4. To Apply Gauss’s Law correctly, it is CRITICAL to know the charge distribution on, over, or within an object
5. Sometimes we would like to know the electric field due to the charge distribution on the surface of a conductor: These calculations are aided by this fact: When excess charge is placed on a solid conductor and is at rest, the excess charge resides entirely on the surface, and not within the interior of the material. PLUS: Don’t forget that within conductors, electrons move easily, in response to electric fields, which surround any object that is charged
6. Logical argument: If there were an electric field in the conductor, then the field would exert a force on every charge within the conductor, giving the free charges a net motion. If the situation is electrostatic in nature (the charges are not moving), then we can only conclude that the electric field inside the conductor is zero. This has been verified experimentally. Plus, where is the safest place to be if your car has a large charge deposited on it from, say… lightning? That’s right, inside. Where the electric field is zero. Yeah.
9. The spark from a lightning strike can reach over five miles (eight kilometers) in length, raise the temperature of the air by as much as 50,000 degrees Fahrenheit (27,700 degrees Celsius), and contain a hundred million electrical volts. Lightning is not confined to thunderstorms. It's been seen in volcanic eruptions, extremely intense forest fires, surface nuclear detonations, heavy snowstorms, and in large hurricanes.
10. Ponder this: Thunderstorms can produce gamma radiation.Scientists first spotted gamma rays in thunderstorms in the early 1990s. The Compton Gamma Ray Observatory unexpectedly detected radiation originating from the ground while peering at distant supernovae.
11. Researchers from Japan found that gamma rays were produced some 70 seconds before a lightning strike. They also determined that gamma bursts, which had been previously measured to last less than a second, could occur for almost a minute.
17. A spherical Gaussian surface (#1) encloses and is centered on a point charge + q . A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is 1. greater 2. the same 3. less, but not zero 4. zero 5. not enough information given to decide Q1
18.
19. Two point charges, + q (in red) and – q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? 1. surface A 2. surface B 3. surface C 4. surface D 5. both surface C and surface D Q2
20. Two point charges, + q (in red) and – q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? 1. surface A 2. surface B 3. surface C 4. surface D 5. both surface C and surface D A2
21. A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity 1. points generally toward the outer surface of the conductor 2. points generally away from the outer surface of the conductor 3. is zero 4. not enough information given to decide Q3
22. A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity 1. points generally toward the outer surface of the conductor 2. points generally away from the outer surface of the conductor 3. is zero 4. not enough information given to decide A3
23. There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field 1. points outward, toward the surface of the conductor 2. points inward, away from the surface of the conductor 3. points parallel to the surface 4. is zero 5. not enough information given to decide Q4
24. There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field 1. points outward, toward the surface of the conductor 2. points inward, away from the surface of the conductor 3. points parallel to the surface 4. is zero 5. not enough information given to decide A4
26. Gauss’s Law is valid for ANY charge distribution and ANY closed surface (regardless of shape): - if we know the charge distribution and if that distribution has enough symmetry to allow to evaluate the integral in Gauss’s law, then we can find the field. - if we know the field, then we can use Gauss’s law to find the charge distribution, such as charges on conducting surfaces.
29. Electric Field of a Hollow Conducting Sphere Copyright Pearson Education, Inc.
30. Electric field of a Solid Insulating Sphere Copyright Pearson Education, Inc.
31. Using a Gaussian Cylinder to Find the Electric Field of a Flat Charged Sheet Copyright Pearson Education, Inc.
32. Problem: Consider a sphere of radius R =8.0 m with a charge Q =3.0 mC uniformly distributed throughout the volume. What is the electric field at a distance r =4.0 m ? Solution: If one draws a Gaussian surface with radius r , the charge contained inside is one eighth of the total charge, because volume goes proportional to r 3 . Thus Gauss law becomes: E =210 N/C
![Relevant Fact of the Day ,[object Object]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)