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Chapter 21 The Electric Field 1: Discrete Charge DistributionsConceptual Problems*1 Similarities: The force between charges and masses varies as 1/r2. The force is directly proportional to the product of the charges or masses.

Differences: There are positive and negative charges but only positive masses. Like charges repel; like masses attract.

The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. 2 Determine the Concept No. In order to charge a body by induction, it must have charges that are free to move about on the body. An insulator does not have such charges. 3 Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B. When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and (c) is correct. 4 (a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged. (b) Bring the insulating rod in contact with the metal sphere; some of the positive charge on the rod will be transferred to the metal sphere. (c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that negatively charged sphere to charge the second metal sphere positively by induction.

1

2

Chapter 21

*5 Determine the Concept Because the spheres are conductors, there are free electrons on them that will reposition themselves when the positively charged rod is brought nearby. (a) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram. (b) When the spheres are separated and far apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the diagram. 6 Determine the Concept The forces acting on +q are shown in the diagram. The force acting on +q due to Q is along the line joining them and directed toward Q. The force acting on +q due to +Q is along the line joining them and directed away from +Q. Because charges +Q and Q are equal in magnitude, the forces due to these charges are equal and their sum (the net force on +q) will be to the right and so (e) is correct. Note that the vertical components of these forces add up to zero. *7 Determine the Concept The acceleration of the positive charge is given by

r r F q0 r a= = E . Because q0 and m are both positive, the acceleration is in the same m m direction as the electric field. (d ) is correct.

*8 r Determine the Concept E is zero wherever the net force acting on a test charge is zero. At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the

The Electric Field 1: Discrete Charge Distributionssquare will be zero. (b) is correct. 9 (a) The zero net force acting on Q could be the consequence of equal collinear charges being equidistant from and on opposite sides of Q. (b) The charges described in (a) could be either positive or negative and the net force on Q would still be zero.

3

(c) Suppose Q is positive. Imagine a negative charge situated to its right and a larger positive charge on the same line and the right of the negative charge. Such an arrangement of charges, with the distances properly chosen, would result in a net force of zero acting on Q. (d) Because none of the above are correct, (d ) is correct. 10 Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the sketch to the right weve assigned 2 field lines to each charge q.

*11 Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right weve assigned 2 field lines to each charge q.

4

Chapter 21

*12 Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right weve assigned 7 field lines to each charge q.

13 Determine the Concept A positive charge will induce a charge of the opposite sign on the near surface of the nearby neutral conductor. The positive charge and the induced charge on the neutral conductor, being of opposite sign, will always attract one another.

(a) is correct.*14 Determine the Concept Electric field lines around an electric dipole originate at the positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. (d ) is correct. *15 Determine the Concept Because 0, a dipole in a uniform electric field will experience a restoring torque whose magnitude is pE x sin . Hence it will oscillate about its equilibrium orientation, = 0. If > a, the charges separated by a would appear to be a single charge (c) 2kq of magnitude 2q. Its field would be given by E x = 2 . xFactor the radicand to obtain:

For a b:

Qinside =and

4 3 b a3 3

(

)

Er (r > b ) =

4k 3 (b a 3 ) 3r 2

=

3 0 r2

(b

3

a3 )

Remarks: Note that E is continuous at r = b.

Cylindrical Symmetry48 Picture the Problem From symmetry, the field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shell. Apply Gausss law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

E dA = S n

10

Qinside

or

2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

2kQinside Qinside = Lr 2rL 0

The Electric Field 2: Continuous Charge Distributions 113For r < R, Qinside = 0 and: For r > R, Qinside = L and:

En (r < R ) = 0

En (r > R ) =

2kL 2k 2k (2R ) = = Lr r r R = 0 r

49 Picture the Problem We can use the definition of surface charge density to find the total charge on the shell. From symmetry, the electric field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the uniformly charged cylindrical shell. (a) Using its definition, relate the surface charge density to the total charge on the shell: Substitute numerical values and evaluate Q:

Q = A = 2RL

Q = 2 (0.06 m )(200 m ) 9 nC/m 2

(

)

= 679 nCE (2 cm ) = 0 E (5.9 cm ) = 0

(b) From Problem 48 we have, for r = 2 cm: (c) From Problem 48 we have, for r = 5.9 cm: (d) From Problem 48 we have, for r = 6.1 cm:

Er =and2

R0 r 1.00 kN/C

E (6.1 cm ) =

(9 nC/m )(0.06 m ) (8.85 10 C /N m )(0.061 m ) =122 2

(e) From Problem 48 we have, for r = 10 cm:

(9 nC/m )(0.06 m ) E (10 cm ) = (8.85 10 C /N m )(0.1 m ) =2

12

2

2

610 N/C

114 Chapter 2250 Picture the Problem From symmetry, the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. Apply Gausss law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

E dA = S n

10

Qinside

or

2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

2kQinside Qinside = Lr 2rL 0

Express Qinside for r < R: Substitute to obtain:

Qinside = (r )V = 0 r 2 LEn (r < R ) =

(

)

0 2k 0 Lr 2 r = Lr 2 0 2 0 R 2

(

)

or, because = R 2

En (r < R ) =

r

Express Qinside for r > R: Substitute to obtain:

Qinside = (r )V = 0 (R 2 L )2k (0 LR En (r > R ) =2

Lr

)=

0 R 22 0 r

or, because = R 2

En (r > R ) =

2 0 r

51 Picture the Problem We can use the definition of volume charge density to find the total charge on the cylinder. From symmetry, the electric field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the uniformly charged cylinder.

The Electric Field 2: Continuous Charge Distributions 115(a) Use the definition of volume charge density to express the total charge of the cylinder: Substitute numerical values to obtain:

Qtot = V = R 2 L

(

)2

Qtot = 300 nC/m 3 (0.06 m ) (200 m ) = 679 nC

(

)

From Problem 50, for r < R, we have: (b) For r = 2 cm:

Er =

2 0

r

(300 nC/m )(0.02 m ) = E (2 cm ) = 2(8.85 10 C /N m )3

r

12

2

2

339 N/C

(c) For r = 5.9 cm:

Er (5.9 cm ) =

(300 nC/m )(0.059 m) = 2(8.85 10 C /N m )3

12

2

2

1.00 kN/C

From Problem 50, for r > R, we have:

Er =

R 22 0 r

(d) For r = 6.1 cm:

Er (6.1 cm ) =

(300 nC/m )(0.06 m ) = 2(8.85 10 C /N m )(0.061 m )3 2

12

2

2

1.00 kN/C

(e) For r = 10 cm:

Er (10 cm ) =

(300 nC/m )(0.06 m ) = 2(8.85 10 C /N m )(0.1 m )3 2

12

2

2

610 N/C

Note that, given the choice of charge densities in Problems 49 and 51, the electric fields for r > R are the same. *52 Picture the Problem From symmetry; the field tangent to the surfaces of the shells must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from

116 Chapter 22the centerline of the infinitely long, uniformly charged cylindrical shells. (a) Apply Gausss law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

E dA = S n

10

Qinside

or

2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

2kQinside Lr

(1)

For r < R1, Qinside = 0 and:

En (r < R1 ) = 0

Express Qinside for R1 < r < R2: Substitute in equation (1) to obtain:

Qinside = 1 A1 = 2 1R1 L En (R1 < r < R2 ) =2k (2 1 R1 L ) Lr R = 1 1 0 r

Express Qinside for r > R2:

Qinside = 1 A1 + 2 A2 = 2 1 R1 L + 2 2 R2 L

Substitute in equation (1) to obtain:

En (r > R2 ) =

2k (2 1 R1 L + 2 2 R2 L ) Lr R + 2 R2 = 1 1 0 r

(b) Set E = 0 for r > R2 to obtain:

1 R1 + 2 R20 ror

=0

1 R1 + 2 R2 = 0R 1 = 2 2 R1

Solve for the ratio of 1 to 2:

The Electric Field 2: Continuous Charge Distributions 117Because the electric field is determined by the charge inside the Gaussian surface, the field under these conditions would be as given above: (c) Assuming that 1 is positive, the field lines would be directed as shown to the right.

En (R1 < r < R2 ) =

1 R10 r

53 Picture the Problem The electric field is directed radially outward. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shell. (a) Apply Gausss law to a cylindrical surface of radius r and length L that is concentric with the inner conductor:

E dA = S n

10

Qinside

or

2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

2kQinside Lr

(1)

For r < 1.5 cm, Qinside = 0 and:

En (r < 1.5 cm ) = 0 Qinside = L= 2RL

Letting R = 1.5 cm, express Qinside for 1.5 cm < r < 4.5 cm: Substitute in equation (1) to obtain:

En (1.5 cm < r < 4.5 cm ) ==

2k r

2k (L ) Lr

Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):

118 Chapter 22En (1.5 cm < r < 4.5 cm ) = 2 8.99 10 9 N m 2 /C 2Express Qinside for 4.5 cm < r < 6.5 cm:

(

) (6 nC/m ) = (108 Nr m/C) r

Qinside = 0and

En (4.5 cm < r < 6.5 cm ) = 0

Letting 2 represent the charge density on the outer surface, express Qinside for r > 6.5 cm: Substitute in equation (1) to obtain:

Qinside = 2 A2 = 2 2 R2 Lwhere R2 = 6.5 cm.

E n (r > R2 ) =

2k (2 2 R2 L ) 2 R2 = Lr 0 r

In (b) we show that 2 = 21.2 nC/m2. Substitute numerical values to obtain:

En (r > 6.5 cm ) =

(21.2 nC/m )(6.5 cm) = (8.85 10 C / N m )r2

12

2

2

156 N m/C r

(b) The surface charge densities on the inside and the outside surfaces of the outer conductor are given by: Substitute numerical values and evaluate 1 and 2:

1 =

and 2 = 1 2R1 6 nC/m = 21.2 nC/m 2 2 (0.045 m )

1 =and

2 = 21.2 nC/m 254 Picture the Problem From symmetry considerations, we can conclude that the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. 1 (a) Apply Gausss law to a En dA = Qinside S 0 cylindrical surface of radius r and length L that is concentric with the or infinitely long nonconducting Q 2rLEn = inside cylinder:

0

where weve neglected the end areas

The Electric Field 2: Continuous Charge Distributions 119because no flux crosses them. Solve for En:

En =

Qinside 2rL 0

(1)

Express dQinside for (r) = ar:

dQinside = (r )dV = ar (2rL )dr = 2ar 2 Ldr

Integrate dQinside from r = 0 to R to obtain:

Qinside

r3 = 2aL r dr = 2aL 3 0 0 2aL 3 = R 3R2

R

Divide both sides of this equation by L to obtain an expression for the charge per unit length of the cylinder: (b) Substitute for Qinside in equation (1) to obtain:

=

Qinside 2aR 3 = L 3

2aL 3 r a 2 3 En (r < R ) = = r 2 0 Lr 3 0

For r > R:

Qinside =

2aL 3 R 3

Substitute for Qinside in equation (1) to obtain:

2aL 3 R aR 3 3 En (r > R ) = = 2rL 0 3r 0

55 Picture the Problem From symmetry; the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. (a) Apply Gausss law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

E dA = S n

10

Qinside

or

120 Chapter 22 2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

Qinside 2rL 0

(1)

Express dQinside for (r) = br2:

dQinside = (r )dV = br 2 (2rL )dr = 2br 3 Ldr

Integrate dQinside from r = 0 to R to obtain:

Qinside

r4 = 2bL r dr = 2bL 4 0 0 bL 4 = R 2R3

R

Divide both sides of this equation by L to obtain an expression for the charge per unit length of the cylinder: (b) Substitute for Qinside in equation (1) to obtain:

=

Qinside bR 4 = L 2

bLEn (r < R ) =

b 3 2 = r 2Lr 0 4 0

r4

For r > R:

Qinside =

bL2

R4

Substitute for Qinside in equation (1) to obtain:

bLEn (r > R ) =

bR 4 2 = 2rL 0 4r 0

R4

56 Picture the Problem From symmetry; the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylindrical shell. Apply Gausss law to a cylindrical surface of radius r and length L that

E dA = S n

10

Qinside

The Electric Field 2: Continuous Charge Distributions 121is concentric with the infinitely long nonconducting cylindrical shell: or

2rLEn =

Qinside 0

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

Qinside 2rL 0

For r < a, Qinside = 0:

E n (r < a ) = 0 Qinside = V = r 2 L a 2 L = L r 2 a 2

Express Qinside for a < r < b:

(

)

Substitute for Qinside to obtain:

En (a < r < b ) = =

L(r 2 a 2 ) 2 0 Lr (r 2 a 2 )2 0 r

Express Qinside for r > b:

Qinside = V = b 2 L a 2 L = L b 2 a 2

(

)

Substitute for Qinside to obtain:

En (r > b ) = =

L(b 2 a 2 ) 2 0 rL (b 2 a 2 )2 0 r

57 Picture the Problem We can integrate the density function over the radius of the inner cylinder to find the charge on it and then calculate the linear charge density from its definition. To find the electric field for all values of r we can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gausss law to each region of the cable to find the electric field as a function of the distance from its centerline. R R (a) Find the charge Qinner on the C Qinner = (r )dV = 2rLdr inner cylinder: r

0

0

= 2CL dr = 2CLR0

R

122 Chapter 22Relate this charge to the linear charge density: Substitute numerical values and evaluate inner: (b) Apply Gausss law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

inner =

Qinner 2CLR = = 2CR L L

inner = 2 (200 nC/m )(0.015 m )= 18.8 nC/m

S

En dA =

1 Qinside 0 Qinside 0

or

2rLEn =

where weve neglected the end areas because no flux crosses them. Solve for En:

En =

Qinside 2rL 0 2CLr C = 2 0 Lr 0

Substitute to obtain, for r < 1.5 cm: Substitute numerical values and evaluate En(r < 1.5 cm):

En (r < 1.5 cm ) =

200 nC/m 2 E n (r < 1.5 cm ) = 8.85 10 12 C 2 /N m 2 = 22.6 kN/C

Express Qinside for 1.5 cm < r < 4.5 cm: Substitute to obtain, for 1.5 cm < r < 4.5 cm:

Qinside = 2CLR 2CRL 2 0 rL CR 0 r

En (1.5 cm < r < 4.5 cm ) = =where R = 1.5 cm.

Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):

En (1.5 cm < r < 4.5 cm ) =

(200 nC/m )(0.015 m ) = (8.85 10 C /N m )r2

12

2

2

339 N m/C r

Because the outer cylindrical shell is a conductor:

E n (4.5 cm < r < 6.5 cm ) = 0

The Electric Field 2: Continuous Charge Distributions 123For r > 6.5 cm, Qinside = 2CLR and:

E n (r > 6.5 cm ) =

339 N m/C r

Charge and Field at Conductor Surfaces*58 Picture the Problem Because the penny is in an external electric field, it will have charges of opposite signs induced on its faces. The induced charge is related to the electric field by E = /0. Once we know , we can use the definition of surface charge density to find the total charge on one face of the penny. (a) Relate the electric field to the charge density on each face of the penny: Solve for and evaluate :

E=

0

=0 E

= 8.85 10 12 C 2 /N m 2 (1.6 kN/C ) = 14.2 nC/m 2

(

)

(b) Use the definition of surface charge density to obtain: Solve for and evaluate Q:

=

Q Q = 2 A r2

Q = r 2 = (14.2 nC/m 2 )(0.01 m ) = 4.45 pC

59 Picture the Problem Because the metal slab is in an external electric field, it will have charges of opposite signs induced on its faces. The induced charge is related to the electric field by E = / 0 . Relate the magnitude of the electric field to the charge density on the metal slab: Use its definition to express :

E=

0

=

Q Q = A L2 Q L 02

Substitute to obtain:

E=

124 Chapter 22Substitute numerical values and evaluate E:

E=

(0.12 m )

2

(

1.2 nC 8.85 10 12 C 2 /N m 2

)

= 9.42 kN/C60 Picture the Problem We can apply its definition to find the surface charge density of the nonconducting material and calculate the electric field at either of its surfaces from /20. When the same charge is placed on a conducting sheet, the charge will distribute itself until half the charge is on each surface. (a) Use its definition to find :

=

Q 6 nC = = 150 nC/m 2 2 A (0.2 m )

(b) Relate the electric field on either side of the sheet to the density of charge on its surfaces:

E=

2 0

=

150 nC/m 2 2 8.85 10 12 C 2 /N m 2

(

)

= 8.47 kN/C

(c) Because the slab is a conductor the charge will distribute uniformly on its two surfaces so that: (d) The electric field just outside the surface of a conductor is given by:

=

Q 6 nC = = 75.0 nC/m 2 2 2 A 2(0.2 m )

75 nC/m 2 = E= 0 8.85 10 12 C 2 /N m 2 = 8.47 kN/C

61 Picture the Problem We can construct a Gaussian surface in the shape of a sphere of radius r with the same center as the shell and apply Gausss law to find the electric field as a function of the distance from this point. The inner and outer surfaces of the shell will have charges induced on them by the charge q at the center of the shell. (a) Apply Gausss law to a spherical surface of radius r that is concentric with the point charge:

S

En dA =

1 Qinside 0

or

4r 2 En =

Qinside 0(1)

Solve for En:

En =

Qinside 4r 2 0

The Electric Field 2: Continuous Charge Distributions 125For r < a, Qinside = q. Substitute in equation (1) and simplify to obtain: Because the spherical shell is a conductor, a charge q will be induced on its inner surface. Hence, for a < r < b: For r > b, Qinside = q. Substitute in equation (1) and simplify to obtain: (b) The electric field lines are shown in the diagram to the right:

En (r < a ) = Qinside = 0and

q kq = 2 2 4r 0 r

E n (a < r < b ) = 0

En (r > b ) =

q kq = 2 2 4r 0 r

(c) A charge q is induced on the inner surface. Use the definition of surface charge density to obtain: A charge q is induced on the outer surface. Use the definition of surface charge density to obtain:

inner =

q q = 2 4a 4a 2

outer =

q 4b 2

62 Picture the Problem We can construct a spherical Gaussian surface at the surface of the earth (well assume the Earth is a sphere) and apply Gausss law to relate the electric field to its total charge. Apply Gausss law to a spherical surface of radius RE that is concentric with the earth:

S

En dA =

1 Qinside 0

or2 4RE En =

Qinside 02 RE En k

Solve for Qinside = Qearth to obtain:

2 Qearth = 4 0 RE En =

126 Chapter 22Substitute numerical values and evaluate Qearth:

Qearth

(6.37 10 m ) (150 N/C) =6 2

8.99 109 N m 2 /C 2

= 6.77 105 C*63 Picture the Problem Let the inner and outer radii of the uncharged spherical conducting shell be a and b and q represent the positive point charge at the center of the shell. The positive point charge at the center will induce a negative charge on the inner surface of the shell and, because the shell is uncharged, an equal positive charge will be induced on its outer surface. To solve part (b), we can construct a Gaussian surface in the shape of a sphere of radius r with the same center as the shell and apply Gausss law to find the electric field as a function of the distance from this point. In part (c) we can use a similar strategy with the additional charge placed on the shell. (a) Express the charge density on the inner surface: Express the relationship between the positive point charge q and the charge induced on the inner surface qinner: Substitute for qinner to obtain:

inner =

qinner A

q + qinner = 0

inner = inner =

q 4a 2 2.5 C = 0.553 C/m 2 2 4 (0.6 m )

Substitute numerical values and evaluate inner: Express the charge density on the outer surface: Because the spherical shell is uncharged: Substitute for qouter to obtain:

outer =

qouter A

qouter + qinner = 0

outer = outer =

qinner 4b 2 2.5 C = 0.246 C/m 2 2 4 (0.9 m )

Substitute numerical values and evaluate outer:

The Electric Field 2: Continuous Charge Distributions 127(b) Apply Gausss law to a spherical surface of radius r that is concentric with the point charge:

S

En dA =

1 Qinside 0

or

4r 2 En =

Qinside 0(1)

Solve for En:

En =

Qinside 4r 2 0

For r < a = 0.6 m, Qinside = q. Substitute in equation (1) and evaluate En(r < 0.6 m) to obtain:

En (r < a ) ==

q kq 8.99 109 N m 2 /C 2 (2.5 C ) = 2 = 4r 2 0 r r2

(

)

(2.25 10

4

N m 2 /C

) r1

2

Because the spherical shell is a conductor, a charge q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:

Qinside = 0and

E n (0.6 m < r < 0.9 m ) = 0

For r > 0.9 m, the net charge inside the Gaussian surface is q and:

En (r > 0.9 m ) =

kq = r2

(2.25 10

4

N m 2 /C

) r1

2

(c) Because E = 0 in the conductor:

qinner = 2.5 Cand

inner = 0.553 C/m 2as before. Express the relationship between the charges on the inner and outer surfaces of the spherical shell: and

qouter + qinner = 3.5 C qouter = 3.5 C - qinner = 6.0 C6 C = 0.589 C/m 2 2 4 (0.9 m )

outer is now given by:

outer =

128 Chapter 22For r < a = 0.6 m, Qinside = q and En(r < 0.6 m) is as it was in (a): Because the spherical shell is a conductor, a charge q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:

En (r < a ) =

(2.25 10

4

N m 2 /C

) r1

2

Qinside = 0and

E n (0.6 m < r < 0.9 m ) = 0

For r > 0.9 m, the net charge inside the Gaussian surface is 6 C and:

En (r > 0.9 m ) =

kq 1 = 8.99 10 9 N m 2 /C 2 (6 C ) 2 = 2 r r

(

)

(5.39 10

4

N m 2 /C

)1

r2

64 Picture the Problem From Gausss law we know that the electric field at the surface of the charged sphere is given by E = kQ R 2 where Q is the charge on the sphere and R is its radius. The minimum radius for dielectric breakdown corresponds to the maximum electric field at the surface of the sphere. Use Gausss law to express the electric field at the surface of the charged sphere: Express the relationship between E and R for dielectric breakdown: Solve for Rmin:

E=

kQ R2

E max =

kQ 2 RminkQ E max

Rmin =

Substitute numerical values and evaluate Rmin:

Rmin =

(8.99 10

N m 2 /C 2 (18 C ) 3 10 6 N/C9

)

= 23.2 cm

The Electric Field 2: Continuous Charge Distributions 12965 Picture the Problem We can use its definition to find the surface charge density just outside the face of the slab. The electric field near the surface of the slab is given by E = face 0 . We can find the electric field on each side of the slab by adding the fields due to the slab and the plane of charge. (a) Express the charge density per face in terms of the net charge on the slab: Substitute numerical values to obtain: Express the electric field just outside one face of the slab in terms of its surface charge density: Substitute numerical values and evaluate Eface:

face =

q 2 L2

face =

80 C = 1.60 C/m 2 2 2(5 m )

Eslab =

face0

Eslab =

1.60 C/m 2 8.85 10 12 C 2 /N m 2

= 1.81 105 N/C(b) Express the total field on the side of the slab closest to the infinite charged plane:

r r r E near = E plane + E slab = Eplane r Eslab r

=

plane2 0

r

face0

r

where r is a unit vector pointing away from the slab.Substitute numerical values and r evaluate E near :

r E near =

2 C/m 2 r 2 8.85 10 12 C 2 /N m 2

(

1.81 10 N/C r5

=

( 0.680 10

(

)

)

5

N/C r

)

130 Chapter 22Express the total field on the side of the slab away from the infinite charged plane: Substitute numerical values and r evaluate E far :

r E far = plane r + face r 0 2 0r E far = 2 C/m 2 r 2 8.85 10 12 C 2 /N m 2

(

+ 1.81 105 N/C r

=

(2.94 10

(

5

N/C r

)

)

)

The charge density on the side of the slab near the plane is:

near =0 Enear = (8.85 10 12 C 2 /N m 2 )(0.680 105 N/C ) = 0.602 C/m 2The charge density on the far side of the slab is:

near =0 Enear = (8.85 10 12 C 2 /N m 2 )(2.94 105 N/C ) = 2.60 C/m 2

General Problems66 Determine the Concept We can determine the direction of the electric field between spheres I and II by imagining a test charge placed between the spheres and determining the direction of the force acting on it. We can determine the amount and sign of the charge on each sphere by realizing that the charge on a given surface induces a charge of the same magnitude but opposite sign on the next surface of larger radius. (a) The charge placed on sphere III has no bearing on the electric field between spheres I and II. The field in this region will be in the direction of the force exerted on a test charge placed between the spheres. Because the charge at the center is negative,

the field will point toward the center.(b) The charge on sphere I (Q0) will induce a charge of the same magnitude but opposite sign on sphere II: + Q0 (c) The induction of charge +Q0 on the inner surface of sphere II will leave its outer surface with a charge of the same magnitude but opposite sign: Q0 (d) The presence of charge Q0 on the outer surface of sphere II will induce a charge of the same magnitude but opposite sign on the inner surface of sphere III: + Q0

The Electric Field 2: Continuous Charge Distributions 131(e) The presence of charge +Q0 on the inner surface of sphere III will leave the outer surface of sphere III neutral: 0 (f) A graph of E as a function of r is shown to the right:

67 Picture the Problem Because the difference between the field just to the right of the origin E x ,right and the field just to the left of the origin E x ,left is the field due to the nonuniform surface charge, we can express E x ,left and the difference between E x ,right and

0 .Express the electric field just to the left of the origin in terms of E x ,right and 0 : Substitute numerical values and evaluate E x ,left :

E x ,left = E x ,right

0

E x ,left = 4.65 105 N/C

3.10 C/m 2 = 1.15 105 N/C 8.85 10 12 C 2 /N m 2

132 Chapter 2268 Picture the Problem Let P denote the point of interest at (2 m, 1.5 m). The electric field at P is the sum of the electric fields due to the infinite line charge and the point charge.

Express the resultant electric field at P: Find the field at P due the infinite line charge:

r r r E = E + Eq

r 2k 2 8.99 109 N m 2 /C 2 ( 1.5 C/m ) E = r= i = ( 6.74 kN/C ) i 4m rExpress the field at P due the point charge: Referring to the diagram above, determine r and r :

(

)

r kq Eq = 2 r rr = 1.12 mand

r = 0.893i 0.446 jSubstitute and evaluate E q (2 m,1.5 m ) :

r

r 8.99 109 N m 2 /C 2 (1.3 C ) E q (2 m,1.5 m ) = j 0.893i 0.446 (1.12 m )2 j = (9.32 kN/C ) 0.893i 0.446

(

)

(

)

(

)

= (8.32 kN/C ) i (4.16 kN/C ) jSubstitute to obtain:

The Electric Field 2: Continuous Charge Distributions 133

r E (2 m,1.5 m ) = ( 6.74 kN/C ) i + (8.35 kN/C ) i (4.17 kN/C ) j = (1.61 kN/C ) i (4.17 kN/C ) j*69 Picture the Problem If the patch is small enough, the field at the center of the patch comes from two contributions. We can view the field in the hole as the sum of the field from a uniform spherical shell of charge Q plus the field due to a small patch with surface charge density equal but opposite to that of the patch cut out. (a) Express the magnitude of the electric field at the center of the hole: Apply Gausss law to a spherical gaussian surface just outside the given sphere: Solve for Espherical shell to obtain:

E = Espherical shell + Ehole

Espherical shell 4r 2 =

(

)

Qenclosed Q = 0 0

Espherical shell = Ehole = 2 0

Q 4 0 r 2

The electric field due to the small hole (small enough so that we can treat it as a plane surface) is: Substitute and simplify to obtain:

E== =

Q + 2 4 0 r 2 0

Q Q 2 4 0 r 2 0 (4 r 2 ) Q 8 0 r 2

(b) Express the force on the patch:

F = qEwhere q is the charge on the patch.

Assuming that the patch has radius a, express the proportion between its charge and that of the spherical shell: Substitute for q and E in the expression for F to obtain:

q Q a2 = or q = Q a 2 4 r 2 4r 2

a2 Q Q 2a 2 = F = 2 Q 4r 8 r 2 32 0 r 4 0 Q2a 2 32 0 r 4

(c) The pressure is the force divided by the area of the patch:

P=

a 2

=

Q2 32 2 0 r 4

134 Chapter 2270 Picture the Problem The work done by the electrostatic force in expanding the soap bubble is W = PdV . From Problem 69:

P=

Q2 32 2 0 r 4

Express W in terms of dr: Substitute for P and simplify:

W = PdV = P 4r 2 dr W= Q2 8 00.2 m

dr r2 0.1 m

Evaluating the integral yields:

Q2 W= 8 0

1 (3 nC)2 1 1 = + 2 2 12 r 0.1m 8 (8.85 10 C / N m ) 0.2 m 0.1 m 0.2 m

= 2.02 10 7 J71 Picture the Problem We can use E = kq/R2, where R is the radius of the droplet, to find the electric field at its surface. We can find R by equating the volume of the bubble at the moment it bursts to the volume of the resulting spherical droplet. Express the field at the surface of the spherical water droplet:

E=

kq R2

(1)

where R is the radius of the droplet and q is its charge. Express the volume of the bubble just before it pops: Express the volume of the sphere into which the droplet collapses: Because the volume of the droplet and the volume of the bubble are equal: Solve for R: Assume a thickness t of 1 m and evaluate R:

V 4r 2 twhere t is the thickness of the soap bubble.

4 V = R 3 3 4 4 r 2t = R 3 3R = 3 3r 2t R = 3 3(0.2 m ) (1 m ) = 4.93 10 3 m2

The Electric Field 2: Continuous Charge Distributions 135Substitute numerical values in equation (1) and evaluate E:

E=

(8.99 10 N m (4.93 109

2

/ C 2 (3 nC ) m

3

)

)

2

= 1.11 10 6 N/C72 Picture the Problem Let the numeral 1 refer to the infinite plane at x = 2 m and the numeral 2 to the plane at x = 2 m and let the letter A refer to the region to the left of plane 1, B to the region between the planes, and C to the region to the right of plane 2. We can use the expression for the electric field of in infinite plane of charge to express the electric field due to each plane of charge in each of the three regions. Their sum will be the resultant electric field in each region. Express the resultant electric field as the sum of the fields due to planes 1 and 2: (a) Express and evaluate the field due to plane 1 in region A:

r r r E = E1 + E 2

(1)

r E1 ( A) = 1 i 2 0

( ) ( )

3.5 C/m 2 i 2(8.85 10 12 C 2 /N m 2 ) = (198 kN/C) i =Express and evaluate the field due to plane 2 in region A:

r E 2 ( A) = 2 i 2 0 =

( ))( )

6 C/m 2 i 2 8.85 10 12 C 2 /N m 2 = ( 339 kN/C ) i

(

Substitute in equation (1) to obtain:

r E ( A) = (198 kN/C )i + ( 339 kN/C )i =

( 141 kN/C) i

136 Chapter 22(b) Express and evaluate the field due to plane 1 in region B:

r E1 (B ) = 1 i 2 0 = 3.5 C/m 2 i 2 8.85 10 12 C 2 /N m 2 = ( 198 kN/C ) i

(

)

Express and evaluate the field due to plane 2 in region B:

r E 2 (B ) = 2 i 2 0 =

( ))( )

6 C/m 2 i 2 8.85 10 12 C 2 /N m 2 = ( 339 kN/C ) i

(

Substitute in equation (1) to obtain:

r E (B ) = ( 198 kN/C )i + ( 339 kN/C )i =

( 537 kN/C)i

(c) Express and evaluate the field due to plane 1 in region C:

r E1 (C ) = 1 i 2 0 3.5 C/m 2 i 2 8.85 10 12 C 2 /N m 2 = ( 198 kN/C ) i =

(

)

Express and evaluate the field due to plane 2 in region C:

r E 2 (C ) = 2 i 2 0 6 C/m 2 i 2 8.85 10 12 C 2 /N m 2 = (339 kN/C ) i =

(

)

Substitute in equation (1) to obtain:

r E (C ) = ( 198 kN/C ) i + (339 kN/C ) i = (141 kN/C ) i

*73 Picture the Problem We can find the electric fields at the three points of interest by adding the electric fields due to the infinitely long cylindrical shell and the spherical shell. In Problem 42 it was established that, for an infinitely long cylindrical shell of radius R, Er (r < R ) = 0, and Er (r > R ) = R 0 r . We know that, for a spherical shell of radius R, Er (r < R ) = 0, and E r (r > R ) = R 2 0 r 2 .

The Electric Field 2: Continuous Charge Distributions 137

Express the resultant electric field as the sum of the fields due to the cylinder and sphere: (a) Express and evaluate the electric field due to the cylindrical shell at the origin:

r r r E = E cyl + E sph

(1)

r E cyl (0,0) = 0because the origin is inside the cylindrical shell.

Express and evaluate the electric field due to the spherical shell at the origin:2 r R 2 12 C/m 2 (0.25 m ) Esph (0,0) = i = (339 kN/C ) i i = 2 2 2 12 0 r 2 8.85 10 C /N m (0.5 m )

( ) ( (

)

)

( )

Substitute in equation (1) to obtain:

r E (0,0 ) = 0 + (339 kN/C ) i=or

(339 kN/C) i

E (0,0 ) = 339 kN/C

and

= 0(b) Express and evaluate the electric field due to the cylindrical shell at (0.2 m, 0.1 m):

138 Chapter 22

r (6 C/m 2 )(0.15 m) R E cyl (0.2 m,0.1 m ) = i= 0 r (8.85 1012 C2 /N m 2 )(0.2 m ) i = (508 kN/C) iExpress the electric field due to the charge on the spherical shell as a function of the distance from its center: Referring to the diagram shown above, find r and r :

r R 2 Esph (r ) = r 0 r 2 where r is a unit vector pointing from (50cm, 0) to (20 cm, 10 cm).

r = 0.316 mand

r r = 0.949i + 0.316 j

Substitute to obtain:

r Esph (0.2 m,0.1 m ) =

( 12 C/m )(0.25 m ) ( 0.949i + 0.316 j ) (8.85 10 C /N m )(0.316 m)2 2

12

2

2

2

= ( 849 kN/C ) 0.949i + 0.316 j = (806 kN/C )i + ( 268 kN/C ) jSubstitute in equation (1) to obtain:

(

)

r E (0.2 m,0.1 m ) = (508 kN/C ) i + (806 kN/C ) i + ( 268 kN/C ) j j = (1310 kN/C ) i + ( 268 kN/C ) or

E (0.2 m,0.1 m ) =and

(1310 kN/C)2 + ( 268 kN/C)2 268 kN/C = 348 1310 kN/C

= 1340 kN/C

= tan 1

(c) Express and evaluate the electric field due to the cylindrical shell at (0.5 m, 0.2 m):

r E cyl (0.5 m,0.2 m ) =

(6 C/m )(0.15 m) (8.85 10 C /N m )(0.5 m ) i = (203 kN/C)i212

2

2

Express and evaluate the electric field due to the spherical shell at

r E sph (0.5 m,0.2 m ) = 0

The Electric Field 2: Continuous Charge Distributions 139(0.5 m, 0.5 m): because (0.5 m, 0.2 m) is inside the spherical shell.

Substitute in equation (1) to obtain:

r E (0.5 m,0.2 m ) = (203 kN/C ) i + 0=or

(203 kN/C) i

E (0.5 m,0.2 m ) = 203 kN/C

and

= 074 Picture the Problem Let the numeral 1 refer to the plane with charge density 1 and the numeral 2 to the plane with charge density 2. We can find the electric field at the two points of interest by adding the electric fields due to the charge distributions of the two infinite planes. Express the electric field at any point in space due to the charge distributions on the two planes:

r r r E = E1 + E 2

(1)

(a) Express the electric field at (6 m, 2 m) due to plane 1:

r 65 nC/m 2 1 = (3.67 kN/C ) E1 (6 m,2 m ) = j= j j 2 0 2(8.85 10 12 C 2 /N m 2 )Express the electric field at (6 m, 2 m) due to plane 2:

r 45 nC/m 2 E 2 (6 m,2 m ) = 2 r = r = (2.54 kN/C )r 2 0 2(8.85 10 12 C 2 /N m 2 ) where r is a unit vector pointing from plane 2 toward the point whose coordinates are (6 m, 2 m).Refer to the diagram below to obtain:

r = sin 30i cos 30 j

140 Chapter 22Substitute to obtain:

r E 2 (6 m,2 m ) = (2.54 kN/C ) sin 30i cos 30 = (1.27 kN/C ) i + ( 2.20 kN/C ) j jSubstitute in equation (1) to obtain:

(

)

r E (6 m,2 m ) = (3.67 kN/C ) + (1.27 kN/C ) i + ( 2.20 kN/C ) j j j = (1.27 kN/C ) i + (1.47 kN/C ) (b) Note that E1 (6 m,5 m ) = E1 (6 m,2 m ) so that:

r

r

r 65 nC/m 2 = (3.67 kN/C ) E1 (6 m, 5 m ) = j= j j 2 0 2(8.85 10 12 C 2 / N m 2 )Note also that E 2 (6 m,5 m ) = E 2 (6 m,2 m ) so that:

r

r

r E 2 (6 m,5 m ) = ( 1.27 kN/C ) i + (2.20 kN/C ) jSubstitute in equation (1) to obtain:

r E (6 m,5 m ) = (3.67 kN/C) + ( 1.27 kN/C ) i + (2.20 kN/C) j j=

( 1.27 kN/C)i + (5.87 kN/C) j

75 Picture the Problem Because the atom is uncharged, we know that the integral of the electrons charge distribution over all of space must equal its charge e. Evaluation of this integral will lead to an expression for 0. In (b) we can express the resultant field at any point as the sum of the fields due to the proton and the electron cloud. (a) Because the atom is uncharged:

e = (r )dV = (r )4 r 2 dr0 0

Substitute for (r):

e = 0e0

2r a

4 r dr = 4 0 r 2 e 2 r a dr2 0

Use integral tables or integration by parts to obtain:

2 2r a r e dr = 0

a3 4

The Electric Field 2: Continuous Charge Distributions 141Substitute to obtain:

a3 e = 4 0 = a 3 0 4

Solve for 0:

0 =

e a3

(b) The field will be the sum of the field due to the proton and that of the electron charge cloud: Express the field due to the electron cloud:

E = Ep + Ecloud = kQ(r ) r2

kq + Ecloud r2

Ecloud (r ) =

where Q(r) is the net negative charge enclosed a distance r from the proton. Substitute to obtain:

E (r ) =

ke kQ(r ) + 2 r2 rr

As in (a), Q(r) is given by:

Q(r ) = 4r ' (r ' )dr '0

Integrate to find Q(r) and substitute in the expression for E to obtain:

ke 2 r / a 2r 2r 2 1 + E (r ) = 2 e + 2 a r a

*76 Picture the Problem We will assume that the radius at which they balance is large enough that only the third term in the expression matters. Apply a condition for equilibrium will yield an equation that we can solve for the distance r. Apply

F = 0 to the proton:

2ke 2 2 r / a e mg = 0 a2r= a 2ke 2 ln 2 mga 2

To solve for r, isolate the exponential factor and take the natural logarithm of both sides of the equation: Substitute numerical values and evaluate r:

2 0.0529 nm 2 8.99 109 N m 2 / C 2 1.60 10 19 C r= = 1.16 nm ln 2 2 27 2 1.67 10 kg 9.81 m/s (0.0529 nm )

(

(

)(

)(

)

)

142 Chapter 22 Thus, even though the unscreened electrostatic force is 40 orders of magnitude larger than the gravitational force, screening reduces it to smaller than the gravitational force within a few nanometers.Remarks: Note that the argument of the logarithm contains the ratio between the gravitational potential energy of a mass held a distance a0 above the surface of the earth and the electrostatic potential energy for two unscreened charges a distance a0 apart. 77 Picture the Problem In parts (a) and (b) we can express the charges on each of the elements as the product of the linear charge density of the ring and the length of the segments. Because the lengths of the segments are the product of the angle subtended at P and their distances from P, we can express the charges in terms of their distances from P. By expressing the ratio of the fields due to the charges on s1 and s2 we can determine their dependence on r1 and r2 and, hence, the resultant field at P. We can proceed similarly in part (c) with E varying as 1/r rather than 1/r2. In part (d), with s1 and s2 representing areas, well use the definition of the solid angle subtended by these areas to relate their charges to their distances from point P. (a) Express the charge q1 on the element of length s1:

q1 = s1 = r1where is the angle subtended by the arcs of length s1 and s2.

Express the charge q2 on the element of length s2: Divide the first of these equations by the second to obtain:

q2 = s2 = r2

q1 r1 r = = 1 q2 r2 r2

Express the electric field at P due to the charge associated with the element of length s1: Express the electric field at P due to the charge associated with the element of length s2: Divide the first of these equations by the second to obtain:

E1 =

kq1 ks1 kr1 k = 2 = 2 = r12 r1 r1 r1 k r2

E2 =

k E1 r r = 1 = 2 E2 k r1 r2

The Electric Field 2: Continuous Charge Distributions 143and, because r2 > r1,

E1 > E 2(b) The two fields point away from their segments of arc.

Because E1 > E2 , the resultant field points toward s2 . E1 =and

(c) If E varies as 1/r:

kq1 ks1 kr1 = = = k r1 r1 r1 kq2 ks2 kr2 = = = k r2 r2 r2

E2 =

Therefore:

E1 = E 2

(d) Use the definition of the solid angle subtended by the area s1 to obtain:

s = 12 4 4r1or

s1 = r12

Express the charge q1 of the area s1: Similarly, for an element of area s2:

q1 = s1 = r12 s 2 = r22and

q 2 = r22Express the ratio of q1 to q2 to obtain:

q1 r12 r2 = = 12 q 2 r22 r2 kq1 E1 r2 r 2q r 2r12 = 1 = 22 1 = 22 =1 E2 kq2 r1 q2 r1 r22 r22

Proceed as in (a) to obtain:

144 Chapter 22Because the two fields are of equal magnitude and oppositely directed:

r E =0

r If E 1/r , then s2 would produce the stronger field at P and E would pointtoward s1.78 Picture the Problem We can apply the condition for translational equilibrium to the particle and use the expression for the electric field on the axis of a ring charge to obtain an expression for |q|/m. Doing so will lead us to the conclusion that |q|/m will be a minimum when Ez is a maximum and so well use the result from Problem 26 that

z = R

2 maximizes Ez.

(a) Apply

F

z

= 0 to the particle:

q E z mg = 0

Solve for |q|/m:

q m

=

g Ez

(1)

Note that this result tells us that the minimum value of |q|/m will be where the field due to the ring is greatest. Express the electric field along the z axis due to the ring of charge:

Ez =

(z

kQz2

+ R2

)

32

Differentiate this expression with respect to z to obtain:

dE x d x = kQ dz dz z 2 + R 2

(z = kQ

(

2

+ R2

)

)

z 2 + R2 = kQ 3 2

(

)

3 2

3 2

(z

z( 3 ) z 2 + R2 22

(

+ R2

)

3

) (2 z ) = kQ (z12

(

d 2 z + R2 dx 3 z2 + R2 z2

( )

)

3 2

+ R2

)

3 2

( (z + R )2

3z 2 z 2 + R 22 3

)

12

The Electric Field 2: Continuous Charge Distributions 145Set this expression equal to zero for extrema and simplify:

(z(z

2

+ R2 + R2

2

( (z + R ) ) 3z (z3 2 2 32 2

)

3z 2 z 2 + R 22 3 2

))

12

= 0, = 0,

+ R2

12

and

z 2 + R 2 3z 2 = 0Solve for x to obtain:

z=

R 2

as candidates for maxima or minima. You can either plot a graph of Ez or evaluate its second derivative at these points to show that it is a maximum at: Substitute to obtain an expression Ez,max:

z=

R 2

E z ,max =

R kQ 2 R 2 + R2 2 27 gR 2 2kQ32

=

2kQ 27 R 2

Substitute in equation (1) to obtain:

q m

=

(b) If |q|/m is twice as great as in (a), then the electric field should be half its value in (a), i.e.:

kQ kQz = 2 27 R z2 + R2

(6

)

32

or

1 = 27 R 4

z2 z2 1 + 2 R R 3

Let a = z2/R2 and simplify to obtain:

a 3 + 3a 2 24a + 1 = 0

The graph of f (a ) = a 3 + 3a 2 24a + 1 shown below was plotted using a spreadsheet program.

146 Chapter 2220 15 10 5 0 f (a ) -5 -10 -15 -20 -25 -30 0.0 1.0 2.0 a 3.0 4.0

Use your calculator or trial-and-error methods to obtain: The corresponding z values are:

a = 0.04188 and a = 3.596

z = 0.205 R and z = 1.90 R

The condition for a stable equilibrium position is that the particle, when displaced from its equilibrium position, experiences a restoring force, i.e. a force that acts toward the equilibrium position. When the particle in this problem is just above its equilibrium position the net force on it must be downward and when it is just below the equilibrium position the net force on it must be upward. Note that the electric force is zero at the origin, so the net force there is downward and remains downward to the first equilibrium position as the weight force exceeds the electric force in this interval. The net force is upward between the first and second equilibrium positions as the electric force exceeds the weight force. The net force is downward below the second equilibrium position as the weight force exceeds the electric force. Thus, the first (higher) equilibrium position is stable and the second (lower) equilibrium position is unstable. You might also find it instructive to use your graphing calculator to plot a graph of the electric force (the gravitational force is constant and only shifts the graph of the total force downward). Doing so will produce a graph similar to the one shown in the sketch to the right. Note that the slope of the graph is negative on both sides of 0.205R whereas it is positive on both sides of 1.90R; further evidence that 0.205R is a position of stable equilibrium and 1.90R a position of unstable equilibrium.

The Electric Field 2: Continuous Charge Distributions 14779 Picture the Problem The loop with the small gap is equivalent to a closed loop and a charge of Ql 2R at the gap. The field at the center of a closed loop of uniform line charge is zero. Thus the field is entirely due to the charge Ql 2R . (a) Express the field at the center of the loop: Relate the field at the center of the loop to the charge in the gap: Use the definition of linear charge density to relate the charge in the gap to the length of the gap:

r r r E center = E loop + E gapr kq E gap = 2 r R

(1)

=or

q Q = l 2R Ql 2R

q=

Substitute to obtain:

r kQl E gap = r 2R 3 r kQl kQl E center = 0 r= r 3 2R 2R 3

Substitute in equation (1) to obtain:

If Q is positive, the field at the origin points radially outward.(b) From our result in (a) we see r that the magnitude of E center is:

Ecenter =

kQl 2R 3

80 Picture the Problem We can find the electric fields at the three points of interest, labeled 1, 2, and 3 in the diagram, by adding the electric fields due to the charge distributions on the nonconducting sphere and the spherical shell.

148 Chapter 22

Express the electric field due to the nonconducting sphere and the spherical shell at any point in space: (a) Because (4.5 m, 0) is inside the spherical shell: Apply Gausss law to a spherical surface inside the nonconducting sphere to obtain: Evaluate E sphere (0.5 m ) :

r r r E = E sphere + E shell

(1)

r E shell (4.5 m,0 ) = 0

r 4 E sphere (r ) = kri 3

r

r 4 E sphere (0.5 m ) = 8.988 10 9 N m 2 /C 2 5 C/m 2 (0.5 m ) i = (94.1kN/C ) i 3

(

)(

)

Substitute in equation (1) to obtain:

r E (4.5 m,0) = (94.1 kN/C ) i + 0

=Find the magnitude and direction of

(94.1 kN/C) i

r E (4.5 m,0) :

E (4.5 m,0 ) = 94.1 kN/Cand

= 0(b) Because (4 m, 1.1m) is inside the spherical shell:

r E shell (4 m,1.1m ) = 0

The Electric Field 2: Continuous Charge Distributions 149Evaluate E sphere (1.1 m ) :3 r 4 8.99 10 9 N m 2 /C 2 5 C/m 2 (0.6 m ) E sphere (1.1 m ) = j = (33.6 kN/C ) j 2 3(1.1 m )

r

(

)(

)

Substitute in equation (1) to obtain:

r E (4.5 m,0) = (33.6 kN/C ) + 0 j j = (33.6 kN/C )

Find the magnitude and direction of

r E (4.5 m,1.1m ) :

E (4.5 m,1.1 m ) = 33.6 kN/Cand

= 90(c) Because (2 m, 3 m) outside the spherical shell:

r kQ E shell (r ) = shell r r2 where r is a unit vector pointing from(4 m, 0) to (2 m, 3 m).

Evaluate Qshell:

Qshell = Ashell = 4 1.5 C/m 2 (1.2 m )

(

)

2

= 27.1 C

Refer to the diagram below to find r and r:

r = 3.61 mand

r = 0.555i + 0.832 j

Substitute and evaluate E shell (2 m,3 m ) :

r

r 8.99 109 N m 2 /C 2 ( 27.1 C ) Eshell (3.61 m ) = r (3.61 m )2 = ( 18.7 kN/C ) 0.555i + 0.832 j

(

)

(

)

= (10.4 kN/C ) i + ( 15.6 kN/C ) j

150 Chapter 22Express the electric field due to the charged nonconducting sphere at a distance r from its center that is greater than its radius: Find the charge on the sphere:

r kQsphere E sphere (r ) = r r2

Qsphere = Vsphere =

4 3 5 C/m 2 (0.6 m ) 3

(

)

= 4.52 CEvaluate E sphere (3.61 m ) :

r

r 8.99 10 9 N m 2 /C 2 (4.52 C ) Esphere (2 m,3 m ) = r = (3.12 kN/C )r (3.61 m )2 = (3.12 kN/C ) 0.555i + 0.832 j

(

)

(

)

= ( 1.73 kN/C ) i + (2.59 kN/C ) jSubstitute in equation (1) to obtain:

r E (2 m,3 m ) = (10.4 kN/C ) i + ( 15.6 kN/C ) + ( 1.73 kN/C ) i + (2.59 kN/C ) j j

j = (8.67 kN/C ) i + ( 13.0 kN/C ) Find the magnitude and direction of E (2 m,3 m ) :

r

E (2 m,3 m ) =and

(8.67 kN/C)2 + ( 13.0 kN/C)2 = tan 1 13.0 kN/C = 304 8.67 kN/C

= 15.6 kN/C

81 Picture the Problem Let the numeral 1 refer to the infinite plane whose charge density is 1 and the numeral 2 to the infinite plane whose charge density is 2. We can find the electric fields at the two points of interest by adding the electric fields due to the charge distributions on the infinite planes and the sphere.

The Electric Field 2: Continuous Charge Distributions 151 r r r r Express the electric field due to the E = E sphere + E1 + E 2 (1)infinite planes and the sphere at any point in space: (a) Because (0.4 m, 0) is inside the sphere: Find the field at (0.4 m, 0) due to plane 1:

r E sphere (0.4 m,0) = 0 r E1 (0.4 m,0) = 1 j 2 0

3 C/m 2 j 2 8.85 10 12 C 2 /N m 2 = (169 kN/C ) j =

(

)

Find the field at (0.4 m, 0) due to plane 2:

r 2 C/m 2 2 = E 2 (0.4 m,0) = i i = (113 kN/C ) i 2 0 2 8.85 10 12 C 2 /N m 2Substitute in equation (1) to obtain:

( ) (

)( )

r E (0.4 m,0 ) = 0 + (169 kN/C ) j + (113 kN/C ) i = (113 kN/C ) i + (169 kN/C ) j

Find the magnitude and direction of

r E (0.4 m,0) :

E (0.4 m,0) =

(113 kN/C)2 + (169 kN/C)2

= 203 kN/Cand

= tan 1 (b) Because (2.5 m, 0) is outside the sphere:

169 kN/C = 56.2 113 kN/C

r kQ Esphere (0.4 m,0 ) = sphere r r2 where r is a unit vector pointing from(1 m, 0.6 m) to (2.5 m, 0).

152 Chapter 22Evaluate Qsphere:

Qsphere = Asphere = 4R 2 = 37.7 C

= 4 ( 3 C/m 2 )(1 m )

2

Referring to the diagram above, determine r and r :

r = 1.62 mand

r = 0.928i + 0.371 jSubstitute and evaluate Esphere (2.5 m,0 ) :

r

r 8.99 10 9 N m 2 /C 2 ( 37.7 C ) Esphere (2.5 m,0 ) = r (1.62 m )2 j = ( 129 kN/C ) 0.928i + 0.371

(

)

(

)

= ( 120 kN/C ) i + ( 47.9 kN/C ) jFind the field at (2.5 m, 0) due to plane 1:

r E1 (2.5 m,0) = 1 j 2 0

3 C/m 2 j 2 8.85 10 12 C 2 /N m 2 = (169 kN/C ) j =

(

)

Find the field at (2.5 m, 0) due to plane 2:

r E 2 (2.5 m,0) = 2 i 2 0

2 C/m 2 i = 2(8.85 10 12 C 2 /N m 2 ) = ( 113 kN/C )i

Substitute in equation (1) to obtain:

r E (0.4 m,0) = ( 120 kN/C ) i + ( 47.9 kN/C ) + (169 kN/C ) + ( 113 kN/C ) i j j j = ( 233 kN/C ) i + (121 kN/C ) Find the magnitude and direction of E (2.5 m,0 ) :

r

The Electric Field 2: Continuous Charge Distributions 153E (2.5 m,0 ) =and

( 233 kN/C )2 + (121 kN/C)2

= 263 kN/C

= tan 1

121 kN/C = 153 233 kN/C

82 Picture the Problem Let P represent the point of interest at (1.5 m, 0.5 m). We can find the electric field at P by adding the electric fields due to the infinite plane, the infinite line, and the sphere. Once weve expressed the field at P in vector form, we can find its magnitude and direction. Express the electric field at P: Find E plane at P:

r r r r E = E plane + E line + E sphere r E plane = i 2 0

r

2 C/m 2 i = 2 8.85 10 12 C 2 /N m 2 = ( 113 kN/C ) i

(

)

Express E line at P:

r

r 2k E line = r r

Refer to the diagram to obtain:

r r = (0.5 m ) i (0.5 m ) j

and

r = (0.707 ) i (0.707 ) j

Substitute to obtain:

r 2 8.99 10 9 N m 2 /C 2 (4 C/m ) (0.707 ) i (0.707 ) E line = j 0.707 m = (102 kN/C ) (0.707 ) i (0.707 ) = (72.1 kN/C ) i + ( 72.1 kN/C ) j j

(

)

[

]

[

]

Letting r represent the distance from the center of the sphere to P,

r 4 E sphere = kr'r' 3

154 Chapter 22apply Gausss law to a spherical surface of radius r centered at (1 m, 0) to obtain an expression r for E sphere at P: Refer to the diagram used above to obtain:

where r' is directed toward the center of the sphere.

r r ' = (0.5 m ) i (0.5 m ) jand

r' = (0.707 ) i (0.707 ) j

Substitute to obtain:

r 4 Esphere = j 8.99 109 N m 2 /C 2 (0.707 m ) 6 C/m 3 (0.707 ) i + (0.707 ) 3 j = ( 113 kN/C) i + = ( 113 kN/C ) i + ( 113 kN/C) j

(

)

(

)[

]

(

)

Substitute and evaluate E :

r

r E = ( 113 kN/C ) i + (72.1 kN/C ) i + ( 72.1 kN/C ) + ( 113 kN/C )i j j + ( 113 kN/C )

= ( 154 kN/C ) i + ( 185 kN/C ) jFinally, find the magnitude and direction r of E :

E=

( 154 kN/C)2 + ( 185 kN/C)2

= 241 kN/Cand

= tan 1 83 Picture the Problem We can find the period of the motion from its angular frequency and apply Newtons 2nd law to relate to m, q, R, and the electric field due to the infinite line charge. Because the electric field is given by E r = 2k r we can express and, hence, T as a function of m, q, R, and . Relate the period T of the particle to its angular frequency :

154 kN/C = 220 185 kN/C

T=

2

(1)

The Electric Field 2: Continuous Charge Distributions 155Apply Newtons 2nd law to the particle to obtain: Solve for :

F=

radial

= qEr = mR 2

qE r mR

Express the electric field at a distance R from the infinite line charge: Substitute in the expression for :

E r = 2k

R

=

2kq 1 = mR 2 R

2kq m

Substitute in equation (1) to obtain:

T = 2R

m 2kq

*84 Picture the Problem Starting with the equation for the electric field on the axis of ring charge, we can factor the denominator of the expression to show that, for x a. Substitute for the ri to obtain:

1 V ( x ) = kq x2 + a2 + (b) For x > a, x a > 0 and: Use Ex = dV/dx to find Ex:

1 x2 + a2

+

1 xa

2 1 = kq x2 + a2 + x a

xa = xa

E x (x > a ) =

d 2 1 = + kq 2 dx x + a 2 x a

(x

2kqx2

+a

2 32

)

+

kq (x a )2

For x < a, x a < 0 and: Use Ex = dV/dx to find Ex:

x a = ( x a ) = a x

E x (x < a ) =

d 2 1 = + kq 2 dx x + a 2 a x

(x

2kqx2

+a

2 32

)

kq (a x )2

Calculations of V for Continuous Charge Distributions44 Picture the Problem We can construct Gaussian surfaces just inside and just outside the spherical shell and apply Gausss law to find the electric field at these locations. We can use the expressions for the electric potential inside and outside a spherical shell to find the potential at these locations. (a) Apply Gausss law to a spherical Gaussian surface of radius r < 12 cm:

E dA =S

r

r

Qenclosed

0

=0

because the charge resides on the outer surface of the spherical surface. Hence

r E (r < 12 cm ) = 0

Apply Gausss law to a spherical Gaussian surface of radius

E 4 r 2 =

(

)

q

0

192 Chapter 23r > 12 cm: and

E (r > 12 cm ) =Substitute numerical values and evaluate E (r > 12 cm ) :

q kq = 2 2 4 r 0 r

E (r > 12 cm ) =

(8.99 10

9

N m 2 /C 2 10 8 C = 6.24 kV/m (0.12 m )2

)(

)

(b) Express and evaluate the potential just inside the spherical shell:

V (r R ) =

kq 8.99 10 9 N m 2 /C 2 10 8 C = = 749 V R 0.12 m

(

)(

)

Express and evaluate the potential just outside the spherical shell:

kq (8.99 109 N m 2 /C 2 )(10 8 C ) V (r R ) = = = 749 V r 0.12 m(c) The electric potential inside a uniformly charged spherical shell is constant and given by:

V (r R ) =

kq 8.99 109 N m 2 /C 2 10 8 C = = 749 V R 0.12 mr E (r < 12 cm ) = 0

(

)(

)

In part (a) we showed that:

45 Picture the Problem We can use the expression for the potential due to a line charge V = 2k ln

r , where V = 0 at some distance r = a, to find the potential at these a

distances from the line. Express the potential due to a line charge as a function of the distance from the line: Because V = 0 at r = 2.5 m:

V = 2k ln

r a

0 = 2k ln

2.5 m , a

Electric Potential 193 0 = lnand

2.5 m , a

2.5 m = ln 1 0 = 1 aThus we have a = 2.5 m and:

r r 4 V = 2 8.99 109 N m 2 /C 2 (1.5 C/m ) ln 2.5 m = 2.70 10 N m/C ln 2.5 m

(

)

(

)

(a) Evaluate V at r = 2.0 m:

2m V = 2.70 10 4 N m/C ln 2 .5 m

(

)

= 6.02 kV(b) Evaluate V at r = 4.0 m:

4m V = 2.70 10 4 N m/C ln 2 .5 m

(

)

= 12.7 kV(c) Evaluate V at r = 12.0 m:

12 m V = 2.70 10 4 N m/C ln 2 .5 m

(

)

= 42.3 kV46 Picture the Problem The electric field on the x axis of a disk charge of radius R is given by E x = 2k 1

. Well choose V() = 0 and integrate from x = to x = x2 + R2 x

x to obtain Equation 23-21. Relate the electric potential on the axis of a disk charge to the electric field of the disk: Express the electric field on the x axis of a disk charge:

dV = E x dx

E x = 2k 1

x2 + R2 x

194 Chapter 23Substitute to obtain:

dV = 2k 1

dx x2 + R2 x dx'

Let V() = 0 and integrate from x = to x = x:

x x' V = 2k 1 x'2 + R 2

= 2k

(x

2

+ R2 x

)

R2 = 2k x 1 + 2 1 x which is Equation 23-21. *47 Picture the Problem Let the charge per unit length be = Q/L and dy be a line element with charge dy. We can express the potential dV at any point on the x axis due to dy and integrate of find V(x, 0). (a) Express the element of potential dV due to the line element dy: Integrate dV from y = L/2 to y = L/2:

dV =

k dy r x2 + y2L2

where r =

kQ dy V ( x,0) = 2 x2 + y2 L L =2 2 kQ x + L 4 + L 2 ln x 2 + L2 4 L 2 L

(b) Factor x from the numerator and denominator within the parentheses to obtain:

2 1+ L + L kQ 4x 2 2x V ( x,0 ) = ln 2 L 1+ L L 4x 2 2x

Use ln

a = ln a ln b to obtain: b2 kQ L2 L ln 1 + L L V (x,0 ) = ln 1 + 2 + L 4x 2x 4 x 2 2 x

Electric Potential 195 L2 L2 12 1 1 2 Let = and use (1 + ) = 1 + 2 8 + ... to expand 1 + : 4x 2 4x 2 L2 1 + 2 4x Substitute to obtain:12

1 L2 1 L2 + ... 1 for x >> L. = 1+ 2 4x2 8 4x2

2

V ( x,0) =

kQ L L ln1 + ln1 L 2x 2 x

Let =

L L and use ln(1 + ) = 1 2 + ... to expand ln1 : 2 2x 2x L L L2 L L L2 ln1 + 2 and ln1 2 for x >> L. 2x 4x 2x 2x 4x 2x

Substitute and simplify to obtain:

V ( x,0) =

kQ L L2 L L2 kQ 2 = 2 2x 4x L 2x 4x x

48 Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits. (a) Express the charge dq on a ring of radius r and thickness dr:

R dq = 2rdr = 2r 0 dr r = 2 0 RdrQ = 2 0 R dr = 2 0 R 20

Integrate from r = 0 to r = R to obtain:

R

196 Chapter 23(b) Express the potential on the axis of the disk due to a circular element of charge dq = 2rdr : Integrate from r = 0 to r = R to obtain:

dV =

kdq 2k 0 Rdr = r' x2 + r 2

V = 2k 0 R 0

R

dr x2 + r 2

R + x2 + R2 = 2k 0 R ln x 49 Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits. (a) Express the charge dq on a ring of radius r and thickness dr:

r2 dq = 2rdr = 2r 0 2 dr R 2 0 3 = r dr R22 0 R 3 Q= r dr = R2 01 2

Integrate from r = 0 to r = R to obtain:

0 R 2r3 x2 + r 2

(b)Express the potential on the axis of the disk due to a circular element of charge dq =

dV =

2 0 3 r dr : R2

kdq 2k 0 = r' R2

dr

Integrate from r = 0 to r = R to obtain:

V=

2k 0 R2

0

R

r 3 dr x2 + r 2

=

2k 0 R2

R2 2x2 3

x2 + R2 +

2x3 3

Electric Potential 19750 Picture the Problem Let the charge per unit length be = Q/L and dy be a line element with charge dy. We can express the potential dV at any point on the x axis due to dy and integrate to find V(x, 0). Express the element of potential dV due to the line element dy:

dV =

k dy r x2 + y2L2

where r = Integrate dV from y = L/2 to y = L/2:

V ( x,0) =

kQ dy 2 x2 + y2 L L

2 2 kQ x + L 4 + L 2 ln = L x 2 + L2 4 L 2

*51 Picture the Problem The potential at any location on the axis of the disk is the sum of the potentials due to the positive and negative charge distributions on the disk. Knowing that the total charge on the disk is zero and the charge densities are equal in magnitude will allow us to find the radius of the region that is positively charged. We can then use the expression derived in the text to find the potential due to this charge closest to the axis and integrate dV from

r=R

2 to r = R to find the potential at xV ( x ) = V+ ( x ) + V- ( x )

due to the negative charge distribution. (a) Express the potential at a distance x along the axis of the disk as the sum of the potentials due to the positively and negatively charged regions of the disk: We know that the charge densities are equal in magnitude and that the

Qr aor

198 Chapter 23total charge carried by the disk is zero. Express this condition in terms of the charge in each of two regions of the disk: Solve for a to obtain:

0a 2 = 0R 2 0a 2

a=

R 2

Use this result and the general expression for the potential on the axis of a charged disk to express V+(x): Express the potential on the axis of the disk due to a ring of charge a distance r > a from the axis of the ring: Integrate this expression from r = R 2 to r = R to obtain:

R2 V+ ( x ) = 2k 0 x 2 + x 2

dV ( x ) = 2k 0where r ' =

r dr r'

x2 + r 2 .

V ( x ) = 2k 0R

2

R

r x2 + r 2

dr

R2 = 2k 0 x 2 + R 2 x 2 + 2 Substitute and simplify to obtain:

R2 V ( x ) = 2k 0 x 2 + x 2k 0 x 2 + R 2 2 R2 R2 = 2k 0 x 2 + x x2 + R 2 + x 2 + 2 2 R2 = 2k 0 2 x 2 + x2 + R2 x 2 (b) To determine V for x >> R, factor x from the square roots and expand using the binomial expansion:

x2 +

R2 2

R2 R2 x + = x1 + 2 2x 2 2 R R4 x1 + 2 4x 32 x 4 2

12

and

Electric Potential 199 R2 x + R = x 1 + 2 x 2 R R4 x 1 + 2 4 2x 8x 2 2 12

Substitute to obtain:

k 0 R 4 R2 R4 R2 R4 x 1 + 2 4 x = V ( x ) 2k 0 2 x1 + 2 4x 32 x 4 2 x 8x 8x3 52 Picture the Problem Given the potential function

V ( x ) = 2k 0 2 x 2 + R 2 2 x 2 + R 2 x found in Problem 51(a), we can find Exfrom dV/dx. In the second part of the problem, we can find the electric field on the axis of the disk by integrating Coulombs law for the oppositely charged regions of the disk and expressing the sum of the two fields. dV Relate Ex to dV/dx:

(

)

Ex =

dx

From Problem 51(a) we have:

R2 V ( x ) = 2k 0 2 x 2 + x2 + R2 x 2 Evaluate the negative of the derivative of V(x) to obtain:2 d 2 x2 + R x2 + R2 x E x = 2k 0 2 dx

2x = 2k 0 R2 2 x + 2 Express the field on the axis of the disk as the sum of the field due to the positive charge on the disk and the field due to the negative charge

x 1 2 2 x +R

E x = E x + E x+

200 Chapter 23on the disk: The field due to the positive charge (closest to the axis) is:

E x+

= 2k 0 1 R

x R2 2 x + 2 rdr2

To determine Ex we integrate the field due to a ring charge:

E x = 2k 0R

(x2

+ r2

)

32

x = 2k 0 R2 2 x + 2 Substitute and simplify to obtain:

x 2 2 x +R

x E x = 2k 0 R2 x2 + 2 2x = 2k 0 R2 x2 + 2

x + 2k 0 1 2 2 x +R x 1 2 2 x +R

x 2 R x2 + 2

53 Picture the Problem We can express the electric potential dV at x due to an elemental charge dq on the rod and then integrate over the length of the rod to find V(x). In the second part of the problem we use a binomial expansion to show that, for x >> L/2, our result reduces to that due to a point charge Q.

Electric Potential 201(a) Express the potential at x due to the element of charge dq located at u:

kdq kdu = r xu or, because = Q/L, kQ du dV = L xu dV =kQ du V (x ) = 2 x u L L kQ L2 ln ( x u ) L 2 L L L = ln x + ln x + 2 2 = x+ kQ ln = L x L 2 L 2L2

Integrate V from u = L/2 to L/2 to obtain:

(b) Divide the numerator and denominator of the argument of the logarithm by x to obtain:

L L 1+ x+ 2 = ln 2 x = ln 1 + a ln 1 L x L 1 a 2 2x where a = L/2x.

Divide 1 + a by 1 a to obtain:

2a 2 1+ a ln = ln1 + 2a + 1 a 1 a = ln1 + ln1 + provided x >> L/2.

L2 L x2 + x 2 L x L x

Expand ln(1 + L/x) binomially to obtain:

L L ln1 + x x provided x >> L/2.

Substitute to express V(x) for x >> L/2:

V (x ) =

kQ L kQ = , the field due to a L x x

202 Chapter 23point charge Q. 54 Picture the Problem The diagram is a cross-sectional view showing the charges on the sphere and the spherical conducting shell. A portion of the Gaussian surface over which well integrate E in order to find V in the region r > b is also shown. For a < r < b, the sphere acts like point charge Q and the potential of the metal sphere is the sum of the potential due to a point charge at its center and the potential at its surface due to the charge on the inner surface of the spherical shell. (a) Express Vr > b: Apply Gausss law for r > b:

Vr >b = Er >b dr

S

r Q E r ndA = enclosed = 0

0

and Er>b = 0 because Qenclosed = 0 for r > b. Substitute to obtain: (b) Express the potential of the metal sphere: Express the potential at the surface of the metal sphere: Substitute and simplify to obtain:

Vr >b = (0 )dr = 0

Va = VQ at its center + Vsurface k ( Q ) kQ = b b

Vsurface =

Va =

kQ kQ 1 1 = kQ a b a b

Electric Potential 20355 Picture the Problem The diagram is a cross-sectional view showing the charges on the inner and outer conducting shells. A portion of the Gaussian surface over which well integrate E in order to find V in the region a < r < b is also shown. Once weve determined how E varies with r, we can find Vb Va from Vb Va = Er dr .

Express the potential difference Vb Va: Apply Gausss law to cylindrical Gaussian surface of radius r and length L: Solve for Er:

Vb Va = Er dr E ndA = E (2rL ) = S

r

q0

r

Er =

q2 0 rL 2 L 0

Substitute for Er and integrate from r = a to b:

Vb Va =

q

b

dr r a

=

2kq b ln L a

56 Picture the Problem Let R be the radius of the sphere and Q its charge. We can express the potential at the two locations given and solve the resulting equations simultaneously for R and Q. Relate the potential of the sphere at its surface to its radius: Express the potential at a distance of 20 cm from its surface:

kQ = 450 V R kQ = 150 V R + 0.2 m

(1)

(2)

204 Chapter 23Divide equation (1) by equation (2) to obtain:

kQ 450 V R = kQ 150 V R + 0.2 mor

R + 0.2 m =3 R

Solve for R to obtain: Solve equation (1) for Q:

R = 0.100 m

Q = (450 V ) Q = (450 V )

R k

Substitute numerical values and evaluate Q:

(8.99 10

(0.1m )9

N m 2 /C 2

)

= 5.01 nC57 Picture the Problem Let the charge density on the infinite plane at x = a be 1 and that on the infinite plane at x = 0 be 2. Call that region in space for which x < 0, region I, the region for which 0 < x < a region II, and the region for which a < x region III. We can integrate E due to the planes of charge to find the electric potential in each of these regions.

(a) Express the potential in region I in terms of the electric field in that region: Express the electric field in region I as the sum of the fields due to the charge densities 1 and 2:

x r r VI = E I dx0

r EI = 1 i 2 i = i i 2 0 2 0 2 0 2 0

=

i 0

Electric Potential 205Substitute and evaluate VI:x VI = dx = x + V (0) 0 0 0

=

x+0= x 0 0

Express the potential in region II in terms of the electric field in that region: Express the electric field in region II as the sum of the fields due to the charge densities 1 and 2:

r r VII = E II dx + V (0)

r E II = 1 i + 2 i = i+ i 2 0 2 0 2 0 2 0

=0

Substitute and evaluate VII:

VII = (0 )dx = 0 + V (0 ) = 00

x

Express the potential in region III in terms of the electric field in that region: Express the electric field in region III as the sum of the fields due to the charge densities 1 and 2:

x r r VIII = E III dx a

r E III = 1 i + 2 i = i+ i 2 0 2 0 2 0 2 0

=

i 0

Substitute and evaluate VIII:

x VIII = dx = x + a 0 0 a 0

=

(a x ) 0

(b) Proceed as in (a) with 1 = and 2 = to obtain:

VI = 0 , VII =

x and VIII = a 0 0

*58

Picture the Problem The potential on the axis of a disk charge of radius R and charge density is given by V = 2k x 2 + R 2

[(

)

12

x .

]

206 Chapter 23Express the potential on the axis of the disk charge:

V = 2k x 2 + R 2

[(

)

12

x

]

Factor x from the radical and use the binomial expansion to obtain:

(x

2

+R

2 12

)

R2 R 2 1 1 1 R 4 = x1 + 2 = x 1 + 2 + 4 + ... x 2 x 2 2 2 x 2 4 R R x 1 + 2 4 8x 2x R2 R4 V = 2k x 1 + 2 4 x 8x 2x R2 R4 = 2k 2 x 8x3 R2 kQ 2k = 2x x provided x >> R.

12

Substitute for the radical term to obtain:

59 Picture the Problem The diagram shows a sphere of radius R containing a charge Q uniformly distributed. We can use the definition of density to find the charge q inside a sphere of radius r and the potential V1 at r due to this part of the charge. We can express the potential dV2 at r due to the charge in a shell of radius r and thickness dr at r > r using dV2 = kdq' r and then integrate this expression from r = r to r = R to find V2. (a) Express the potential V1 at r due to q: Use the definition of density and the fact that the charge density is uniform to relate q to Q:

V1 =

kq' r q' Q = 4 3 3 4 3 r 3 R

=

Electric Potential 207Solve for q:

q' =

r3 Q R3k r3 kQ 2 3 Q = r R r R3

Substitute to express V1:

V1 =

(b) Express the potential dV2 at r due to the charge in a shell of radius r and thickness dr at r > r:

dV2 =

kdq' r

Express the charge dq in a shell of radius r and thickness dr at r > r:

3Q dq' = 4r' 2 dr ' = 4r' 2 dr' 3 4R 3Q = 3 r' 2 dr' RdV2 = 3kQ r'dr' R3R

Substitute to obtain:

(c) Integrate dV2 from r = r to r = R to find V2: (d) Express the potential V at r as the sum of V1 and V2:

V2 =

3kQ 3kQ 2 (R r 2 ) r'dr' = 3 3 R r 2R

V = V1 + V2 = kQ 2 3kQ 2 2 r + 3 R r R3 2R

(

)

kQ r2 3 2 = 2R R 60 Picture the Problem We can equate the expression for the electric field due to an infinite plane of charge and V/x and solve the resulting equation for the separation of the equipotential surfaces. Express the electric field due to the infinite plane of charge: Relate the electric field to the potential:

E=

2 0V x

E=

208 Chapter 23Equate these expressions and solve for x to obtain: Substitute numerical values and evaluate x :

x =

2 0 V

x =

2 8.85 10 12 C 2 /N m 2 (100 V ) 3.5 C/m 2

(

)

= 0.506 mm61 Picture the Problem The equipotentials are spheres centered at the origin with radii ri = kq/Vi. Evaluate r for V = 20 V:

r20 V =

(8.99 10

9

N m 2 /C 2 20 V

)(

1 9

10 8 C

)

= 0.499 mEvaluate r for V = 40 V:

r40 V

(8.99 10 == 0.250 m

9

N m 2 /C 2 40 V

)(

1 9

10 8 C

)

Evaluate r for V = 60 V:

r60 V

(8.99 10 == 0.166 m

9

N m 2 /C 2 60 V

)(

1 9

10 8 C

)

Evaluate r for V = 80 V:

r80 V =

(8.99 10

9

N m 2 /C 2 80 V

)(

1 9

10 8 C

)

= 0.125 mEvaluate r for V = 100 V:

r100 V =

(8.99 10

9

N m 2 /C 2 100 V

)(

1 9

10 8 C

)

= 0.0999 m

Electric Potential 209The equipotential surfaces are shown in cross-section to the right:

The equipotential surfaces are not equally spaced.62 Picture the Problem We can relate the dielectric strength of air (about 3 MV/m) to the maximum net charge that can be placed on a spherical conductor using the expression for the electric field at its surface. We can find the potential of the sphere when it carries its maximum charge using V = kQmax R . (a) Express the dielectric strength of a spherical conductor in terms of the charge on the sphere: Solve for Qmax:

Ebreakdown =

kQmax R2

Qmax =

Ebreakdown R 2 k

Substitute numerical values and evaluate Qmax:

Qmax =

(3 MV/m )(0.16 m )28.99 109 N m 2 /C 2

= 8.54 C(b) Because the charge carried by the sphere could be either positive or negative:

Vmax =

kQmax R 8.99 109 N m 2 /C 2 (8.54 C ) = 0.16 m

(

)

= 480 kV*63 Picture the Problem We can solve the equation giving the electric field at the surface of a conductor for the greatest surface charge density that can exist before dielectric breakdown of the air occurs. Relate the electric field at the surface of a conductor to the surface charge density:

E=

0

210 Chapter 23Solve for under dielectric breakdown of the air conditions: Substitute numerical values and evaluate max:

max =0 Ebreaddown max = (8.85 10 12 C 2 /N m 2 )(3 MV/m )= 26.6 C/m 2

64 Picture the Problem Let L and S refer to the larger and smaller spheres, respectively. We can use the fact that both spheres are at the same potential to find the electric fields near their surfaces. Knowing the electric fields, we can use =0 E to find the surface charge density of each sphere. Express the electric fields at the surfaces of the two spheres: Divide the first of these equations by the second to obtain:

ES =

kQS kQ and EL = 2L 2 RS RL

kQS ES R2 Q R2 = S = S L 2 EL kQL QL RS 2 RL Q R kQL kQS and S = S = RL RS QL RL2 ES RS RL RL = = 2 EL RL RS RS

Because the potentials are equal at the surfaces of the spheres: Substitute to obtain:

Solve for ES:

ES =

RL 12 cm (200 kV/m ) EL = RS 5 cm

= 480 kV/mUse = 0 E to find the surface charge density of each sphere:

12 cm =0 E12 cm = (8.85 1012 C 2 /N m 2 )(200 kV/m ) = 1.77 C/m 2and

5 cm =0 E5 cm = (8.85 1012 C 2 /N m 2 )(480 kV/m ) = 4.25 C/m 2

Electric Potential 21165 Picture the Problem The diagram is a cross-sectional view showing the charges on the concentric spherical shells. The Gaussian surface over which well integrate E in order to find V in the region r b is also shown. Well also find E in the region for which a < r < b. We can then use the relationship V = Edr to find Va and Vb and their difference.b

Express Vb: Apply Gausss law for r b:

Vb = E r a dr

S

r Q E r ndA = enclosed = 0

0

and Erb = 0 because Qenclosed = 0 for r b. Substitute to obtain:

Vb = (0)dr = 0

b

Express Va:

Va = Er a drb

a

Apply Gausss law for r a:

Er a 4 r 2 =and

(

)

q

0= kq r2

Er a =

q 4 0 ra

2

Substitute to obtain:

Va = kq

dr kq kq = r2 a b b

The potential difference between the shells is given by:

1 1 Va Vb = Va = kq a b

*66 Picture the Problem We can find the potential relative to infinity at the center of the sphere by integrating the electric field for 0 to . We can apply Gausss law to find the

212 Chapter 23electric field both inside and outside the spherical shell. The potential relative to infinity the center of the spherical shell is: Apply Gausss law to a spherical surface of radius r < R to obtain: Using the fact that the sphere is uniformly charged, express Qinside in terms of Q: Substitute for Qinside to obtain: Solve for Er < R:

V = Er < R dr + Er > R dr0R

R

(1)

Q E dA = E (4r ) = S nr R to obtain: Solve for Er>R to obtain:

Q E dA = E (4r ) = S nr >R

=

Q 0

Er > R =

Q kQ = 2 2 4 0 r rR

Substitute for ErR in equation (1) and evaluate the resulting integral:

V=

kQ dr r dr + kQ 2 3 R 0 r RR

kQ r 2 3kQ 1 = 3 + kQ = R 2 0 2R r R

67 Picture the Problem (a) The field lines are shown on the figure. The charged spheres induce charges of opposite sign on the spheres near them so that sphere 1 is negatively charged, and sphere 2 is positively charged. The total charge of the system is zero.

(b)

V1 = V2 because the spheres are connected. From the direction of the electric field lines it follows that V3 > V1.

Electric Potential 213 If 3 and 4 are connected, V3 = V4 and the conditions of part (b) can(c) only be satisfied if all potentials are zero. Consequently the charge

on each sphere is zero.

General Problems68 Picture the Problem Because the charges at either end of the electric dipole are point charges, we can use the expression for the Coulomb potential to find the field at any distance from the dipole charges. Using the expression for the potential due to a system of point charges, express the potential at the point 9.21010 m from each of the two charges: Because q+ = q:

V=

kq+ kq + d d k = (q+ + q ) d

q+ + q = 0 , V = 0 and (b) is correct.

69 Picture the Problem The potential V at any point on the x axis is the sum of the Coulomb potentials due to the two point charges. Once we have found V, we can use E = grad V to find the electric field at any point on the x axis.

r

(a) Express the potential due to a system of point charges: Substitute to obtain:

V =i

kqi ri

V ( x ) = Vcharge at + a + Vcharge at -a = = kq x +a2 2

+

kq x + a22

2kq x2 + a2

214 Chapter 23(b) The electric field at any point on the x axis is given by:

r d 2kq E ( x ) = grad V = i dx x 2 + a 2

=

(x

2kqx2

+a

2 32

)

i

70 Picture the Problem The radius of the sphere is related to the electric field and the potential at its surface. The dielectric strength of air is about 3 MV/m. Relate the electric field at the surface of a conducting sphere to the potential at the surface of the sphere: Solve for r:

Er =

V (r ) r

r=

V (r ) Er V (r ) Emax

When E is a maximum, r is a minimum: Substitute numerical values and evaluate rmin: *71 Picture the Problem The geometry of the wires is shown to the right. The potential at the point whose coordinates are (x, y) is the sum of the potentials due to the charge distributions on the wires. (a) Express the potential at the point whose coordinates are (x, y):

rmin =

rmin

10 4 V = = 3.33 mm 3 MV/m

V ( x, y ) = Vwire at a + Vwire at a r r = 2k ln ref + 2k ( ) ln ref r r 2 1 r r = 2k ln ref ln ref r r 2 1 r ln 2 = 2 0 r1 where V(0) = 0.

Electric Potential 215Because r1 =

(x + a )2 + y 2 and

r2 =

(x a )2 + y 2 :

V ( x, y ) = ln 2 0

(x a )2 + y 2 2 2 (x + a ) + y

On the y-axis, x = 0 and:

a2 + y2 ln V (0, y ) = 2 0 a 2 + y 2 =

ln (1) = 0 2 0

(b) Evaluate the potential at

( 1 a,0) = (1.25 cm, 0) : 4

V(

1 4

a ,0 ) = ln 2 0 =

( 1 a a )2 4 2 1 (4 a + a)

Equate V(x,y) and V ( 1 a,0 ) : 4

3 ln 2 0 5

3 = 5

(x 5)2 + y 2 (x + 5)2 + y 2

Solve for y to obtain:

y = 21.25 x x 2 25

A spreadsheet program to plot y = 21.25 x x 2 25 is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell A2 A3 B2 B4

Content/Formula 1.25 A2 + 0.05 SQRT(21.25*A2 A2^2 25) SQRT(21.25*A2 A2^2 25) A x 1.25 1.30 1.35 1.40 1.45 1.50 19.65 19.70 19.75 B y_pos 0.00 0.97 1.37 1.67 1.93 2.15 2.54 2.35 2.15 C y_neg 0.00 0.97 1.37 1.67 1.93 2.15 2.54 2.35 2.15

Algebraic Form1 4

a

x + x

y = 21.25 x x 2 25 y = 21.25 x x 2 25

1 2 3 4 5 6 7 370 371 372

216 Chapter 23373 374 375 376 19.80 19.85 19.90 19.95 1.93 1.67 1.37 0.97 1.93 1.67 1.37 0.97

The following graph shows the equipotential curve in the xy plane for

V ( 1 a ,0 ) = 4

3 ln . 2 0 5 10 8 6 4 y (cm) 2 0 -2 -4 -6 -8 -10 0

5

10 x (cm)

15

20

72 Picture the Problem We can use the expression for the potential at any point in the xy plane to show that the equipotential curve is a circle. (a) Equipotential surfaces must satisfy the condition:

V=

r ln 2 2 0 r1 2 0 V

Solve for r2/r1:

r2 =e r1

= C or r2 = Cr1

where C is a constant. Substitute for r1 and r2 to obtain: Expand this expression, combine like terms, and simplify to obtain:

(x a