparison to the size of the electrode, so we can model

= 1.13 x lo5 N / C 17 Q - 80 x 1 0 - ~ c Eplme = - = - -

2&, 2&,A 2(8.85 x C z / N m2)(0.20 x 0.20 m)

26.17. Motion: Solve:

The very wide charged electrode is a plane of charge. 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-'' C2 / N m2)(1000 N / C) = 1.39 x 10-12C = 1.39 x nC

Assess: to use the distance z = 5.0 cm given in the problem.

Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have

26.18. Model: the same as that of a point charge Q located at the center of the sphere. Visualize: spherical shell of radius R. Solve:

A spherical shell of charge Q and radius R has an electric field ourside the sphere that is exactly

In the case of a metal ball, the charge resides on its surface. This can then be visualized as a charged

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 = +(lo cm) + 2.0 cm = 7.0 cm = 0.07 m. So,

26.19. Model: exactly the same as that of a point charge Q located at the center of the sphere. Visualize: 2 cm 2 cm

It-rit 6.0 cm -+I+--+

An insulating sphere of charge Q and radius R has an electric field outside the sphere that is


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 0 ~ N m 2 /C2)(10x10-9 C)

(0.04 m)* El = = 5.625 x lo4 N / C

(9.0 x lo9 N m2 / C2)(15 x C) E2 = = 8.438 x lo4 N / C

(0.04 m)'

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 1.41 X

1 0 s N / C .

26.20. Solve:

Model: The electric field inside a capacitor is E = &/&,A. Thus, the charge needed to produce a field of strength E is

The electric field is uniform in a region of space between closely spaced capacitor plates.

Q = &,AE = (8.85 x C2 / N m2)(0.04 m 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. Solve:

Model: The electric field strength inside the capacitor is E = Q/&,A. Thus, the area is

The electric field in a region of space between two charged circular disks is uniform.

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

Assess: spacing and depends only on the diameter of the plates.

As long as the spacing is much less than the plate dimensions, the electric field is independent of the

26.22. Solve:

Model: The electric field in a region of space between the plates of a parallel-plate capacitor is uniform. 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)*](3 x 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,

(0.10 X lo9 kg)(9.8 N / kg)

4 (1.0 x 10’0)(1.60 x C) (F,,,)y = qE -mg = 0 N 3 E = 3 = =6.13x105 N / C

Thus the required field is 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

M R = l c m

Solve: (a) The two disks form a parallel-plate capacitor with surface charge density

From Equation 26.29, the field strength inside a capacitor is

q E ,

3.18 x lo-’ C / m2 8.85 x lo-’’ C2 / N m2

E = - = =3.6Ox1O6 N / C

(b) The electric field points toward the negative plate, so in the coordinate system of the figure E = -3.60 x 106 j N / C . The field exerts forces = qFotonE = eE on the proton, causing an acceleration with a y-component that is

F eE (1.6 X C)(-3.60 X lo6 N / C) = -3.45 x 1014 m / s2 a =-=Y=

’ m m 1.67 x 1 0-27 kg

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)(0.001 m) = 8 . 3 0 ~ lo5 m / s

Assess: acceleration g. That is why we did not explicitly consider g in our calculations.

The acceleration of the proton in the electric field is enormous in comparison to the gravitation

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

ir-------Ax=1.2cm------.I Solve: A constant-acceleration kinematic equation of motion is

v? - v: - (5.0 x 10’ m / s)2 - (0 m / s)’ vf = v,z + 2a(Ax) * u = - - = 1.042 x 10” m / s2

2Ax ~(1.2 x IO-‘ m)

The Electric Field 26-1 1

The net force on the electron in the electric field is F = qE = mu. Thus,

ma (9.1 1 x kg)(1.042 x 1017 m / s2) E = - = =5.93x1Os N / C

4 1.60 x 10-l~ c Hence, the field strength is 5.93 X lo5 N / C .

26.26. Model: Solve:

The infinite charged plane produces a uniform electric field. (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 ,

2 2 ( 1 . 0 ~ 1 0 ~ m/s ) ‘ -(Orn/s)’ a=-- - v o - 2A.r ~ ( 2 . 0 x IO-’ m)

= 2.5 x 10” m / s2

3(8.85 x IO-” C’ / N m2)(9.1 1 x kg)(2.5 x 10” m / s’) a q = = 2.52 x C / m’

i . 6 0 ~ 1 0 - ~ ~ c (b) Using the kinematic equation of motion v1 = vo + adt,

v -v,, - 1 . 0 ~ 1 0 ~ m / s - o m / s At=-- = 4.0 x s

2.50 x 10” m / s2 U

26.27. Model: the plane. Visualize: v

The infinite negatively charged plane produces a uniform electric field that is directed toward

I

I Known I I 7 = -2.0 x C/m2 I I I

vo = 2.0 x 106 m / s v, = 0 mJs

“0

\I / v, =om/s Lh- - - *-- - - -

mp = 1.67 X lo-” kg

Find AX

- - - * - - - - a t

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

-mv; a&=- m 2Ax 2qE

= - qE - -2 -vo

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

&=-- = 0.185 m -mi’;&, ’ - -(1.67 x kg)(2.0 x lo6 m / s)’(8.85 x lo-’’ C’ / N m‘)

q17 (1.60 x C)(-2.0 x lo6 C / m’)

26.28. Model: dipole.

Because r >> s, we can use Equation 26.12 for the electric field in the plane perpendicular to the

26- 12 Chapter 26

Solve: (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 3 . The magnitude of the dipole field at the position of Q is

- -

- The magnitude of Fhpole on is

- FQ on has the same magnitude, so

4m0 r3 Le, direction of

(b) The electric field of charge Q exerts a torque on the dipole. The field E‘ is perpendicular to the dipole 3 , so 8= 90”. The torque is given by Equation 26.34:

26.29. Equation 26.12 for the electric field in the plane that bisects the dipole. Visualize:

Model: The size of a molecule is ~ 0 . 1 nm. The proton is 2.0 nm away, so r >> s and we can use

++ - &pole Fori proton

- Solve: dipole field at the position of the proton is

You can see from the diagram that F,,leoapmfon is opposite to the direction of j . The magnitude of the

5.0 x C m (2.0 x 10-~ rn13

Edpole = 1 P = ( 9 . 0 x 1 O 9 Nm2/C2) = 5.624 x lo5 N/C 4m0 1’

The magnitude of Fdpole on is

F,,leonpmmn = eEapole = ( 1 . 6 0 ~ C)(5.624 x IO5 N/C) = 9.0 x

N, direction opposite 3) .

N -

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

26.30. Solve: maximum torque is exerted when sin0 = 1 or 0 = 90”. Thus,

Model: The external electric field exerts a torque on the dipole moment of the water molecule. From Equation 26.34, the torque exerted on a dipole moment by an electric field is z = pEsin0. The

zmax = pE = (6.2 x C m)(5.0 x 10’ N I C) = 3.10 x lo-*’ N m

26.31. 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:

Model: The electric field is that of three point charges q,, q2 and q3.

The electric field produced by q, is

1 lqll ( 9 . 0 ~ 1 0 ~ Nm’ /C2)(5.0x10” C) El =- -= = 112,500 N / C

47r&0 ‘12 (0.02 m)’


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

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

(0.02 m)' + (0.04 m)' E3 = -- =

4 m 0 r,'

& points away from q3 and makes an angle 4 = tan-'(4 / 2) = 63.43" with the x-axis. So,

E, = E3 cos42 - E3 s ink = (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

E,, = 4- = ,/(132,600 N / C)' + (-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

the +x-axis).

= (133,200 N / C, 5.23' below

26.32. Model: Visualize: Please refer to Figure €96.32. 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 is that of three point charges q,, q2 and q3.

The electric field produced by ql is

1 l q 1 1 (9.0 x IO9 N m' / C2)(5.0x C) E, = -- - - = 112,500 N / C

47r&0 '12 (0.02 m)' - El points toward ql , 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 q3 is

1 1q31 ( 9 . 0 ~ IO9 N m' / C')(lO.O x C) = 45,000 N / C - E, = -- -

4m0 r3, (0.02 m)* + (0.04 m)'

E, points away from q3 and makes an angle @ = tan-'(2 / 4) = 26.57' with the x-axis. So,

E, = -E3 cos& + E3 s ink = (-40,2502 + 20,130j) N / C


E,,, = E, + E, + E, = (-96,5001 - 92,4005) N / C


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. Visualize: Please refer to Figure P26.33. Assume the charges are in the x-y plane. 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:

Model: The electric field is that of three point charges q,, q2 and q3.

The electric field produced by ql is

1 14,1 E, = -- - - (9.0 x IO9 N m' / C2)(10.0 x 10"

= 100,000 N / C 4 7 7 ~ ~ r,' (0.03 m)'

26-14 Chapter 26

El points away from q l , so in component form E, = -100,ooO~ N / C. The electric field produced by q2

is E2 = 28,125 N / C . & points away from q2, so 2, = -28,1251 N / C . Finally, the electric field produced by q3 is

(9.0 x lo9 N m2 / C2)(10.0 x lo-' C)

(0.03 m)' + (0.04 m)' 1 1q 1 E, = -2 =

4m0 r: = 36,000 N / C

E, points toward q3 and makes an angle @ = t d ( 3 / 4) = 36.87" with the x-axis. So,

E, = E3 cos@; + E3 sin& = (28,8001 + 21,600j) N / C


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


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

the +x-axis).

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

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 -., E3 I f

L

Solve: (a) The electric field produced by q1 points toward q1 and is given by

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

E, =-(--)j. 1 Q - 4m, L2

The electric field produced by q3 is

E, points away from q, and makes an angle 4 = tar-'(L/L) = 45" with the x-axis. So



- 1 Q Enel = El + E2 + E3 = - - -

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

26.35. Visualize: v

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

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

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

E =-

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- 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 E, net points to the left and up, E4,,, has a reversed y-component and points to the left and down. Thus,

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

26.36. Visualize: Y E (N/C)

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

I

E2

-6

l j p / - q P . . E‘ .f x _ - - 8 ,,

800,Ooo -

200.000 -

0 I I , X 5 10 15

800,Ooo -

200.000 -

0 I I , X 5 10 15

The figure shows E, and E, due to the individual charges. The total field is E = E, + E,. Solve: and add. Thus,

(a) From symmetry, the y-components of the two electric fields cancel out. The x-components are equal

E = 2 ( E , ) x ’ 1 - - 2E,cosei

The field strength and angle are

E, =&= 141 X X case = - = 4m0r2 4m0(x2 + ?/4) r

- 2 4 X * E = 47CEo(x’ + s?/4)3i2

(b) The electric field at position x is

N I C 18x

[x’ + (0.003 m)2]3’2

E = (9.0 x lo9 N m’ / C’)(2)(1.0 x C)x - - [x’ + (0.003 m)’]’’’


where x has to be in meters. We can now evaluate I? for different values of x:

2 0.002 768,000 4 0.004 576,000 6 0.006 358,000

10 0.010 158,000

(c) We can use the values of part (b) as a start in drawing the graph. Also note that E + 0 NIC as x + 0 m and as x + -. The graph is shown in the figure above.

26.37. Visualize: bisecting axis. The field at a point on the x-axis is Edpole = E+ + E-. Solve: electric field will cancel, that they have equal ycomponents, and that I?hple points in the -y-direction. Thus,

Model: The electric field of a dipole is that of two opposite charges +q that make the dipole. Please refer to Figure 26.7. The figure shows a dipole aligned on the y-axis, so the x-axis is the

From the symmetry of the situation we can see that the x-components of the two contributions to the

- - - .

(Ehple)x = O N / C (Edple), , =-2E+sine

where 0 is the angle E+ makes with the x-axis. From the geometry of the figure, S I 2

sine = [xz + (s I 2)*]1/2

For a point charge +q. the field is

1 9 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)”’ = x3. n u s

- 1 9s;

4Z&0 x3 Ehple - - -

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. Visualize: Y

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

I E

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

26- 1 8 Chapter 26

Solve: the net field will have only an x-component. The x-components of E, and E, are equal, so

(a) We note from the symmetry that the y-component of El and E3 cancel. Since i2 has no y-component,

(Ena)x = E2-2E1cos€J = O N / C Note that the signs of q1 and q3 were used in writing this equation. The electric field strength due to q2 is

The electric field strength due to q1 is

1 1911 - 1 4 El = -- - -- 4n&, ‘1‘ 4m0 x 2 +d‘

From geometry, X cos@=-= ‘1 Jm

Assembling these pieces, the net field is

1 2q 1 2q -----

X + E,,, - = - 2q [ -- (.2 +d?)3’2]’ 4m, x2

(b) For x << d, the observation point is very close to q2 = +2q. Furthennore, at x = 0 m the fields 2, 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: Visualize:

The rods are lines of charge with a uniform

A 43

linear charge density.

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,

( 4 ) d - = tan30" 3 d = - tan30" = (0.05 m)tan30° = 0.02887 m

* E = 9 0 N m 2 / C = 54,000 N / C

3 L

(0.02887 m)J(0.02887 m)' + (0.05 m)'

E, is along the -y direction, E, makes an angle of 30" below the -x axis, and E, makes an angle of 30" below the +x axis. In component form,

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

E, = (54,000 N / C)(cos30°~ - 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: Visualize: Solve: line charge is

The electric field is that of two infinite lines of charge extending out of the page. Please refer to Figure P26.40. The line charges lie on the x-axis.

(a) From Equation 26.15, the electric field strength due to an infinite line of charge at a distance r from the

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

Ele, makes an angle $ above the +x-axis and Enght makes the same angle 4 above the -x-axis. From the geometry of the figure,

1 d -

- - - We now add these two vectors to find E,,, = E,,, + E,,,, . The x-components cancel to give

Thus the field strength is

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

0 f4 f 3.2 f 1.88 f 1.30 f 0.98

8 , I I I I I Yld -4 -3 -2 -1 1 2 3 4 0

26.41. Visualize: Solve: line charge is

Model:

(a) From Equation 26.15, the electric field strength due to an infinite line of charge at a distance r from the

The electric field is that of two infinite lines of charge extending out of the page. Please refer to Figure P26.41. The line charges lie on the x-axis.

E = - - 1 21ill 4ncO r

For the left and right line charges, the electric fields are 1 2il 1 4il

Elen = Enghl = - 4n&0 ,/- = 41cEo J-

d l 2 d

Thus the field strength is


+ + + +

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

~ T I E , d y (in units of d)

+ +, + + + + + + +

X

Y

+ + + + + + + + + + + + + + ? x

0 0.5 1 .o 2.0 3.0 4.0

X 4 1.6 0.47 0.22 0.12

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

- 1 2A, 47r&,, r

E = - - 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

1 22.

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

The field strength at this point in space is

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

E=--- 1 214 4 n t 0 r

c3 Solve: infinite line of charge at a distance r from the line charge is

The wire must be negative to attract the proton. From Equation 26.16, the electric field strength due to an

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 .67~10-~ ' kg)4n2(1.0x n~)~(l .Ox lo6 s-')* - = 2.29 x C / m

(9.0 x lo9 N m2 / C2)2(1.60 x C) I4 = (4neO> 2q

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


The electric field is that of a Line of charge of length L.

X

It X - 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 ) = 3 E, = c ( E j ) x = 1 - x F Aq 41cE0(x-x;) 4m0 (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,

- 1 Q ; E, = - ( Q / L ) 7 - dx' - ( Q / L ) [ 1 I L l 2 =(p/L) L * E = - 4Z&, - L / z ( x - x f ) 2 4Zt0 x - x ' 4n€, X I - L2/4 4 7 ~ ~ X I - L'/4

The electric field strength at x is


(b) For x >> L , E=-- 1 Q

47r&0 x 2 That is, the line charge behaves like a point charge. (e) Substituting into the above formula

=9.82x104 N / C 3.0 x lo4 C

(3.0 x lo-’ m)‘ - (4 x 5.0 x lo-* m) E = ( 9 . 0 ~ lo9 N m’ / C 2 )

26.45. 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:

Model: The electric field is that of a line charge of length L.

11‘

The electric field created by segment i at point P is

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. Visualize: Solve:

Model:

From Example 26.5, the on-axis field of a ring of charge Q and radius R is

Assume that the ring of charge is thin and that the charge lies along circle of radius R. Please refer to Figure 26.15.

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 iz =

26-24 Chapter 26

If this is used in the expression for E,, we get

Ez =-- zQ 1 Q 4iT&,Z3 4Z&, z2

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

26.47. Model: Solve:

Assume that the ring of charge is thin and that the charge lies along circle of radius R. (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 - 3 2 (2’ + 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.

Segment i I 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. 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 A = QL, where L is the rod’s length. This allows us to relate charge Aq to the arc length As through

Aq = il As = (QL)As


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 A8 = 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,

(9.0 x lo9 N mz I C 2 ) 2 z ( 3 0 x C ) E = = 1.70 x io5 N I c (0.10 m)'

26.49. Model: of radius R.

Assume that the quarter-circle plastic rod is thin and that the charge lies along the quarter-circle

Visualize: Y

X

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, r 2 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

1 Aq cose, =-- ' ' ( - 2Q)RA6'cos8i = (EiIx = q ~ 4n0 R2 IrR

(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: Visualize: Solve:

Assume that the plastic sheets are planes of charge. Please refer to Figure P26.50.

At point 1 the electric field due to the left sheet and the right sheet are

* E",, = E,,, + E,,, = -- v o E EO

26.51. capacitor is a uniform field, so the electron and proton will have constant acceleration. Visualize:

Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate

~~ - - - c---

- - 4- - .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

a d X , = d + L x , X , =-

a, 1 + a,/%


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 ap /as = rn,/mP. Using this, the point where the two charges pass is

1 cm = 0.9995 cm - - d

XI = 1 + m, /mP 1 + 9.1 1 x kgl1.67 x kg

26.52. Model: apply to the motion of the proton. Visualize: Y

The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations

Known xo =yo = 0 m

x’, = 2.0 cm 1) = 1.0 x C/m’

Find

v&= 1.0 x 1 0 6 d s v o y = o d s

X

1 + + + + + + + + + + + + + 1 ) I -Yo

Solve: From Equation 26.29, the electric field between the parallel plates f? = ( V / E , ) ~ . The force on the proton is

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

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 lo6 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 1 . 0 ~ 1 0 ~ m / s

A X = ~ . O X ~ O - * m = v o x ( t l - t o ) + O m ~ ( t , - t o ) = = 2.0 x s

(1.60 x C)(1.0 x 10- C / m2)(2.0 x lo-’ s)’ * A y = + = 2.2 mm

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

26.53. Model: apply to the motion of the electron. Visualize: vertical velocity should just become zero as the electron reaches the plate. Solve:

The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations

Please refer to Figure 26.53. The condition for the electron to not hit the negative plate is that its

The force on the electron inside the capacitor is - - F = mz = qE = - 42

m - (1 .60~10-’~ c)( i .ox104 N / C )

* a v = = -1.756 x lOI5 m / s2 9 . 1 1 ~ 1 0 - ~ ’ kg

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

voy = ,/- = ,/-2(-1.756 x IOl5 m / s2)(0.02 m) = 8.381 x lo6 m / s

8.381 x lo6 m / s sin 45"

3 Yo = =1.19x107 m / s

26.54. Model: kinematic equations apply. Visualize: Solve:

Assume that the electric field inside the capacitor is constant, so constant-acceleration

Please refer to Figure P26.54. (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 0.04 m = o m + vo cos450(t, - to ) + o m

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

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

v,, = voy + a?(?) * o m 1 s = vo sin450+ (:)(5+) -

2 m vo sin450

s(t1 - to)

2(9.1 x kg)(5.0 x lo6 m I s)sin45" = 3550 N /C - * E = - --

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

(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,' sin2 45" + 2ay(y1 - y o )

From part (a),

qE (-1.60 X C)(3550 N / c) a =-= = -6.242 x 1014 m / s2 " . r n 9.1 x kg

(5.0 x lo6 m / s)' =0.010m = 1.0cm

* y l - y o =-4(-6.242x1Ol4 m / s 2 )

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


26.55. Model: capacitor is a uniform field, so the electrons will have a constant acceleration. Visualize:

The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate

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 / s vyo = vo sin45” = 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 y , = yo + v,,t, + + a / , = vyotI + + a t ; = o m

From the x-equation, t, = x I / v x o = 1.742 x s . Using this in the y-equation gives

2v t a = -yo1 - - -6.59 x 1015 m / s2 t:

But the acceleration of an electron in an electric field is

ma (9.1 1 X kg)(-6.59 x 1OIs m / s 2 ) % E = - - = - = 37,500 N / C <,, - 4 E -eE 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. uniform, so that the protons will have a constant acceleration. Visualize:

Model: Assume that the plates form a parallel-plate capacitor. The electric field inside the capacitor is

1: 1: v‘

@-.a

2.0 cm ____)/

Solve: equation of motion vi = v i + 2a(x, - xo). We have

a = ”; -

(a) The acceleration needed to slow the protons to 2.0 x IO’ m/s can be obtained using the kinematic

7 1 - (2.0 x 10’ m / s)’ - (2.0 x lo6 m / s)’ = -9.9 x 10’’ m Is’ -

2(x , - xo 1 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

ma (1.67 x kg)(-9.9 x 1OI3 m / s2) E =-= = -1.033 x lo6 N / C

4 +i.6 x 10-l9 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,

q = E,E = (8.85 x C2 / Nm2)(1.033 x lo6 N / C) = 9.14 x 10" C / m'

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),

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

r n ( 4 4) - - ( E =

2q(x, - xo 1 2(1.6 x C)(0.02 m)

3 q = g o E = (8.85 x lo-'' C2 / 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 / m2, 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: uniform, so the electrons have constant acceleration. Visualize:

The two electrodes form a parallel plate capacitor. The electric field inside the electrodes is

Electrodes X

-2cm

Charged drop Solve: electrodes. The electric field exerts a force on the ink drops which is

(a) The ink drops are deflected up or down as they experience an electric field between the two parallel

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 = 1.131 x lo-" kg 1 4 = (8 x 105)(1.60 x C) = 1.28 x C

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

3 m m vx 2.0cm

tans=-- ( vy - - ) = H . 1 5 * v y =(H0.15)v, =(H. l5) (2Om/s)=3i3m/s

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

VlY - V o y ay = - a, = +3.0m/s-O m / s

t, - to t, -10

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

xi = xo + vox (t, - to ) + +a, ( t, - to)' xi - xo = 6 mm = vox ( t, - to ) + 0 m

k3.0 m / s V O X 2 0 m / s 3.0 x lo4 s

3 1, -to = - 6.0 x m 6.0 x m = 3.0 x lo4 s j a, = = k1.0 x 10' m / s' -

The Electric Field 26-3 1

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

(1.131 xlO-" kg)(l.Ox lo4 m / s2) E = - = =8.84x105 N / C

4 1.28 x 10-l3 c (b) The surface charge density is

q = &,E = (8.85 x lo-'' C2 / Nm2)(8.84 x lo5 N / C) = 7.82 x lod C / m2

The area of each plate is

Aplare = (6.0 x Thus the charge on the plates is Q = +T,V$$,, = f1.88 x lo-'' C = M.188 nC. Assess: in the calculations.

m)(4.0 x m) = 24.0 x 10" m2

Because of the large acceleration due to the electric field, the acceleration due to gravity can be ignored

26.58. Model: An orbiting electron Visualize: +

V

experiences a force that causes centripetal acceleration.

Solve: motion is simply the electric force F given by Coulomb's law:

The electron orbits at a radius r = R + h = 3(2 mm) + 1 mm = 2 mm. The force that causes the circular

Klapkm)qe1,1 - Keapbae ( 9 ~ 10' N m2 / C2)(1.6 x C)(1.0 x C) F = _-- - = 3.60 x 10-l3 N

r' r2 (2.0 x m)'

For circular motion,

I

26.59. Visualize: Model: 0- An orbiting proton experiences a force that causes centripetal acceleration.

Solve: motion is simply the electric force F given by Coulomb's law:

The proton orbits at a radius r = 5 mm + 1 mm = 6 mm = 6.0 x m. The force that causes the circular

The speed of the proton is

2m 21~(6.0 x m) T 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)

(9.0 x 10' N m2 / C2)(1.60 x lo-'' C) =) I Q d = = 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. visualize:

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

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 Hz


The electron orbiting the proton experiences a force given by Coulomb’s law. 6 Solve: The force that causes the circular motion is

1 qpIqe1 m e ( 4 n 2 r 2 f 2 ) F = - - - - 4 m 0 r’ r r

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

(1.60 x 10-l~ c)(i.60 x 10-l~ c) (9.0 x lo9 N m‘ / C’) ] = 1.86 x lo-’ m = 18.6 nm

(9.1 x kg)41c2(1.0 x lo1* s”)‘

26.62. electric force between the positron and the electron. Solve: causes the circular motion is

Model: The orbital motion of the positron-electron system about their center of mass is due to the

The distance between the charges is twice the radius of their orbits about the center of mass. The force that

where we used v = 2@. The frequency is

= L U X 1 0 ’ ~ HZ 9.1 x kg)16~’(0.5 x 10” m)

26.63. Visualize: +4 Dipole

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

I. i r


Visualize:

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 Newton’s third law, Fdple on ,on - -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?)’ 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: the plane perpendicular to the dipole is 1150 N/C. What is the charge separation? (b) Solving for s,

(a) Charges k2.0 nC form an electric dipole. The electric field strength 2.5 cm from the dipole in

= 9.98 x lo4 m = 1.0 x m = 1.0 mm (1 150 N / C)(0.025 m)' S =

(9 x io9 N m2 / c2)(2.0 x 10-~ C)

. 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,

(9 x io9 N m2 / c2)2(2.0 x io-' c / m) r = = 0.144 m = 14.4 cm

25,000 N / C

26.67. Solve: 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 ,

(a) At what distance along the axis of a charged disk is the electric field strength half its value at

R ' -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

C2 / N m2)(0.020 m)2 (1.67 x lo-" kg)(2.Ox 10l2 m / s2)(8.85 x = 7.39 x lo-" c

1.60 x 10-1~ c Q =

26.69. Model: are uniform. Visualize: Solve:

The long charged wire is an infinite line of charge. The charges on the wire and the plastic rod

Please refer to Figure F'26.69. The plastic stirrer is located on the x-axis. The electric field of the infinite line of charge at a distance x from its axis is

1 2 1 E, =-- 47r&0 x

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

lox lo-' ln(4) = 4 . 1 6 ~ IO" N ( 0.06 m ] = (9.0 x lo9 N m' / C')2(1.0 x lo-' C / m)

‘The Electric Field 26-35

26.70. Visualize: Y

Model: The rod is thin and is assumed to be a line of charge of length L.

L12 h

-L/2 0 L / 2

Solve: (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

(a) The kversus-y graph over the length of the rod is shown in the figure.

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

aLZ 2 0 4

+L IZ LIZ

Q = j d q = alyldy = 2 aydy = 20ri/ = - 0 -LIZ

Thus the constant a is

4Q L2

a=-

(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

x - 1 X4YIAY -- 1 h Y case = - 1 4 AEx =- 47E0 (x’ + y’) 4?E0 (2 + yZ) JTx’+y? 4?E0 ( x Z + ),’)3’2

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

2ax LIZ

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

26-36 Chapter 26

26.71. Visualize:

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

E

l

I I I I I z

L 2L 3L 4L 0

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

2 A 217Ax

47rEO< 4ZEO< E,=-=-

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 ) : = Z E , I

112 The distance < = (x,’ + z‘)”’ and cos8, = z/r = z/(x,’ + z ’ ) . Thus,

2qAx z 2 qz Ax =-x- 2)”2 (.; + z ’ ) ” 2 2 47rEE,(X’ + Z 4 m 0 , X, +..

E: = E

If we now let Ax + dx, the sum becomes an integral ranging from x = MI2 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.


(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.

Z/L Ezl: 0 0.25 0.50 1 .o 2.0 3.0 4.0

0.5 0.35 0.25 0.15 0.08 0.05 0.04


Model the infinitely long sheet of charge with width L as a uniformly charged sheet.

E

X

.h Chargestnpi

- r, = d -1,

\ Charge strip i

I 0

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

I I I I X

L 2L 3L 4L

- 2 i l ? 2qAx ? E, =- I = 1 4lT&,< 4K&,(d - X i )

In this situation all fields E, point in the same direction. Their x-components all add to give a net field in the +.r-di rect ion :

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

2r7 Ll? ' q 2q 2 d + L = -[-ln(d - x)]-,,? = A [ - l n ( d - L/2) + ln(d + L/2)] = -1n - " 'f

( 2 d - L ) E , =-

~ K E " -I.,2 (d - s) 4m0 41CEO

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 l L 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: circle of radius R. Visualize:

The electric field is that of a positively charged ring. The ring is thin and the charge lies along

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 = - sQ 4m0 R’


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

(c) Substituting into the above expression,

(1.0 x C)(1.6 x C)(9.0 x lo9 N / m2 C') = 2 . 0 ~ 1 0 ' ~ HZ

21c ( 9 . 1 1 ~ 1 0 - ~ ' kg)(l.Ox1O6 m)'

This electric field polarizes charge foilo on ball^ :: Q 1 Edipok

h

26.74. Model: The electric field is that of the vialii:

'ball

ball.

\ Polarized /charge

Fball on fall - &,,D'PO'e I +

t-----+

Net E,,= 0

2R Please refer to Figure CP26.74. Solve: (a) The electric field at the location of the piece of foil is

in the aluminum foil.

(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 the superposition of the ball's field E,, and the polarization field E,I, that arises when the foil is polarized by the

charged ball. That is, E,,, = Eball + E,,,,. 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,,, = -EM, , causing E,,, = 0 N/C.

(c) In part (b) we noted that E,, = Ebdl + E,,, = 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 R2 q = qA = q& = -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 foil’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 I O 9 N m2 / C2)(100 x C ) 2 = 0.0140m = 1.40cm 2n(2700 kg / m3)(9.8 m / s’)

chapter 26

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