# ch22 - 1 We note that the symbol q2 is used in the problem...

• Homework Help
• 94

This preview shows pages 1–8. Sign up to view the full content.

1. We note that the symbol q 2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for 1 2 q q = . The following two sketches are for the cases q 1 > q 2 (left figure) and q 1 < q 2 (right figure).

This preview has intentionally blurred sections. Sign up to view the full version.

2. (a) We note that the electric field points leftward at both points. Using G G F q E = 0 , and orienting our x axis rightward (so ˆ i points right in the figure), we find G F = + × F H G I K J = − × 16 10 40 6 4 10 19 18 . # . # C N C i N i c h which means the magnitude of the force on the proton is 6.4 × 10 –18 N and its direction ˆ ( i) is leftward. (b) As the discussion in §22-2 makes clear, the field strength is proportional to the crowdedness of the field lines. It is seen that the lines are twice as crowded at A than at B , so we conclude that E A = 2 E B . Thus, E B = 20 N/C.
3. The following diagram is an edge view of the disk and shows the field lines above it. Near the disk, the lines are perpendicular to the surface and since the disk is uniformly charged, the lines are uniformly distributed over the surface. Far away from the disk, the lines are like those of a single point charge (the charge on the disk). Extended back to the disk (along the dotted lines of the diagram) they intersect at the center of the disk. If the disk is positively charged, the lines are directed outward from the disk. If the disk is negatively charged, they are directed inward toward the disk. A similar set of lines is associated with the region below the disk.

This preview has intentionally blurred sections. Sign up to view the full version.

4. We find the charge magnitude | q | from E = | q |/4 πε 0 r 2 : ( )( ) 2 2 10 0 9 2 2 1.00 N C 1.00m 4 1.11 10 C. 8.99 10 N m C q Er = π = = × × ε
5. Since the magnitude of the electric field produced by a point charge q is given by 2 0 | |/ 4 E q r πε = , where r is the distance from the charge to the point where the field has magnitude E , the magnitude of the charge is ( ) ( ) 2 2 11 0 9 2 2 0.50m 2.0 N C 4 5.6 10 C. 8.99 10 N m C q r E ε = π = = × ×

This preview has intentionally blurred sections. Sign up to view the full version.

6. With x 1 = 6.00 cm and x 2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q 1 = – q 2 = – 2.00 × 10 –7 C, and the magnitudes and directions of the individual fields are given by: 5 1 1 2 0 1 5 2 2 2 0 1 | | ˆ ˆ i (3.196 10 N C)i 4 ( ) ˆ ˆ i (3.196 10 N C)i 4 ( ) q E x x q E x x πε πε = − = − × = − = − × G G Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i E E E = + = − × G G G
7. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is E q R = 4 0 2 π ε where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze , so E Ze R = = × × × = × 4 8 99 10 94 160 10 6 64 10 307 10 0 2 9 2 2 19 15 2 21 π ε . .

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern