79. From the second measurement (at (2.0, 0) we see that the charge must be somewhere on the x axis. A line passing through (3.0, 3.0) with slope tan -1 (3 4) will intersect the x axis at x = 1.0. Thus, the location of the particle is specified by the coo
23. (a) and (b) We note that the charge on C3 is q3 = 12 C 8.0 C = 4.0 C. Since the charge on C4 is q4 = 8.0 C, then the voltage across it is q4/C4 = 2.0 V. Consequently, the voltage V3 across C3 is 2.0 V C3 = q3/V3 = 2.0 F. Now C3 and C4 are in parallel
24. For maximum capacitance the two groups of plates must face each other with maximum area. In this case the whole capacitor consists of (n 1) identical single capacitors connected in parallel. Each capacitor has surface area A and plate separation d so
27. We assume the charge density of both the conducting cylinder and the shell are uniform, and we neglect fringing effect. Symmetry can be used to show that the electric field is radial, both between the cylinder and the shell and outside the shell. It i
28. As we approach r = 3.5 cm from the inside, we have Einternal = 2 4 0 r = 1000 N/C .
And as we approach r = 3.5 cm from the outside, we have Eexternal = 2 4 0 r + 2 = 3000 N/C . 4 0 r
Considering the difference (Eexternal Einternal ) allows us to find
29. We denote the inner and outer cylinders with subscripts i and o, respectively. (a) Since ri < r = 4.0 cm < ro,
i 5.0 106 C/m E (r ) = = = 2.3 106 N/C. 12 2 2 2 2 0 r 2 (8.85 10 C / N m ) (4.0 10 m)
(b) The electric field E (r ) points radially outward
30. (a) In Eq. 23-12, = q/L where q is the net charge enclosed by a cylindrical Gaussian surface of radius r. The field is being measured outside the system (the charged rod coaxial with the neutral cylinder) so that the net enclosed charge is only that w
26. We reason that point P (the point on the x axis where the net electric field is zero) cannot be between the lines of charge (since their charges have opposite sign). We reason further that P is not to the left of line 1 since its magnitude of charge (
57. (a) The magnitude of the dipole moment is . p = qd = 150 109 C 6.20 106 m = 9.30 1015 C m. (b) Following the solution to part (c) of Sample Problem 22-5, we find U (180 ) U ( 0 ) = 2 pE = 2 ( 9.30 1015 C m ) (1100 N/C ) = 2.05 1011 J.
c
hc
h