. 3. We use = E A , where A = Aj = 140m j .
2 2 i j (a) = ( 6.00 N C ) (1.40 m ) = 0. 2 (b) = ( 2.00 N C ) (1.40 m ) = 3.92 N m 2 C. j j 2 (c) = ( 3.00 N C ) + ( 400 N C ) k (1.40 m ) = 0 . i j
b
g
(d) The total flux of a uniform field through a closed s
Students Material Experiment 06 Force in a magnetic field
Experiment 06
6-1
Force in a magnetic field
Material
Multimeter:
Arzana: Analog 30
Bu Hasa: METRAmax 3
Sensor-CASSY
CASSY Lab
30 Ampere-box
Force sensor with sensor plug
Support for conductor lo
25. 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 = 1.60 1019 C gives the number of electrons: 4.5 1014, which
26. 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). The charge on the equivalent capacitor is the same as th
27. (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) Capacitors 2 and 4 are also in series: q2 = q4 = (c) q3 =
28. Initially the capacitors C1, C2, and C3 form a combination equivalent to a single capacitor which we denote C123. This obeys the equation C + C2 + C3 1 1 1 =+ =1 . C123 C1 C2 + C3 C1 (C2 + C3 ) Hence, using q = C123V and the fact that q = q1 = C1 V1 ,
29. The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference V across the capacitors is the same and the total energy is U= 1 1 2 ( C1 + C2 )V 2 = ( 2.0 106 F + 4.0 106
12. Eq. 23-6 (Gauss law) gives = qenc . (a) Thus, the value = 2.0 105 N m 2 /C for small r leads to qcentral = 0 = (8.85 1012 C2 /N m 2 )(2.0 105 N m 2 /C) = 1.77 106 C 1.8 106 C . (b) The next value that takes is = 4.0 105 N m 2 /C , which implies qenc =
11. (a) Let A = (1.40 m)2. Then j j = 3.00 y A
(
)(
)
j j + 3.00 y A
y =0
(
)( )
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 m 2 C = 7.29 1011 C .
(c) The electric field can be re
9. Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the upper face, and El be the magnitude of the field at the lower face. Since the field is downward, the flux through the upper face is negative and the flux throug
8. We note that only the smaller shell contributes a (non-zero) field at the designated point, since the point is inside the radius of the large sphere (and E = 0 inside of a spherical charge), and the field points towards the x direction. Thus, with R =
7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d, with a proton of charge q = +1.6 1019 C situated at the inside center of the cube. The cube has six faces, and we expect an equal a
6. The flux through the flat surface encircled by the rim is given by = a 2 E. Thus, the flux through the netting is
= = a 2 E = (0.11 m) 2 (3.0 103 N/C) = 1.1 104 N m 2 /C .
Students Material Experiment 07 Faradays Law of Induction
Experiment 07
Material
7-1
Faradays Law of Induction
Sensor-CASSY
Power-CASSY
CASSY Lab
Coil, 250 turns
Coil, 1000 turns
2 round bar magnets with clips
Stand base, V-shape, 28 cm
Stand rod, 50 cm
T