43. (a) During the time interval t, the light emitted from galaxy A has traveled a distance ct. Meanwhile, the distance between Earth and the galaxy has expanded from r to r' = r + r t. Let ct = r = r
29. Consider an infinitesimal segment of the rod, located between x and x + dx. It has length dx and contains charge dq = dx = cx dx. Its distance from P1 is d + x and the potential it creates at P1 i
33. We apply Eq. 24-41: V = 2.00 yz 2 x V = 2.00 xz 2 Ey = y V = 4.00 xyz Ez = z Ex = which, at (x, y, z) = (3.00, 2.00, 4.00), gives (Ex, Ey, Ez) = (64.0, 96.0, 96.0) in SI units. The magnitude of th
22. Initially the capacitors C1, C2, and C3 form a combination equivalent to a single capacitor which we denote C123. This obeys the equation 1 1 1 C1 + C2 + C3 = C123 . Hence, using q = C123V and the
18. Eq. 23-6 (Gauss law) gives = qenc. (a) The value = 9.0 105 (in SI units) for small r leads to qcentral = 7.97 106 C or roughly 8.0 C. (b) The next (non-zero) value that takes is +4.0 105 (in SI un
34. (a) According to the result of problem 28, the electric potential at a point with coordinate x is given by V= At x = d we obtain (in SI units) V= Q d+L (8.99 109 )(43.6 1015 ) 0.135 ln ln 1 + = 4
23. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q1 = q3 = C1C3V (1.00 F ) ( 3.00 F ) (12.0V ) = = 9.00 C. C1 + C3 1.00 F+3.00 F
(b) Cap
44. Assuming the current is along the wire (not radial) we find the current from Eq. 26-4: i= | J | dA =
R 0
1 kr 2 2 rdr = 2 kR4 = 3.50 A
where k = 2.75 1010 A/m4 and R = 0.00300 m. The rate of therm
19. (a) Consider a Gaussian surface that is completely within the conductor and surrounds the cavity. Since the electric field is zero everywhere on the surface, the net charge it encloses is zero. Th
35. The electric field (along some axis) is the (negative of the) derivative of the potential V with respect to the corresponding coordinate. In this case, the derivatives can be read off of the graph
42. The slopes of the lines yield P1 = 8 mW and P2 = 4 mW. Their sum (by energy conservation) must be equal to that supplied by the battery: Pbatt = ( 8 + 4 ) mW = 12 mW.
21. The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from 1 1 1 C2 + C3 = + = . Ceq C2 C3 C2 C3 Thus, Ceq = C2C3/(C2 + C3). T
18. Eq. 23-14 applies to each of these capacitors. Bearing in mind that = q/A, we find the total charge to be qtotal = q1 + q2 = 1 A1 + 2 A2 = o E1 A1 + o E2 A2 = 3.6 pC where we have been careful to
8. (a) The total surface area bounding the bathroom is
A = 2 ( 2.5 3.0 ) + 2 ( 3.0 2.0 ) + 2 ( 2.0 2.5 ) = 37 m 2 . The absolute value of the total electric flux, with the assumptions stated in the pr
30. The magnitude of the electric field is given by
| E |= V 2(5.0V) = = 6.7 102 V m. 0.015m x
At any point in the region between the plates, E points away from the positively charged plate, directly
19. (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
40. (a) From P = V 2/R we find R = V 2/P = (120 V)2/500 W = 28.8 . (b) Since i = P/V, the rate of electron transport is
i P 500 W = = = 2.60 1019 / s. e eV (1.60 1019 C)(120 V)
13. (a) Let A = (1.40 m)2. Then
j j = 3.00 y A
(
)( )
y =0
j j + 3.00 y A
(
)( )
y =1.40
= ( 3.00 )(1.40 )(1.40 ) = 8.23 N m 2 C.
2
(b) The charge is given by
qenc = 0 = 8.85 1012 C2 / N m 2 8.23 N
31. We use Eq. 24-41: V = (2.0V / m2 ) x 2 3.0V / m2 ) y 2 = 2(2.0V / m2 ) x; x x V E y ( x, y) = = (2.0V / m2 ) x 2 3.0V / m2 ) y 2 = 2(3.0V / m2 ) y . y y E x ( x, y) =
c c
h h
We evaluate at x = 3
20. We note that the total equivalent capacitance is C123 = [(C3)1 + (C1 + C2)1]1 = 6 F. (a) Thus, the charge that passed point a is C123 Vbatt = (6 F)(12 V) = 72 C. Dividing this by the value e = 1.6
41. (a) From P = V 2/R = AV 2 / L, we solve for the length: L= AV 2 (2.60 106 m2 )(75.0 V) 2 . = = 585 m. P (5.00 107 m)(500 W)
(b) Since L V 2 the new length should be
FV I L = LG J HV K
2
F 100 V IJ
15. (a) The charge on the surface of the sphere is the product of the surface charge density and the surface area of the sphere (which is 4r 2 , where r is the radius). Thus,
1.2 m q = 4r = 4 2
2 2
(8
32. We use Eq. 24-41. This is an ordinary derivative since the potential is a function of only one variable. E=
FG dV IJ i = d (1500x ) i = (3000x) i H dx K dx
2
= ( 3000 V / m2 ) (0.0130 m)i = ( 39 V
45. (a) Using Table 26-1 and Eq. 26-10 (or Eq. 26-11), we have
| E |= | J |= (1.69 108 m )
2.00A = 1.69 102 V/m. 6 2 2.00 10 m
(b) Using L = 4.0 m, the resistance is found from Eq. 26-16: R = L/A = 0.
20. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric q E dA = 2rE = enc . with the metal tube. Then by symmetry
A
0
(a) For r < R, qenc = 0, so E = 0. (b) For r > R,
36. (a) Consider an infinitesimal segment of the rod from x to x + dx. Its contribution to the potential at point P2 is
dV = 1 ( x )dx 1 = 2 2 4 0 x + y 4 0 cx x + y2
2
dx.
Thus, (in SI units) V= dVP