QUIZ 3

2. (a) - 7.7x10¹⁶ N*m²/C

4. (a) E*pi*r² . (b) zero, since the field points parallel to the surface and makes an angle of 90 degrees with the normal.

14. E = k*Q/R². = Q/(4*pi*epsilon*R² ) = (sigma)/(epsilon) from the definition of area charge density. .

16. Lines are radial inward, pointing toward the center.

18. (a) 0 (b) 0 (c) 5140 N/C, toward center. (d) 772 N/C, toward center (e) same answers (f) 458, 5310, 5140, 772, all toward center for distances 0.250, 2.90 , 3.10, 8.00 meters, all toward the center.

22. (a) k*Q/r² (b) 0 (c) k*Q/r²

(d)

The shell does not influence the field due to Q alone, except in the shell material, where the field is 0. The charge Q does affect the shell in that it polarizes the shell. We discussed this deeply in 9/17 Friday review lecture. There will be an induced surface charge of –Q uniformly distributed over the inside surface of the shell, and an induced surface charge of +Q uniformly distributed over the outside surface of the shell.

24. See discussions--Quiz 3. E(between) = sigma/epsilon, E(outside) = 0.
(c) If the plates were non-conducting thin sheets, the results would not change.

26. 1.4x10^-9 C.

36. (a) 0 (b) 4.5 x10⁴ N/C, toward the central axis. (c) 2.0 x10⁴ N/C, away from central axis.

46. See discussions. Inside: E_x = (sigma)*x/epsilon; Outside: E_x = + or - (sigma)*d/(2*epsilon). The minus sign is for the region left of the plane and the plus sign is for the region right of the plane.