Chapter

Chapter: Chapter 23

Learning ObjectivesLO 23.1.0 Solve problems related to electric 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 through the surface.LO 23.1.3 Identify that an area vector 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 elements (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 through a surface by integrating the dot product of the electric field vector and the area vector (for patch elements) over the surface, in 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 through a closed surface, algebraic sign included, by integrating the dot product of the electric field vector and the area vector (for patch 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 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 for the magnitude of the electric field of a charged particle by using Gauss law.LO 23.2.5 Identify that for a charged 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 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 an isolated conducting surface with a uniform surface charge density .LO 23.3.6 For a uniformly charged conducting surface, apply the relationship between the charge density 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 magnitude outside a line of charge or a cylindrical surface (such as a plastic rod) with a uniform linear charge density .LO 23.4.2 Apply the relationship between linear charge density 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 .

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 .LO 23.5.2 For points near a large, flat nonconducting surface with a uniform charge density , 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 density , 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 shells charge is concentrated 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 shells center.LO 23.6.4 Identify the magnitude of the electric field for points enclosed by a spherical 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. Gausss 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: DDifficulty: ESection: 23-1Learning Objective 23.1.1

2. The electric flux 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: ADifficulty: ESection: 23-1Learning 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.


4. To calculate the flux through a curved surface,A) the area vector has to be perpendicular to the surface somewhereB) you must divide the surface into pieces that are tiny enough to be almost flatC) the surface must be sphericalD) the surface cannot be curved very much; then you can treat it as though it were flatE) actually the flux through a curved surface cannot be calculated.

Ans: BDifficulty: ESection: 23-1Learning Objective 23.1.4

5. When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25 Nm2/C. When the paper is turned 25 with respect to the field the flux through it is: A) 0 Nm2/CB) 11 Nm2/CC) 12 Nm2/CD) 23 Nm2/CE) 25 Nm2/C


6. The flux of the electric field (24 N/C) + (30 N/C) + (16 N/C) through a 2.0 m2 portion of the yz plane is: A) 32 N m2/C B) 34 N m2/C C) 42 N m2/C D) 48 N m2/C E) 60 N m2/C

Ans: DDifficulty: MSection: 23-1Learning 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: A) 0 Nm2/CB) 4.2 Nm2/CC) 21 Nm2/CD) 280 Nm2/CE) can't tell without knowing the areas of the sides and bottom

Ans: CDifficulty: MSection: 23-1Learning 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 negative and outward flux is positive.E) Inward flux through a closed surface is positive and outward flux is negative.


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: A) 0 Nm2/CB) 4.2 Nm2/CC) 21 Nm2/CD) 280 Nm2/CE) can't tell without knowing the length of the cylinder


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


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: A) 0 Nm2/CB) 25 106 Nm2/CC) 2.2 105 Nm2/CD) 2.8 106 Nm2/CE) can't tell unless the lecture hall dimensions are given


12. A point particle with charge q is placed inside a cube but not at its center. The electric flux through any one side of the cube:A) is zeroB) is q/0C) is q/40D) is q/60E) cannot be computed using Gauss' law

Ans: EDifficulty: ESection: 23-2Learning Objective 23.2.1

13. A particle with charge 5.0C is placed at the corner of a cube. The total electric flux through all sides of the cube is:A) 0 Nm2/CB) 7.1 104 Nm2/CC) 9.4 104 Nm2/CD) 1.4 105 Nm2/CE) 5.6 105 Nm2/C

Ans: BDifficulty: MSection: 23-2Learning 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/0B) q/40C) q/40D) q/60E) q/160


15. A 3.5-cm radius hemisphere contains a total charge of 6.6 107 C. The flux through the rounded portion of the surface is 9.8 104 Nm2/C. The flux through the flat base is: A) 0 Nm2/CB) +2.3 104 Nm2/C C) 2.3 104 Nm2/C D) 9.8 104 Nm2/C E) +9.8 104 Nm2/C


16. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux through the outer surface of the shell is: A) 0 B) Q/0C) 2Q/0D) Q/40E) Q/20


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.

left endright endrounded surface

cylinder 1:+2 109 Nm2/C+4 109 Nm2/C6 109 Nm2/C

cylinder 2:+3 109 Nm2/C2 109 Nm2/C+6 109 Nm2/C

cylinder 3:2 109 Nm2/C5 109 Nm2/C+3 109 Nm2/C

cylinder 4:+2 109 Nm2/C5 109 Nm2/C3 109 Nm2/C

A) 1, 2, 3, 4B) 4, 3, 2, 1C) 3, 4, 2, 1D) 3, 1, 4, 2E) 4, 3, 1, 2


18. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux through the inner surface of the shell is: A) 0 B) Q/0C) 2Q/0D) Q/40E) Q/20


19. Consider Gauss law: . Which of the following is true? A) must be the electric field due to the enclosed charge B) If q = 0 then everywhere on the Gaussian surface C) If the three particles inside have charges of +q, +q and 2q, then the integral is zero D) On the surface is everywhere parallel to E) If a charge is placed outside the surface, then it cannot affect at any point on the surface

Ans: CDifficulty: ESection: 23-2Learning 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 enclosed 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


21. The outer surface of the cardboard center of a paper towel roll: A) is a possible Gaussian surface B) cannot be a Gaussian surface because it encloses no charge C) cannot be a Gaussian surface since it is an insulator D) cannot be a Gaussian surface since it is not closedE) none of the above


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) C) q/40r2D) (q + Q)/40r2E)


23. A conducting sphere of radius 5.0 cm carries a net charge of 7.5 C. What is the surface charge density on the sphere?A) 2.4 x 10-2 C/m2 B) 1.4 x 10-2 C/m2 C) 9.5 x 10-4 C/m2 D) 2.4 x 10-4 C/m2E) 6.0 x 10-5 C/m2


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


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


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 CB) 3 CC) 0 CD) +3 CE) +7 C


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 CB) 3 CC) 0 CD) +3 CE) +7 C


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) q/40r2E) Q/40r2


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, QB) q, Q qC) Q, 0D) q, Q + qE) q, 0


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) AB) B C) CD) DE) E


31. A conducting sphere of radius 0.01 m has a charge of 1.0 109 C deposited on it. The magnitude of the electric field just outside the surface of the sphere is: A) 0 N/CB) 450 N/CC) 900 N/CD) 4500 N/CE) 90,000 N/C

Ans: EDifficulty: MSection: 23-3Learning 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: A) 2.7 109 C/m2; +2.7 109 C/m 2B) +2.7 109 C/m2; 2.7 109 C/m 2C) 5.3 109 C/m2; +5.3 109 C/m 2D) +5.3 109 C/m2; 5.3 109 C/m 2E) 0 C/m2; 0 C/m2


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


34. A long line of charge with 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) and cB) and c + C) and c D) + c and c E) c and c +


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


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, 5B) 5, 4, 3, 2, 1C) 1 and 4 and 5 tie, then 2 and 3 tieD) 2 and 3 tie, then 1 and 4 tie, then 5E) 2 and 3 tie, then 1 and 4 and 5 tie


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.


Ans: EDifficulty: MSection: 23-5Learning 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.



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) AB) BC) CD) DE) E


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


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) q/40r2D) (q + Q)/40r2E)


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) Q/40r2D) 0E)


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: A) Q/40R2B) Q/0R2C) 3Q/40R2D) Q/80R2E) none of these


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 = 2R on the x axis. The magnitude of the electric field at x = R/2 on the x axis is: A) Q/720R2B) Q/80R2C) 7Q/180R2D) 11Q/180R2E) none of these

Ans: ADifficulty: HSection: 23-6Learning Objective 23.6.5

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