Chapter: Chapter 23 Learning Objectives LO 23.1.0 Solve problems related to electric e lectric flux. LO 23.1.1 Identify that Gauss’ law relates the electric field at points on a closed surface (real or imaginary, said to be a Gaussian surface) to the net charge enclosed by that surface. LO 23.1.2 Identify that the amount of electric field piercing a surface (not skimming along the surface) is the electric flux Φ through the surface. LO 23.1.3 Identify that an area vector v ector for a flat surface is a vector that is perpendicular to the surface and that has a magnitude equal to the area of the surface. LO 23.1.4 Identify that any surface can be divided into area el ements (patch elements) that are each small enough and flat enough for an area vector ⃗ to be assigned to it, with the vector perpendicular to the element and having a magnitude equal to the area of the element. LO 23.1.5 Calculate the flux Φ through a surface by integrating the dot product p roduct of the electric and the area vector ⃗ (for patch elements) over the surface, in field vector magnitude-angle notation and unit-vector notation. LO 23.1.6 For a closed surface, explain the algebraic signs associated with inward flux and outward flux. LO 23.1.7 Calculate the net flux Φ through a closed surface, algebraic algebraic sign included, by and the area vector ⃗ (for patch integrating the dot product of the electric field vector elements) over the full surface. LO 23.1.8 Determine if a closed surface can be broken up into parts (such as the sides of a cube) to simplify the integration that yields the net flux through the surface. LO 23.2.0 Solve problems related to Gauss’ law. LO 23.2.1 Apply Gauss’ law to relate the net flux Φ through a closed surface (real or imaginary) to the net charge qenc enclosed by the surface. LO 23.2.2 Identify how the algebraic sign of the net enclosed charge corresponds to the direction (inward or outward) of the net flux through a Gaussian surface. LO 23.2.3 Identify that charge outside a Gaussian surface makes no contribution to the net flux through the closed surface. LO 23.2.4 Derive the expression ex pression for the magnitude of the electric field of a charged particle by using Gauss’ law. LO 23.2.5 Identify that for a charged cha rged particle or uniformly charged sphere, Gauss’ law is applied with a Gaussian surface that is a concentric sphere in order to utilize the spherical symmetry, to simplify the calculation. LO 23.3.0 Solve problems related to a charged isolated conductor. LO 23.3.1 Apply the relationship between surface charge density σ and and the area over which the charge is uniformly spread. LO 23.3.2 Identify that if excess charge (positive or negative) is placed on an isolated conductor, that charge moves to the surface and none is in the interior. LO 23.3.3 Identify the value of the electric field inside an isolated conductor. LO 23.3.4 For a conductor with a cavity that contains a charged object, determine the charge on the cavity wall and on the external surface. LO 23.3.5 Explain how Gauss’ law is used to find the electric field magnitude E near near an isolated
conducting surface with a uniform surface charge density σ . LO 23.3.6 For a uniformly charged conducting surface, apply the relationship between the charge density σ and the electric field magnitude E at points near the conductor, and identify the direction of the field vectors. LO 23.4.0 Solve problems related to applying Gauss' law: cylindrical symmetry. LO 23.4.1 Explain how Gauss’ law is used to derive the electric field ma gnitude outside a line of charge or a cylindrical surface (such as a plastic rod) with a uniform linear charge density λ. LO 23.4.2 Apply the relationship between linear charge density λ on a cylindrical conducting surface and the electric field magnitude E at radial distance r from the central axis. LO 23.4.3 Explain how Gauss’ law can be used to find the electric field magnitude inside a cylindrical nonconducting surface (such as a plastic rod) with a uniform volume charge density ρ. LO 23.5.0 Solve problems related to applying Gauss' law: planar symmetry. LO 23.5.1 For interior and exterior points, apply Gauss’ law to derive the electric field magnitude E near a large, flat, nonconducting surface with a uniform surface charge density σ . LO 23.5.2 For points near a large, flat nonconducting surface with a uniform charge densityσ , apply the relationship between the charge density and the electric field magnitude E and also specify the direction of the field. LO 23.5.3 For points near two large, flat conducting surfaces with a uniform charge d ensity σ , apply the relationship between the charge density and the electric field magnitude E and also specify the direction of the field. LO 23.6.0 Solve problems related to applying Gauss' law: spherical symmetry. LO 23.6.1 Identify that a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge is conce ntrated at the center of the shell. LO 23.6.2 Identify that if a charged particle is enclosed by a shell of uniform charge, there is no electrostatic force on the particle from the shell. LO 23.6.3 For a point outside a spherical shell with uniform charge, apply the relationship between the electric field magnitude E , the charge q on the shell, and the distance r from the shell’s center. LO 23.6.4 Identify the magnitude of the electric field for points enclosed by a sph erical shell with uniform charge. LO 23.6.5 For a uniform spherical charge distribution (a uniform ball of charge), determine the magnitude and direction of the electric field at interior and exterior points. Multiple Choice 1. Gauss’s law: A) can always be used to calculate the electric field. B) relates the electric field throughout space to the charges distributed through that space. C) only applies to point charges. D) relates the electric field at points on a closed surface to the net charge enclosed by that surface. E) relates the surface charge density to the electric field.
Ans: D Difficulty: E Section: 23-1 Learning Objective 23.1.1 2. The electric flux Φ through a surface: A) is the amount of electric field piercing the surface. B) is the electric field multiplied by the area. C) does not depend on the area involved. D) is the line integral of the electric field around the edge of the surface. E) is the amount of electric field skimming along the surface. Ans: A Difficulty: E Section: 23-1 Learning Objective 23.1.2 3. The area vector for a flat surface: A) is parallel to the surface and has a magnitude equal to the length of a side of the surface. B) is perpendicular to the surface and has a magnitude equal to the length of a side of the surface. C) is parallel to the surface and has a magnitude equal to the area of the surface. D) is perpendicular to the surface and has a magnitude equal to the area of the surface. E) none of the above. Ans: D Difficulty: E Section: 23-1 Learning Objective 23.1.3 4. To calculate the flux through a curved surface, A) the area vector has to be perpendicular to the surface somewhere B) you must divide the surface into pieces that are tiny enough to be almost flat C) the surface must be spherical D) the surface cannot be curved very much; then you can treat it as though it were flat E) actually the flux through a curved surface cannot be calculated. Ans: B Difficulty: E Section: 23-1 Learning Objective 23.1.4 5. When a piece of paper is held with one face perpendicular to a uniform electric field the flux 2 through it is 25 Nm /C. When the paper is turned 25 with respect to the field the flux through it is: 2 A) 0 Nm /C
2
B) 11 Nm /C 2 C) 12 Nm /C 2 D) 23 Nm /C 2 E) 25 Nm /C Ans: D Difficulty: E Section: 23-1 Learning Objective 23.1.5 6. The flux of the electric field (24 N/C)̂ + (30 N/C) ̂ + (16 N/C) of the yz plane is: 2 A) 32 N m /C 2 B) 34 N m /C 2 C) 42 N m /C 2 D) 48 N m /C 2 E) 60 N m /C
2
through a 2.0 m portion
Ans: D Difficulty: M Section: 23-1 Learning Objective 23.1.5 7. A cylindrical wastepaper basket with a 0.15-m radius opening is in a uniform electric field of 300 N/C, perpendicular to the opening. The total flux through the sides and bottom is: 2 A) 0 Nm /C 2 B) 4.2 Nm /C 2 C) 21 Nm /C 2 D) 280 Nm /C E) can't tell without knowing the areas of the sides and bottom Ans: C Difficulty: M Section: 23-1 Learning Objective 23.1.5 8. Which statement is correct? A) The flux through a closed surface is always positive. B) The flux through a closed surface is always negative. C) The sign of the flux through a closed surface depends on an arbitrary choice of sign for the surface vector. D) Inward flux through a closed surface is neg ative and outward flux is positive. E) Inward flux through a closed surface is po sitive and outward flux is negative. Ans: D Difficulty: E
Section: 23-1 Learning Objective 23.1.6 9. A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 N/C, perpendicular to the ends. The total flux through the cylinder is: 2 A) 0 Nm /C 2 B) 4.2 Nm /C 2 C) 21 Nm /C 2 D) 280 Nm /C E) can't tell without knowing the length of the cylinder Ans: A Difficulty: E Section: 23-1 Learning Objective 23.1.7 10. A point charge is placed at the center of a spherical Gaussian surface. The electric flux E is changed if: A) the sphere is replaced by a cube of the same volume B) the sphere is replaced by a cube of one-tenth the volume C) the point charge is moved off center (but still inside the original sphere) D) the point charge is moved to just outside the sphere E) a second point charge is placed just outside the sphere Ans: D Difficulty: E Section: 23-2 Learning Objective 23.2.1 11. A physics instructor in an anteroom charges an electrostatic generator to 25 C, then carries it into the lecture hall. The net electric flux through the lecture hall walls is: 2 A) 0 Nm /C – 6 2 B) 25 10 Nm /C 5 2 C) 2.2 10 Nm /C 6 2 D) 2.8 10 Nm /C E) can't tell unless the lecture hall dimensions are given Ans: D Difficulty: E Section: 23-2 Learning Objective 23.2.1 12. A point particle with charge q is placed inside a cube but not at its center. flux through any one side of the cube: A) is zero B) is q/0
The electric
C) is q/40 D) is q/60 E) cannot be computed using Gauss' law Ans: E Difficulty: E Section: 23-2 Learning Objective 23.2.1 13. A particle with charge 5.0 C is placed at the corner of a cube. through all sides of the cube is: 2 A) 0 Nm /C 4 2 B) 7.1 10 Nm /C 4 2 C) 9.4 10 Nm /C 5 2 D) 1.4 10 Nm /C 5 2 E) 5.6 10 Nm /C
The total electric flux
Ans: B Difficulty: M Section: 23-2 Learning Objective 23.2.1 14. A point particle with charge q is at the center of a Gaussian surface in the form of a cube. The electric flux through any one face of the cube is: A) q/0 B) q/40 C) q/40 D) q/60 E) q/160 Ans: D Difficulty: M Section: 23-2 Learning Objective 23.2.1 – 7
15. A 3.5-cm radius hemisphere contains a total charge of 6.6 10 C. The flux through the 4 2 rounded portion of the surface is 9.8 10 Nm /C. The flux through the flat base is: 2 A) 0 Nm /C 4 2 B) +2.3 10 Nm /C 4 2 C) – 2.3 10 Nm /C 4 2 D) – 9.8 10 Nm /C 4 2 E) +9.8 10 Nm /C Ans: C Difficulty:
M
Section: 23-2 Learning Objective 23.2.1 16. Charge Q is distributed uniformly throughout a spherical insulating shell. The net ele ctric flux through the outer surface of the shell is: A) 0 B) Q/0 C) 2Q/0 D) Q/40 E) Q/20 Ans: B Difficulty: E Section: 23-2 Learning Objective 23.2.1 17. The table below gives the electric flux through the ends and round surfaces of four Gaussian surfaces in the form of cylinders. Rank the cylinders according to the charge inside, from the most negative to the most positive.
cylinder 1: cylinder 2: cylinder 3: cylinder 4:
left end right end – 9 2 – 9 2 +2 10 Nm /C +4 10 Nm /C – 9 2 – 9 2 +3 10 Nm /C – 2 10 Nm /C – 9 2 – 9 2 – 2 10 Nm /C – 5 10 Nm /C – 9 2 – 9 2 +2 10 Nm /C – 5 10 Nm /C
rounded surface – 9 2 – 6 10 Nm /C – 9 2 +6 10 Nm /C – 9 2 +3 10 Nm /C – 9 2 – 3 10 Nm /C
A) 1, 2, 3, 4 B) 4, 3, 2, 1 C) 3, 4, 2, 1 D) 3, 1, 4, 2 E) 4, 3, 1, 2 Ans: E Difficulty: E Section: 23-2 Learning Objective 23.2.2 18. Charge Q is distributed uniformly throughout a spherical insulating shell. The net ele ctric flux through the inner surface of the shell is: A) 0 B) Q/0 C) 2Q/0 D) Q/40 E) Q/20 Ans:
A
Difficulty: E Section: 23-2 Learning Objective 23.2.3 19.
Consider Gauss law:
∮ ⃗ . Which of the following is true?
must be the electric field due to the enclosed charge A) everywhere on the Gaussian surface B) If q = 0 then C) If the three particles inside have charges of +q, +q and 2q, then the integral is zero is everywhere parallel to ⃗ D) On the surface at any point on the surface E) If a charge is placed outside the surface, then it cannot affect Ans: C Difficulty: E Section: 23-2 Learning Objective 23.2.5 20. Choose the INCORRECT statement: A) Gauss' law can be derived from Coulomb's law B) Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge en closed within the surface C) Coulomb's law can be derived from Gauss' law and symmetry D) Gauss' law applies to a closed surface of any shape E) According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface Ans: E Difficulty: E Section: 23-2 Learning Objective 23.2.5 21. A) B) C) D) E)
The outer surface of the cardboard center of a paper towel roll: is a possible Gaussian surface cannot be a Gaussian surface because it encloses no charge cannot be a Gaussian surface since it is an insulator cannot be a Gaussian surface since it is not closed none of the above
Ans: D Difficulty: E Section: 23-2 Learning Objective 23.2.5 22. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the cavity, a distance r from the center, is:
A) B)
( ) 2
C)
q/40r
D)
(q + Q)/40r
E)
( )
2
Ans: C Difficulty: E Section: 23-3 Learning Objective 23.3.0 23. A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere? -2 2 A) 2.4 x 10 C/m -2 2 B) 1.4 x 10 C/m -4 2 C) 9.5 x 10 C/m -4 2 D) 2.4 x 10 C/m -5 2 E) 6.0 x 10 C/m Ans: D Difficulty: E Section: 23-3 Learning Objective 23.3.1 24. A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A) a positive charge B) a negative charge C) no appreciable charge D) a charge whose sign depends on what part of the inner surface it touched E) a charge whose sign depends on where the small hole is located in the conductor Ans: C Difficulty: E Section: 23-3 Learning Objective 23.3.2 25. A particle with charge +Q is placed outside a large neutral conducting sheet. At any point in the interior of the sheet the electric field produced by charges on the surface is directed: A) toward the surface B) away from the surface C) toward Q
D) away from Q E) none of the above Ans: C Difficulty: M Section: 23-3 Learning Objective 23.3.3 26. 10 C of charge are placed on a spherical conducting shell. A particle with a charge of – 3 C is placed at the center of the cavity. The net charge on the inner surface of the shell is: A) – 7 C B) – 3 C C) 0 C D) +3 C E) +7 C Ans: D Difficulty: E Section: 23-3 Learning Objective 23.3.4 27. 10 C of charge are placed on a spherical conducting shell. A particle with a charge of – 3C is placed at the center of the cavity. The net charge on the outer surface of the shell is: A) – 7 C B) – 3 C C) 0 C D) +3 C E) +7 C Ans: E Difficulty: E Section: 23-3 Learning Objective 23.3.4 28. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the interior of the conductor a distance r from the center is: A) 0 B) C) D) E)
2 q/40r 2 Q/40r
Ans: A Difficulty:
E
Section: 23-3 Learning Objective 23.3.4 29. A spherical conducting shell has charge Q. A particle with charge q is placed at the center of the cavity. The charge on the inner surface of the shell and the charge on the outer surface of the shell, respectively, are: A) 0, Q B) q, Q – q C) Q, 0 D) – q, Q + q E) – q, 0 Ans: D Difficulty: E Section: 23-3 Learning Objective 23.3.4 30. Which of the following graphs represents the magnitude of the electric field as a function of the distance from the center of a solid charged conducting sphere of radius R?
A) A B) B C) C D) D E) E Ans: E Difficulty: E Section: 23-3 Learning Objective 23.3.5 – 9
31. A conducting sphere of radius 0.01 m has a charge of 1.0 10 C deposited on it. The magnitude of the electric field just outside the surface of the sphere is: A) 0 N/C B) 450 N/C C) 900 N/C
D) 4500 N/C E) 90,000 N/C Ans: E Difficulty: M Section: 23-3 Learning Objective 23.3.6 32. A 300-N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet. The area charge density on the left and right faces, respectively, are: – 9 2 – 9 2 A) – 2.7 10 C/m ; +2.7 10 C/m – 9 2 – 9 2 B) +2.7 10 C/m ; – 2.7 10 C/m – 9 2 – 9 2 C) – 5.3 10 C/m ; +5.3 10 C/m – 9 2 – 9 2 D) +5.3 10 C/m ; – 5.3 10 C/m 2 2 E) 0 C/m ; 0 C/m Ans: A Difficulty: E Section: 23-3 Learning Objective 23.3.6 33. Charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 20 N/C. The electric field 4 cm from the wire is: A) 120 N/C B) 80 N/C C) 40 N/C D) 10 N/C E) 5 N/C Ans: D Difficulty: E Section: 23-4 Learning Objective 23.4.2 34. A long line of charge with λℓ charge per unit length runs along the cylindrical axis of a cylindrical conducting shell which carries a charge per unit length of c. The charge per unit length on the inner and outer surfaces of the shell, respectively are: A) λℓ and c B) − λℓ and c + λℓ C) − λℓ and c – λℓ D) λℓ + c and c – λℓ E) λℓ – c and c + λℓ Ans: B Difficulty: E Section: 23-4
Learning Objective 23.4.2 35. Charge is distributed uniformly on the surface of a large flat plate. The electric field 2 cm from the plate is 30 N/C. The electric field 4 cm from the plate is: A) 120 N/C B) 60 N/C C) 30 N/C D) 15 N/C E) 7.5 N/C Ans: C Difficulty: E Section: 23-5 Learning Objective 23.5.1 36. Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A) 1, 2, 3, 4, 5 B) 5, 4, 3, 2, 1 C) 1 and 4 and 5 tie, then 2 and 3 tie D) 2 and 3 tie, then 1 and 4 tie, then 5 E) 2 and 3 tie, then 1 and 4 and 5 tie Ans: C Difficulty: M Section: 23-5 Learning Objective 23.5.2 37. Two large insulating parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A) B) C) D)
1, 2, 3, 4, 5 5, 4, 3, 2, 1 1 and 4 and 5 tie, then 2 and 3 tie 2 and 3 tie, then 1 and 4 tie, then 5
E)
2 and 3 tie, then 1 and 4 and 5 tie
Ans: E Difficulty: M Section: 23-5 Learning Objective 23.5.2 38. Two large conducting parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A) 1, 2, 3, 4, 5 B) 5, 4, 3, 2, 1 C) 1 and 4 and 5 tie, then 2 and 3 tie D) 2 and 3 tie, then 1 and 4 tie, then 5 E) 2 and 3 tie, then 1 and 4 and 5 tie Ans: C Difficulty: M Section: 23-5 Learning Objective 25.5.3 39. A solid insulating sphere of radius R contains a positive charge that is distributed with a volume charge density that does not depend on angle but does increase linearly with distance from the sphere center. Which of the graphs below correctly gives the magnitude E of the electric field as a function of the distance r from the center of the sphere?
A) A B) B C) C D) D E) E
Ans: D Difficulty: M Section: 23-6 Learning Objective 23.6.1 40. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A point charge q is placed at the center of the cavity. The force on the charge q is: A) B) C) D) E)
( ) ( )
0
Ans: E Difficulty: E Section: 23-6 Learning Objective 23.6.2 41. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field at a point outside the shell, a distance r from the center, is: A) B) C) D) E)
( ) 2 q/40r 2 (q + Q)/40r ( )
Ans: D Difficulty: M Section: 23-6 Learning Objective 23.6.3 42. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a point r < R1 is: A) B) C) D) E)
( ) 2 Q/40r
0
( )
Ans: D Difficulty: E Section: 23-6 Learning Objective 23.6.4 43. Charge Q is distributed uniformly throughout an insulating sphere of radius R. The magnitude of the electric field at a point R/2 from the center is: 2 A) Q/40 R 2 B) Q/0 R 2 C) 3Q/40 R 2 D) Q/80 R E) none of these Ans: D Difficulty: M Section: 23-6 Learning Objective 23.6.5 44. Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at the origin. A particle with a positive charge Q is placed at x = 2 R on the x axis. The magnitude of the electric field at x = R/2 on the x axis is: 2 A) Q/720 R 2 B) Q/80 R 2 C) 7Q/180 R 2 D) 11Q/180 R E) none of these Ans: A Difficulty: H Section: 23-6 Learning Objective 23.6.5