1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1
c6.3 10 kghc7.0 m s h = 4.9 10 m =
-7 2 2
-7
9.0 m s
2
kg.
(b) The magnitude of the (only) force on particle 1 is
q q q F = m1a1 = k 1 2 2
3. Eq. 21-1 gives Coulomb's Law, F = k
k | q1 | q2 | r= = F
q1 q2 r2
, which we solve for the distance:
(8.99 10 N m
9
2
C2 ) ( 26.0 10-6 C ) ( 47.0 10-6 C ) 5.70N
= 1.39m.
10. With rightwards positive, the net force on q3 is
F3 = F13 + F23 = k q1q3
( L12 + L23 )
2
+k
q2 q3 . L2 23
We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For exampl
11. (a) Eq. 21-1 gives
( 20.0 10 C ) = 1.60N. qq F12 = k 1 22 = ( 8.99 109 N m 2 C2 ) 2 d (1.50m )
-6
2
(b) A force diagram is shown as well as our choice of y axis (the dashed line).
The y axis is meant to bisect the line between q2 and q3 in order to ma
12. (a) According to the graph, when q3 is very close to q1 (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q1 has the
10. We place the origin of our coordinate system at point P and orient our y axis in the direction of the q4 = 12q charge (passing through the q3 = +3q charge). The x axis is perpendicular to the y axis, and thus passes through the identical q1 = q2 = +5q
13. (a) There is no equilibrium position for q3 between the two fixed charges, because it is being pulled by one and pushed by the other (since q1 and q2 have different signs); in this region this means the two force arrows on q3 are in the same direction
14. Since the forces involved are proportional to q, we see that the essential difference between the two situations is Fa qB + qC (when those two charges are on the same side) versus Fb -qB + qC (when they are on opposite sides). Setting up ratios, we ha
16. (a) For the net force to be in the +x direction, the y components of the individual forces must cancel. The angle of the force exerted by the q1 = 40 C charge on q3 = 20 C is 45, and the angle of force exerted on q3 by Q is at where
= tan -1
FG 2.0IJ
17. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q3 must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q3 could
18. (a) We note that cos(30) = 2 3 , so that the dashed line distance in the figure is
r = 2d / 3 . We net force on q1 due to the two charges q3 and q4 (with |q3| = |q4| = 1.60 10-19 C) on the y axis has magnitude
1
2
| q1q3 | 3 3 | q1q3 | cos(30) = . 2 4
19. The charge dq within a thin shell of thickness dr is A dr where A = 4r2. Thus, with = b/r, we have
q = dq = 4 b
z
z
r2
r1
r dr = 2b r22 - r12 .
c
h
With b = 3.0 C/m2, r2 = 0.06 m and r1 = 0.04 m, we obtain q = 0.038 C = 3.8 10-8 C.
20. If is the angle between the force and the x-axis, then cos = x . x + d2
2
We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2 o
21. (a) The magnitude of the force between the (positive) ions is given by
bqgbqg = k q F=
4 0r 2
2
r2
where q is the charge on either of them and r is the distance between them. We solve for the charge: q=r F = 5.0 10-10 m k
c
h
3.7 10-9 N = 3.2 10-19 C.
24. (a) Eq. 21-1 gives . c8.99 10 N m C hc100 10 Ch F= . c100 10 mh
9 2 2 -16 -2 2 2
= 8.99 10-19 N .
(b) If n is the number of excess electrons (of charge e each) on each drop then n=- . q -100 10-16 C =- = 625. 160 10-19 C . e
26. The volume of 250 cm3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm3. This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten protons (each with charge q = +e) in each molecule of H2O, so Q
25. The unit Ampere is discussed in 21-4. The proton flux is given as 1500 protons per square meter per second, where each proton provides a charge of q = +e. The current through the spherical area 4 R2 = 4 (6.37 106 m)2 = 5.1 1014 m2 would be i = 51 1014
27. Since the graph crosses zero, q1 must be positive-valued: q1 = +8.00e. We note that it crosses zero at r = 0.40 m. Now the asymptotic value of the force yields the magnitude and sign of q2: q1 q2 =F 4o r2 q2 =
1.5 x 10-25 2 r = 2.086 x 10-18 C = 13e .
1. We note that the symbol q2 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 q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figure) an
2. (a) We note that the electric field points leftward at both points. Using F = q0 E , and orienting our x axis rightward (so ^ points right in the figure), we find i
. F = +16 10-19 C -40
c
h FGH
N i = -6.4 10-18 N i C
IJ K
which means the magnitude of
6. With x1 = 6.00 cm and x2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q1 = q2 = 2.00 107 C, and the magnitudes and directions of the individual fields are given by:
E1 = - E2 = - | q1 | ^
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 4 0 R 2
where q is the magnitude of th
9. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q2= - 4.00 q1 located at x2 = 70 cm has a greater magnitude than q1 = 2.1 10-8 C located at x1 = 20 cm, a point of zero field must b
12. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 22-3 to compute magnitude of the field due to q3 = +2e. (a) The magnitude of the net electric field is | Enet |= 1 2e 1 2e 1 4e 4(1.60 10-1
11. The x component of the electric field at the center of the square is given by
Ex = = | q3 | | q1 | | q2 | | q4 | 1 cos 45 + - - 2 2 2 4 0 (a / 2) (a / 2) (a / 2) (a / 2) 2
1 1 1 (| q1 | + | q2 | - | q3 | - | q4 |) 2 4 0 a / 2 2 = 0. Similarly, the y c