openstax university physics volume ii unit 2: electricity...

OpenStax University Physics Volume II Unit 2: Electricity and Magnetism Chapter 6: Gauss’s Law

of 28

University Physics Volume II Unit 2: Electricity and Magnetism

Chapter 6: Gauss’s Law Conceptual Questions 1. Discuss how to orient a planar surface of area A in a uniform electric field of magnitude 0E to obtain (a) the maximum flux and (b) the minimum flux through the area. Solution a. If the planar surface is perpendicular to the electric field vector, the maximum flux would be obtained. b. If the planar surface were parallel to the electric field vector, the minimum flux would be obtained. 2. What are the maximum and minimum values of the flux in the preceding question? Solution maximum minimum 0Φ = 3. The net electric flux crossing a closed surface is always zero. True or false? Solution False. The net electric flux crossing a closed surface is always zero if and only if the net charge enclosed is zero. 4. The net electric flux crossing an open surface is never zero. True or false? Solution False. Consider a region of uniform electric field with an open surface that is U shaped. If the flat parts of the U are perpendicular to the electric field, the net flux is zero. 5. Two concentric spherical surfaces enclose a point charge q. The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces. Solution

Since the electric field vector has a 2

1r

dependence, the fluxes are the same since 24A rπ= .

6. Compare the electric flux through the surface of a cube of side length a that has a charge q at its center to the flux through a spherical surface of radius a with a charge q at its center. Solution Both are /enc oq ε . 7. (a) If the electric flux through a closed surface is zero, is the electric field necessarily zero at all points on the surface? (b) What is the net charge inside the surface? Solution a. no; b. zero 8. Discuss how Gauss’s law would be affected if the electric field of a point charge did not vary as 21 .r Solution Gauss’s law would have some dependence on r and this would violate conservation of flux. 9. Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge q. Solution

Both fields vary as 2

1r

. Because the gravitational constant is so much smaller than 0

14πε

, the

gravitational field is orders of magnitude weaker than the electric field. Also, the gravitational


of 28

flux through a closed surface is zero or positive; however, the electric flux is positive, negative, or zero, depending on the definition of flux for the given situation. 10. Discuss whether Gauss’s law can be applied to other forces, and if so, which ones. Solution Gauss’s law can be applied to Newton’s universal law of gravitation, since both are inverse square forces. The law cannot be applied to nonconservative forces, such as friction, tension, or air resistance. 11. Is the term in Gauss’s law the electric field produced by just the charge inside the Gaussian surface? Solution No, it is produced by all charges both inside and outside the Gaussian surface. 12. Reformulate Gauss’s law by choosing the unit normal of the Gaussian surface to be the one directed inward. Solution In the below figure, replace the positive charge at the center with a negative charge. Then the electric field would point inward and in the opposite direction of the unit normal. The minus sign would come out after the dot product of the electric field and the unit normal, leaving

.

13. Would Gauss’s law be helpful for determining the electric field of two equal but opposite charges a fixed distance apart? Solution No, since the situation does not have symmetry, making Gauss’s law challenging to simplify. 14. Discuss the role that symmetry plays in the application of Gauss’s law. Give examples of continuous charge distributions in which Gauss’s law is useful and not useful in determining the electric field. Solution Gauss’s law is useful in determining the electric field in and around continuous charge distributions on spheres, concentric spherical shells, cylinders, concentric cylindrical shells, and solid and concentric geometrical objects such as cubes. Gauss’s law is not useful when the charge distribution is not continuous. If the charge density is different on opposite sides of a geometrical object, there is no symmetry and Gauss’s law should not be used. 15. Discuss the restrictions on the Gaussian surface used to discuss planar symmetry. For example, is its length important? Does the cross-section have to be square? Must the end faces be on opposite sides of the sheet?


of 28

Solution Any shape of the Gaussian surface can be used. The only restriction is that the Gaussian integral must be calculable; therefore, a box or a cylinder are the most convenient geometrical shapes for the Gaussian surface. 16. Is the electric field inside a metal always zero? Solution The answer is no. The argument for the vanishing of the electric field inside a metal required static conditions on the conduction electrons. When a metal wire carries current the conduction electrons are not static. Therefore, the electric field will not be zero inside a metal in which conduction electrons are not static. Therefore, often- quoted simplistic rule that, “the electric field inside a conductor is zero,” applies only to static situations. If electric field were zero in all situations, then there will be no electric current in a metal wire. We will find that the conduction of electricity requires non-zero electric field inside a conductor. 17. Under electrostatic conditions, the excess charge on a conductor resides on its surface. Does this mean that all the conduction electrons in a conductor are on the surface? Solution No. If a metal was in a region of zero electric field, all the conduction electrons would be distributed uniformly throughout the metal. 18. A charge q is placed in the cavity of a conductor as shown below. Will a charge outside the conductor experience an electric field due to the presence of q?

Solution Yes, a charge +q is induced on the outside surface, generating an electric field external to the conductor. 19. The conductor in the preceding figure has an excess charge of –5.0 Cµ . If a 2.0- Cµ point charge is placed in the cavity, what is the net charge on the surface of the cavity and on the outer surface of the conductor? Solution Since the electric field is zero inside a conductor, a charge of 2.0 Cµ− is induced on the inside surface of the cavity. This will put a charge of 2.0 Cµ+ on the outside surface leaving a net charge of 3.0 Cµ− on the surface.


of 28

Problems 20. A uniform electric field of magnitude 41.1 10 N/C× is perpendicular to a square sheet with sides 2.0 m long. What is the electric flux through the sheet? Solution

21. Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an angle of to the field. Find the flux for both directions of the unit normal to the sheet. Solution

4 2 4 2cos 1.1 10 N/C(4.0 m )cos 60 2.2 10 N m CEA θΦ = ⋅ → = × = × ⋅E A

electric field in direction of unit normal; 4 2cos 2.2 10 N m CEA θΦ = ⋅ → = − × ⋅E A

electric field opposite to

unit normal 22. Find the electric flux through a rectangular area between two parallel plates where there is a constant electric field of 30 N/C for the following orientations of the area: (a) parallel to the plates, (b) perpendicular to the plates, and (c) the normal to the area making a angle with the direction of the electric field. Note that this angle can also be given as

Solution a. 2 230.0 N C(0.03)(0.02)m 0.018 N m C= ⋅ ; b. 0⋅ =E A

; c.

2 230.0 N C(0.03)(0.02)m cos30 0.016 N m C= ⋅ 23. The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be when the area is parallel to the plate. Find the charge density on the sheet. Solution

5 213 2

02

3 10 N m / C 0.012 N C 2 2.12 10 C m(0.05 m)

E Eσ ε−

−× ⋅= = ⇒ = = ×

24. Two large rectangular aluminum plates of area face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, 20 Cµ± . The charges on the plates face each other. Find the flux through a circle of radius 3 cm between the plates when the normal to the circle makes an angle of with a line perpendicular to the plates. Note that this angle can also be given as Solution

( )6

84 2 12

0

20.0 10 C 1.51 10 N C150.0 10 m 8.85 10

E σε

−

− −

×= = = ×

× ×;

if the electric field is in the same direction of the area normal. If 185θ = , then . 25. A square surface of area 22 cm is in a space of uniform electric field of magnitude

310 N/C . The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) 30° , (b) 90° , and (c) 0° . Note that these angles can also be given as 180 θ°+ . Solution


of 28

a. 3 4 2 2cos30 1.0 10 N C(2.0 10 m )cos30 0.17 N m CEA −Φ = = × × = ⋅ ; b. cos90 0EAΦ = = ; c. 3 4 2 2cos 0 1.0 10 N C(2.0 10 m) cos 0 0.20 N m CEA −Φ = = × × = ⋅

26. A vector field is pointed along the z-axis, 2 2 ˆ.x yα

=+

v z (a) Find the flux of the vector field

through a rectangle in the xy-plane between a x b< < and c y d< < . (b) Do the same through a rectangle in the yz-plane between a z b< < and c y d< < . (Leave your answer as an integral.) Solution

a. 2 2 2 2ˆ ˆ ˆ

b d

S a cdA dA dxdy

x y x yα α

Φ = ⋅ = ⋅ =+ +∫ ∫ ∫ ∫v n z z

;

b. 2 2ˆ ˆ ˆ 0

SdA dA

x yα

Φ = ⋅ = ⋅ =+∫ ∫v n z x

27. Consider the uniform electric field What is its electric flux through a circular area of radius 2.0 m that lies in the xy-plane? Solution

4 23.8 10 N m C× ⋅

28. Repeat the previous problem, given that the circular area is (a) in the yz-plane and (b) above the xy-plane. Solution a. 3ˆ ˆˆˆ (4.0 3.0 ) 10 N/C 0dA dAΦ = ⋅ = + × ⋅ =∫ ∫E n j k i

; b. The normal to the circular area is put at 45

degrees to the xy-plane: 31 3.0 10 N Cˆ ˆˆ ˆ ,

2 2×

= + ⋅ =n (i k), E n

32 4 23.0 10 N Cˆ (2.0m) 2.7 10 N m C

2dA π×

Φ = ⋅ = = × ⋅∫E n

29. An infinite charged wire with charge per unit length λ lies along the central axis of a cylindrical surface of radius r and length l. What is the flux through the surface due to the electric field of the charged wire? Solution

30. Determine the electric flux through each surface whose cross-section is shown below.


of 28

Solution

1 2 3 40 0 0

2 4: 0, : , : , : ,q q qS S S Sε ε ε− −

Φ = Φ = Φ = Φ =

5 60 0

2 3: , :q qS Sε ε−

Φ = Φ =

31. Find the electric flux through the closed surface whose cross-sections are shown below.


of 28

Solution

a. 8

3 2enc12

0

3.0 10 C 3.39 10 N m C8.85 10

qε

−

−

×Φ = = = × ⋅

×; b. 0Φ = ;


of 28

c. 6

5 2enc12

0

2.0 10 C 2.25 10 N m C8.85 10

qε

−

−

− ×Φ = = = − × ⋅

×;

d. 6 2 4 2

2enc12

0

2.0 10 C m (4.0 10 m ) 90.4 N m C8.85 10

qε

− −

−

× ×Φ = = = ⋅

×

32. A point charge q is located at the center of a cube whose sides are of length a. If there are no other charges in this system, what is the electric flux through one face of the cube? Solution

e c

0

n

0

1ˆ 6S

qdA qε ε

Φ = ⋅ = ⇒Φ =∫ E n

because the flux is divided equally among the six sides

33. A point charge of 10 Cµ is at an unspecified location inside a cube of side 2 cm. Find the net electric flux though the surfaces of the cube. Solution

34. A net flux of passes inward through the surface of a sphere of radius 5 cm. (a) How much charge is inside the sphere? (b) How precisely can we determine the location of the charge from this information? Solution

a. enc4 2 12 4 2 8

0

ˆ 1.0 10 N m /C 8.85 10 (1.0 10 N m /C) 8.85 10 CS

dA qε

− −Φ = ⋅ = × ⋅ = ⇒ × × ⋅ = ×∫ E n

;

b. anywhere inside the sphere 35. A charge q is placed at one of the corners of a cube of side a, as shown below. Find the magnitude of the electric flux through the shaded face due to q. Assume 0q > .

Solution Make a cube with q at the center, using the cube of side a. This would take four cubes of side a to make one side of the large cube. The shaded side of the small cube would be 1/24th of the

total area of the large cube; therefore, the flux through the shaded area would be 0

124

qε

Φ = .

36. The electric flux through a cubical box 8.0 cm on a side is What is the total charge enclosed by the box? Solution


of 28

37. The electric flux through a spherical surface is What is the net charge enclosed by the surface? Solution

38. A cube whose sides are of length d is placed in a uniform electric field of magnitude

3 4.0 10 N/CE = × so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube? Solution zero, since flux in equals flux out 39. Repeat the previous problem, assuming that the electric field is directed along a body diagonal of the cube. Solution zero, also because flux in equals flux out 40. A total charge 6 5.0 10 C−× is distributed uniformly throughout a cubical volume whose edges are 8.0 cm long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with 12.0-cm edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are 10.0 cm long and 5.0 cm long. (d) What is the electric flux through a spherical surface of radius 3.0 cm that is also concentric with the charge distribution? Solution

a. 6

3 23

5.0 10 C 9.76 10 C m(0.08m)

ρ−

−×= = × ; b.

6 5 2enc

120

5.0 10 C 5.65 10 N m C8.85 10

qε

−

−

×Φ = = = × ⋅

×; c. The

same flux for the 10-cm long cube. The fraction of charge enclosed for the 5-cm long cube is 3 6

6 6 5 2enc3 12

0

5 1.22 10 C5.0 10 C 1.22 10 C, 1.38 10 N m C8 8.85 10

qε

−− −

−

×× = × Φ = = = × ⋅

×;

d. 6

3 3 2 6 5 2enc 12

4 1.10 10 C(0.03m) (9.76 10 C m ) 1.10 10 C, 1.24 10 N m C3 8.85 10

q π−

− −−

×= × = × Φ = = × ⋅

×

41. Recall that in the example of a uniform charged sphere, Rewrite the answers in terms of the total charge Q on the sphere. Solution

42. Suppose that the charge density of the spherical charge distribution shown in the following figure is 0( )r r Rρ ρ= for r R≤ and zero for .r R> Obtain expressions for the electric field both inside and outside the distribution.


of 28

Solution

42

0 00( )4 ( ) ,

r rq r R r drR

ρ π ρ π′ ′= =∫ 20 0 ˆ( 4 )r Rρ ε=E r

for ;r R≤ 3 2

0 0 ˆ( 4 )R rρ ε=E r

for .r R≥

43. A very long, thin wire has a uniform linear charge density of What is the electric field at a distance 2.0 cm from the wire? Solution

67

0 0 0

50.0 10 C m2 4.50 10 N C2 2 (0.02m)

lEA E rl Er

λ λπε πε πε

−×= = ⇒ = = = ×

44. A charge of is distributed uniformly throughout a spherical volume of radius 10.0 cm. Determine the electric field due to this charge at a distance of (a) 2.0 cm, (b) 5.0 cm, and (c) 20.0 cm from the center of the sphere. Solution

a. 6

63 3

0 0

ˆ ˆ( ) ( )30.0 10 (0.02m) ˆ5.40 10 N C( )4 4 (0.1m)

qrRπε πε

−− − ×= = = × −

r rE r

;

b. 6

73 3

0 0

ˆ ˆ( ) ( )30.0 10 (0.05m) ˆ1.34 10 N C( )4 4 (0.1m)

qrRπε πε

−− − ×= = = × −

r rE r

;

c. 6

62 2

0 0

ˆ ˆ( ) ( )30.0 10 ˆ6.74 10 N C( )4 4 (0.2m)

qrπε πε

−− − ×= = = × −

r rE r

45. Repeat your calculations for the preceding problem, given that the charge is distributed uniformly over the surface of a spherical conductor of radius 10.0 cm. Solution

a. 0; b. 0; c. 6

62 2

0 0

ˆ ˆ( ) ( )30.0 10 ˆ6.74 10 N C( )4 4 (0.2m)

qrπε πε

−− − ×= = = × −

r rE r

46. A total charge Q is distributed uniformly throughout a spherical shell of inner and outer radii 1 2 and ,r r respectively. Show that the electric field due to the charge is


of 28

1

3 31

1 22 3 30 2 1

220

( );

ˆ ( );4

ˆ ( ).4

r r

r rQ r r rr r r

Q r rr

πε

πε

= ≤

−= ≤ ≤ −

= ≥

E 0

E r

E r

Solution

a. enc 0 0q EA= ⇒ = ; b. 3 3

13 3 2

enc 10

4 ( )4 3( ) 43

r rq r r E r

ρ πρ π π

ε

−= − ⇒ = ,

3 3 3 31 12 2 3 3

3 3 0 0 2 12 1

( )4 3 4( )3

r r r rQ QEr r r rr r

ρ ρε πεπ

− −= ⇒ = = − −

; The electric field is directed outward.

c. 20 04

Q QEA Erε πε

= ⇒ = ; The electric field is directed outward.

47. When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of 3.0 Cµ+ charge put on a 5.0-cm aluminum spherical ball at the following two points in space: (a) a point 1.0 cm from the center of the ball (an inside point) and (b) a point 10 cm from the center of the ball (an outside point). Solution

a. 0; b. 6

62

0

3.0 10 C 2.70 10 N C4 (0.1m)

Eπε

−×= = ×

48. A large sheet of charge has a uniform charge density of What is the electric field due to this charge at a point just above the surface of the sheet? Solution

6 25

0 0

10.0 10 C m 5.65 10 N C2 2

E σε ε

−×= = = ×

49. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (b) A 10-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (c) A 150-cm wooden rod is glued to a 150-cm plastic rod to make a 300-cm long rod, which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is 1 cm, and we seek an electric field at a point that is 4 cm from the center of the rod. (d) Same rod as (c), but we seek electric field at a point that is 500 cm from the center of the rod. Solution a. Yes, the length of the rod is much greater than the distance to the point in question. b. No, The length of the rod is of the same order of magnitude as the distance to the point in question. c. Yes, the length of the rod is much greater than the distance to the point in question. d. No. The length of the rod is of the same order of magnitude as the distance to the point in question.


of 28

50. A long silver rod of radius 3 cm has a charge of on its surface. (a) Find the electric field at a point 5 cm from the center of the rod (an outside point). (b) Find the electric field at a point 2 cm from the center of the rod (an inside point) Solution

a. 6 4

3 25.0 10 C (0.01m) 5.0 10 C m 2.65 10 C m2 2 (0.03m) 2 (0.03m)Rλσπ π π

− −−× ×

= = = = × ,

3 2

80 1 (0.03m)(2.65 10 C m ) 1ˆ ˆ ˆ( ) 1.8 10 N C( )(0.05m)o o

Rr

σε ε

−×= = − = × −E r r r

; b. 0

51. The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod. (a) How much charge per unit length exists on the copper rod? (b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly? Solution

a. 11 20 0 00

0

3.0 (2.0)1 ˆ 5.31 10 C m1.0

R E rr R

σ ε εσε

−= ⇒ = = = ×E r

,

11 2 122 5.31 10 C m ( 2 (0.01 m)) 3.33 10 C mRλ σ π π− −= = × = × ;

b. 12

2enc

0 0

3.33 10 C m (0.05 m) 0.019 N m Cqε ε

−×Φ = = = ⋅

52. A long copper cylindrical shell of inner radius 2 cm and outer radius 3 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 4 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, and (e) 7 cm. Solution

a. 0; b. 12

11 24.0 10 C m 6.37 10 C m2 2 (0.01 m)Rλσπ π

−−×

= = = × ,

11 20 1 (0.01 m)(6.37 10 C m ) 1ˆ ˆ ˆ4.8 N C

0.015 mo o

Rr

σε ε

−×= = =E r r r

directed outward; c. 0;

d. 12

11 24.0 10 C m 2.12 10 C m2 2 (0.03 m)Rλσπ π

−−×

= = = × ,

11 20 1 (0.03 m)(2.12 10 C m ) 1ˆ ˆ ˆ2.05 N C

0.035 mo o

Rr

σε ε

−×= = =E r r r

;

e. 11 2

0 1 (0.03 m)(2.12 10 C m ) 1ˆ ˆ ˆ1.03N C 0.07 mo o

Rr

σε ε

−×= = =E r r r

53. Charge is distributed uniformly with a density ρ throughout an infinitely long cylindrical volume of radius R. Show that the field of this charge distribution is directed radially with respect to the cylinder and that


of 28

02

0

( );2

( ).2

rE r R

RE r Rr

ρε

ρε

= ≤

= ≥

Solution 2

0 0

2 ( )2

r l rE rl E r Rρπ ρπε ε

= ⇒ = ≤ ; 2 2

0 0

2 ( )2

R l RE rl E r Rr

ρπ ρπε ε

= ⇒ = ≥

54. Charge is distributed throughout a very long cylindrical volume of radius R such that the charge density increases with the distance r from the central axis of the cylinder according to

,rρ α= where α is a constant. Show that the field of this charge distribution is directed radially with respect to the cylinder and that

2

03

0

( );3

( ).3

rE r R

RE r Rr

αε

αε

= ≤

= ≥

Solution

, 3 2

0 0

22 ( )3 3

r l rE rl E r Rπα απε ε

= ⇒ = ≤,

55. The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the ball’s center and has magnitude 24.0 10 N/C.× How much charge is on the surface of the ball? Solution

56. Charge is distributed throughout a spherical shell of inner radius 1r and outer radius 2r with a volume density given by 0 1 ,r rρ ρ= where 0ρ is a constant. Determine the electric field due to this charge as a function of r, the distance from the center of the shell. Solution

10 ( )r r≤ , 11

2 2 2 21enc 0 1 0 1 0 14 ( ) 2 ( ) 2 ( )

r r

rr

rq r dr r r r r rr

π ρ π ρ π ρ ′ ′ ′= = = − ′ ∫ ,

2 2 2 22 1 0 1 1 0 1

1 220 0

2 ( ) ( )4 ( )2

r r r r r rE r E r r rr

π ρ ρπε ε

− −= ⇒ = ≤ ≤ ,

2 2 2 22 2 2 1 0 2 1 1 0 2 1

enc 1 0 2 1 220 0

2 ( ) ( )2 ( ), 4 ( )2

r r r r r rq r r r E r E r rr

π ρ ρπ ρ πε ε

− −= − = ⇒ = ≥


of 28

57. Charge is distributed throughout a spherical volume of radius R with a density 2 ,rρ α=where α is a constant. Determine the electric field due to the charge at points both inside and outside the sphere. Solution

, 5 3

2

0 0

44 ( )5 5

r rE r E r Rπα απε ε

= ⇒ = ≤,

58. Consider a uranium nucleus to be sphere of radius 157.4 10 mR −= × with a charge of 92e distributed uniformly throughout its volume. (a) What is the electric force exerted on an electron when it is 15 3.0 10 m−× from the center of the nucleus? (b) What is the acceleration of the electron at this point? Solution

a. 19

24 3

15 3

92.0 (1.6 10 C) 8.7 10 C m4 (7.4 10 m)3

ρπ

−

−

× ×= = ×

×,

15 15

3.0 10 m33.0 10 m 24 3 2 24 3 19enc 0

0

(8.7 10 C m )4 4 (8.7 10 C m ) 9.8 10 C3rq r drπ π

−−

×× −= × = × = ×∫ ,

19

2015 2

0

9.8 10 C 9.8 10 N C4 (3.0 10 m)

Eπ ε

−

−

×= = ×

×;

b. 19 20 32 231

156.8 N(1.6 10 C) 9.8 10 N C 156.8N 1.7 10 m s9.1 10 kg

F qE a−−= = × × = = = ×

×

59. The volume charge density of a spherical charge distribution is given by 0( ) ,rr e αρ ρ −= where 0ρ and α are constants. What is the electric field produced by this charge distribution? Solution

integrate by parts: 2

20 0 2 30

0

( ) 2 24 ( ) 4 [ ( )]r

r r renc

r rq e r dr eα αρ π πρα α α

′ ′− − ′ ′′ ′= = − + +∫

2 20

0 2 3 3 2 2 3 30

( ) 2 2 2 ( ) 2 2 2 4 ( ) ( )r rr r r re E er

α αρπρα α α α ε α α α α

− − = − + + + ⇒ = − + + +

60. An uncharged conductor with an internal cavity is shown in the following figure. Use the closed surface S along with Gauss’ law to show that when a charge q is placed in the cavity a total charge –q is induced on the inner surface of the conductor. What is the charge on the outer surface of the conductor?


of 28

Solution

enc

0

ˆS

qdAε

⋅ =∫ E n ; Since the electric field at the Gaussian surface is inside the conductor it must be

zero; therefore, q− is induced on the inside surface of the cavity. Put a Gaussian surface outside the conductor. Since the conductor was initially uncharged the enclosed charge is q, so there must be a charge of q+ on the outside surface. 61. An uncharged spherical conductor S of radius R has two spherical cavities A and B of radii a and b, respectively as shown below. Two point charges and are placed at the center of the two cavities by using non-conducting supports. In addition, a point charge is placed outside at a distance r from the center of the sphere. (a) Draw approximate charge distributions in the metal although metal sphere has no net charge. (b) Draw electric field lines. Draw enough lines to represent all distinctly different places.

Solution

62. A positive point charge is placed at the angle bisector of two uncharged plane conductors that make an angle of See below. Draw the electric field lines.


of 28

Solution

63. A long cylinder of copper of radius 3 cm is charged so that it has a uniform charge per unit length on its surface of 3 C/m. (a) Find the electric field inside and outside the cylinder. (b) Draw electric field lines in a plane perpendicular to the rod. Solution

a. Outside: 0 0

3.0C m22

lE rl Er

λπε πε

= ⇒ = ; Inside ; b.

64. An aluminum spherical ball of radius 4 cm is charged with 5 Cµ of charge. A copper spherical shell of inner radius 6 cm and outer radius 8 cm surrounds it. A total charge of 8 Cµ− is put on the copper shell. (a) Find the electric field at all points in space, including points inside the aluminum and copper shell when copper shell and aluminum sphere are concentric. (b) Find the electric field at all points in space, including points inside the aluminum and copper shell when the centers of copper shell and aluminum sphere are 1 cm apart. Solution

a. 6

2 enc2 2

0 0 0

5 8 3.0 10 C4 8 cm4 4

q C CE r E rr r

µ µπε πε πε

−− − ×= ⇒ = = > , 0 6 cm 8 cmE r= ≤ ≤ ,


of 28

6

20

5.0 10 C 4 cm 6 cm4

E rrπε

−×= ≤ ≤ ,

0 4 cmE r= < ; b. Put the center of the copper shell at the origin, then 8 cmr > same as (a);

0 6 cm 8 cmE r= ≤ ≤ , any point between the sphere and shell 6

20

5.0 10 C4 ( )

Erπε

−×=

′ where r′

is the radial distance to the point as measured from the center of the sphere. Inside the sphere 0E = .

65. A long cylinder of aluminum of radius R meters is charged so that it has a uniform charge per unit length on its surface of λ . (a) Find the electric field inside and outside the cylinder. (b) Plot electric field as a function of distance from the center of the rod. Solution

a. 0 0

2 2

lE rl E r Rr

λ λπε πε

= ⇒ = ≥ E inside equals 0; b.

66. At the surface of any conductor in electrostatic equilibrium, Show that this equation is consistent with the fact that at the surface of a spherical conductor. Solution

2 20 04

q kqEr r

σε πε

= = =

67. Two parallel plates 10 cm on a side are given equal and opposite charges of magnitude The plates are 1.5 mm apart. What is the electric field at the center of the region

between the plates? Solution

9 24

0 0

5.0 10 C (0.1m) 5.65 10 N CE σε ε

−×= = = ×

68. Two parallel conducting plates, each of cross-sectional area 2400 cm , are 2.0 cm apart and uncharged. If electrons are transferred from one plate to the other, what are (a) the charge density on each plate? (b) The electric field between the plates? Solution

a. 12 19 7

6 22 2 2

1.0 10 e(1.6 10 C e) 1.6 10 C 4.0 10 C m400.0(10 m) 0.04m

σ− −

−−

× × ×= ± = = ± × ;

b.6 2

5

0 0

4.0 10 C m 4.52 10 N CE σε ε

−×= = = ×


of 28

69. The surface charge density on a long straight metallic pipe is σ . What is the electric field outside and inside the pipe? Assume the pipe has a diameter of 2a.

Solution

0 0

2 2 l aa E rl E r ar

λ σλ π σ πε ε

= = ⇒ = ≥ , 0E = inside since enclosed = 0 q

70. A point charge is placed at the center of a spherical conducting shell of inner radius 3.5 cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b) What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor? Solution

a. 12

10 22

5.0 10 C 3.25 10 C m4 (0.035m)

σπ

−−×

= = × ;

b. 12

10 22

5.0 10 C 2.49 10 C m4 (0.04m)

σπ

−−− ×

= = × ; c. 125.0 10 C−×

71. A solid cylindrical conductor of radius a is surrounded by a concentric cylindrical shell of inner radius b. The solid cylinder and the shell carry charges +Q and –Q, respectively. Assuming that the length L of both conductors is much greater than a or b, determine the electric field as a function of r, the distance from the common central axis of the cylinders, for (a) (b)

and (c) Solution

a. ; b. 0 0

22

Q QE rL ErL

πε πε

= ⇒ = ; c. since r would be either inside the second

shell or if outside then q enclosed equals 0.


of 28

Additional Problems 72. A vector field (not necessarily an electric field; note units) is given by Calculate where S is the area shown below. Assume that

Solution

73. Repeat the preceding problem, with Solution

74. A circular area S is concentric with the origin, has radius a, and lies in the yz-plane. Calculate for

Solution

75. (a) Calculate the electric flux through the open hemispherical surface due to the electric field (see below). (b) If the hemisphere is rotated by around the x-axis, what is the flux

through it?

Solution

a.


of 28

2222 20 00

0

sin2

E r d E rπ

π θ ϕ π= =∫ ; b. zero, since the flux through the upper half cancels the flux

through the lower half of the sphere 76. Suppose that the electric field of an isolated point charge were proportional to 21 r σ+ rather than 21 .r Determine the flux that passes through the surface of a sphere of radius R centered at the charge. Would Gauss’s law remain valid? Solution

2 22

0 0

4 44

q qE r rr rσ σπ π

πε ε+= = ; No, since the flux would depend on the radius of the sphere,

which violates the continuity of flux. 77. The electric field in a region is given by where 2.0 m,b = and 2.0.c = What is the net charge enclosed by the shaded volume shown below?

Solution

e

0

ncˆ S

dA qε

Φ = ⋅ =∫ E n

; There are two contributions to the surface integral: one at the side of the

rectangle at 0x = and the other at the side at 2.0 mx = ; 2 2 2enc

0

2

200.0 N m C 200.0 N m C(0)[1.5m ] (2.0m)[1.5m ] 1.5m2.0 m 2.0m 2.0(2.)m

100 Nm C

qE Eε

⋅ ⋅− + = = − + +

= −

where the minus sign indicates that at 0x = , the electric field is along positive x and the unit normal is along negative x. At 2x = , the unit normal and the electric field vector are in the same direction: 10

enc 0 8.85 10 Cq ε −= Φ = − × . 78. Two equal and opposite charges of magnitude Q are located on the x-axis at the points +a and –a, as shown below. What is the net flux due to these charges through a square surface of side 2a that lies in the yz-plane and is centered at the origin? (Hint: Determine the flux due to each charge separately, then use the principle of superposition. You may be able to make a symmetry argument.)


of 28

Solution Make a cube around the positive charge. The flux through one side is

0

ˆ6QdAε

⋅ =∫E n

. The flux due to the negative charge is the same, so total flux is 03

Qε

Φ =

79. A fellow student calculated the flux through the square for the system in the preceding problem and got 0. What went wrong? Solution didn’t keep consistent directions for the area vectors, or the electric fields 80. A piece of aluminum foil of 0.1 mm thickness has a charge of that spreads on both wide side surfaces evenly. You may ignore the charges on the thin sides of the edges. (a) Find the charge density. (b) Find the electric field 1 cm from the center, assuming approximate planar symmetry. Solution

a. 6

3 22

20.0 10 C 1.0 10 C m2(0.1m)

σ−

−×= = × ; b.

37

0 0

1.0 10 5.65 10 N C2 2

E σε ε

−×= = = × from each side:

81. Two pieces of aluminum foil of thickness 0.1 mm face each other with a separation of 5 mm. One of the foils has a charge of and the other has . (a) Find the charge density at all surfaces, i.e., on those facing each other and those facing away. (b) Find the electric field between the plates near the center assuming planar symmetry. Solution

a. 6

3 22

30.0 10 C 3.0 10 C m(0.1m)

σ−

−×= = × , 3 23 10 C/m−+ × on one and 3 23 10 C/m−− × on the other; b.

3 28

0 0

3.0 10 C m 3.39 10 N CE σε ε

−×= = = ×

82. Two large copper plates facing each other have charge densities on the surface facing the other plate, and zero in between the plates. Find the electric flux through a 3 cm × 4 cm rectangular area between the plates, as shown below, for the following orientations of the area. (a) If the area is parallel to the plates, and (b) if the area is tilted from the parallel direction. Note, this angle can also be


of 28

Solution

a.

211 11 2

0 08 2

4.0C m 4.52 10 N C 4.52 10 N C(0.03)(0.04)m

5.42 10 N m C;

E EAσε ε

= = = × ⇒ = ×

= × ⋅

b. 8 2 8 2cos 5.42 10 N m C(cos30 ) 4.70 10 N m CEA θΦ = = × ⋅ = × ⋅ 83. The infinite slab between the planes defined by 2z a= − and 2z a= contains a uniform volume charge density ρ (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?

Solution Construct a Gaussian cylinder along the z-axis with cross-sectional area A.

,

84. A total charge Q is distributed uniformly throughout a spherical volume that is centered at

1O and has a radius R. Without disturbing the charge remaining, charge is removed from the spherical volume that is centered at 2O (see below). Show that the electric field everywhere in the empty region is given by

30

,4

QRπε

=rE

where r is the displacement vector directed from 1 2 to .O O


of 28

Solution

The electric field at an arbitrary point inside the sphere without the hole is 0

( ) 3ρε′

′ =rE r . The

actual distribution is a sphere of charge density ρ plus a hole of charge density ρ− . By

superposition the electric field in the hole is 30 0 0 0

( )( )3 3 3 4

QR

ρ ρ ρε ε ε π ε′ ′− −′ = + = =

r r r r rE r

.]

85. A non-conducting spherical shell of inner radius 1a and outer radius 1b is uniformly charged with charged density 1ρ inside another non-conducting spherical shell of inner radius 2a and outer radius 2b that is also uniformly charged with charge density 2ρ . See below. Find the electric field at space point P at a distance r from the common center such that (a) 2 r b> , (b)

2 2 a r b< < , (c) 1 2 b r a< < , (d) 1 1 a r b< < , and (e) 1 r a< .

Solution

a. 3 3 3 3

3 3 3 31 1 1 2 2 22 1 1 1 2 2 2

2 20 0

4 [ ( ) ( ) ( ) ( )3 43

b a b a b a b ar b E r Er

π ρ ρ ρ ρπε ε

− + − − + −> = ⇒ = ;

b. 3 3 3 3

3 3 3 31 1 1 2 22 1 1 1 2 2

2 2 20 0

4 [ ( ) ( )] ( ) ( )3 43

b a r a b a r aa r b E r Er

π ρ ρ ρ ρπε ε

− + − − + −< < = ⇒ = ;

c. 3 3

3 31 1 12 1 1 1

1 2 20 0

4 ( ) ( )3 43

b a b ab r a E r Er

πρ ρπε ε

− −< < = ⇒ = ;


of 28

d. 3 3

3 31 12 1 1

1 1 20 0

4 ( ) ( )3 43

r a r aa r b E r Er

πρ ρπε ε

− −< < = ⇒ = ; e. 0

86. Two non-conducting spheres of radii and are uniformly charged with charge densities and respectively. They are separated at center-to-center distance a (see below). Find the

electric field at point P located at a distance r from the center of sphere 1 and is in the direction θ from the line joining the two spheres assuming their charge densities are not affected by the presence of the other sphere. (Hint: Work one sphere at a time and use the superposition principle.)

Solution The electric field at P is equal to the vector sum of the electric fields from the two charges. Put the origin at the center of sphere 1.

P outside both spheres: 3 3

1 1 2 22 2 2

0 0

ˆ ˆ ˆ( )3 3 ( 2 cos )

R Rr r a ra

ρ ρε ε θ

= + −+ −

E r r a

;

P inside sphere 1: 3

1 2 22 2

0 0

ˆ ˆ ˆ( )3 3 ( 2 cos )

r Rr a ra

ρ ρε ε θ

= + −+ −

E r r a

;

P inside sphere 2: 3 2 2 1 2

1 1 22

0 0

( 2 cos )ˆ ˆ ˆ( )3 3

R r a rar

ρ ρ θε ε

+ −= + −E r r a

87. A disk of radius R is cut in a non-conducting large plate that is uniformly charged with charge density σ (coulomb per square meter). See below. Find the electric field at a height h above the center of the disk. ( , or h R h l w>> << ). (Hint: Fill the hole with σ± .)

Solution

Electric field due to plate without hole: 02

E σε

= .

Electric field of just hole filled with 2 2

0

12

zR z

E σε

σ −−

+−

=

.

Thus, 2 2

02hE

R hσε

=+

net .


of 28

88. Concentric conducting spherical shells carry charges Q and –Q, respectively (see below). The inner shell has negligible thickness. Determine the electric field for (a) (b) (c) and (d)

Solution

a. 0E = ; b. 204

QErπε

= ; c. 0E = ; d. 0E = since q enclosed equals 0

89. Shown below are two concentric conducting spherical shells of radii 1R and 2R , each of finite thickness much less than either radius. The inner and outer shell carry net charges and respectively, where both and are positive. What is the electric field for (a) (b)

and (c) (d) What is the net charge on the inner surface of the inner shell, the outer surface of the inner shell, the inner surface of the outer shell, and the outer surface of the outer shell?

Solution

a. 0E = ; b. 12

04qE

rπε= ; c. 1 2

204

q qErπε

+= ; d. 1 1 1 20 , , , q q q q− +

90. A point charge of is placed at the center of an uncharged spherical conducting shell of inner radius 6.0 cm and outer radius 9.0 cm. Find the electric field at (a)


of 28

4.0 cmr = , (b) 8.0 cmr = , and (c) 12.0 cmr = . (d) What are the charges induced on the inner and outer surfaces of the shell? Solution

a. 8

52

0

5.0 10 C 2.81 10 N C4 (0.04 m)

Eπε

−×= = × ; b. 0E = ;

c. 8

42

0

5.0 10 C 3.12 10 N C4 (0.12 m)

Eπε

−×= = × ; d. 8 85.0 10 C and 5.0 10 C, respectively−− × ×

Challenge Problems 91. The Hubble Space Telescope can measure the energy flux from distant objects such as supernovae and stars. Scientists then use this data to calculate the energy emitted by that object. Choose an interstellar object which scientists have observed the flux at the Hubble with (for example, Vega1), find the distance to that object and the size of Hubble’s primary mirror, and calculate the total energy flux. (Hint: The Hubble intercepts only a small part of the total flux.) Solution Given the referenced link, using a distance to Vega of 15237 10× m and a diameter of 2.4 m for the primary mirror, we find that at a wavelength of 555.6 nm, Vega is emitting 242.44 10 J/s× at that wavelength. Note that the flux through the mirror is essentially constant. 92. Re-derive Gauss’s law for the gravitational field, with directed positively outward. Solution Let a mass M be enclosed by a spherical surface of radius R:

22 2 2 2 2

ˆ ˆ ˆ ˆ ˆ 4 4S

GM GM GM GM GMdA dA dA R GMr R R R R

π π= = ⋅ = ⋅ = = =∫ ∫ ∫g r r, g n r n where M is

the total mass inside S, G is Newton’s constant, and the unit normal is in the direction of r̂ . 93. An infinite plate sheet of charge of surface charge density σ is shown below. What is the electric field at a distance x from the sheet? Compare the result of this calculation with that of worked out in the text.

1 http://adsabs.harvard.edu/abs/2004AJ....127.3508B


of 28

Solution The symmetry of the system forces E

to be perpendicular to the sheet and constant over any

plane parallel to the sheet. To calculate the electric field, we choose the cylindrical Gaussian surface shown. The cross-section area and the height of the cylinder are A and 2x, respectively, and the cylinder is positioned so that it is bisected by the plane sheet. Since E is perpendicular to each end and parallel to the side of the cylinder, we have EA as the flux through each end and there is no flux through the side. The charge enclosed by the cylinder is ,Aσ so from Gauss’s

law, 0

2 ,AEA σε

= and the electric field of an infinite sheet of charge is

0

,2

E σε

= in agreement with the calculation of in the text.

94. A spherical rubber balloon carries a total charge Q distributed uniformly over its surface. At 0t = , the radius of the balloon is R. The balloon is then slowly inflated until its radius reaches

2R at the time Determine the electric field due to this charge as a function of time (a) at the surface of the balloon, (b) at the surface of radius R, and (c) at the surface of radius 2R. Ignore any effect on the electric field due to the material of the balloon and assume that the radius increases uniformly with time. Solution

a. 0 0

( ) 1Rt tr t R Rt t

= + = +

,

22enc

20 0 0 2

00

4 1

4 1

q t Q QEA E R Et tR

t

πε ε

ε π

= = + = ⇒ =

+

; b. zero,

since enclosed 0q = ;

c. 2 20 04 (2 ) 8Q QE

R Rπε πε= =

95. Find the electric field of a large conducting plate containing a net charge q. Let A be area of one side of the plate and h the thickness of the plate (see below). The charge on the metal plate will distribute mostly on the two planar sides and very little on the edges if the plate is thin.


of 28

Solution

There is Q/2 on each side of the plate since the net charge is Q: 2QA

σ = ,

0 0 0

2ˆ 22P P P P

S

A QdA E A E A E A EA

σ σε ε ε∆

⋅ = ∆ + ∆ = ∆ = ⇒ = =∫ E n

This file is copyright 2016, Rice University. All Rights Reserved.

openstax university physics volume ii unit 2: electricity...

Documents