ANSWERS TO CHAPTER 21 HOMEWORK
1.(P.569 Ex.4) The magnitude of the Coulomb force is
F = kQ1Q2/r2
= (9.0 x 109N.m2/C2)(1.60 x 10 -19C)(1.60
x 10-19C)/(5.0 x 10-12m)2 = 9.2N
| 2. Using the symbols in the figure, we find the magnitudes of
the three individual forces.
F12 = F21 = kQ1Q2/r122
= kQ1Q2/L2
= (9.0 x 109N.m2/C2)(70
x 10-6C)(48 x 10-6C)/(0.35m)2
= 2.47 x 102N
F13 = F31 = kQ1Q2/r132
= kQ1Q3/(2L)2
= (9.0 x 109N.m2/C2)(70
x 10-6C)(80 x 10-6C)/[2(0.35m)]2
= 1.03 x 102N
F23 = F32 = kQ2Q3/r23
= kQ2Q3/L2
=(9.0 x 109N.m2/C2)(48
x 10-6C)(80 x 10-6C)/(0.35m)2
= 2.82 x 102N.
The directions of the forces are determined from the signs of the
charges and are indicated on the diagram. For the net forces, we get
F1 = F13
- F12 = 1.03 x 102N - 2.47 x 102N
= -1.4 x102N (left)
F2 = F21 +
F23 = 2.47 x 102N + 2.82 x 102N = +5.3
x 102N (right)
F3 = -F31
- F32 = -1.03 x102N - 2.82 x 102N
= -3.9 x 102N (left)
Note that the sum for the three charges is zero.
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| 3. The magnitudes of the individual forces on the charges are
F12 = kQ2Q/l2
= 2kQ2/l2;
F13 = kQ3Q/(lÖ2)2
= 3kQ2/2l2;
F14 = kQ4Q/l2
= 4kQ2/l2;
F23 = k2Q3Q/l2
= 6kQ2/l2;
F24 = k2Q4Q/(lÖ2)2
= 4kQ2/l2
F34 = k3Q4Q/l2
= 12kQ2/l2.
The directions of the forces are determined from the signs of the
charges and are indicated on the diagram.
For charge Q we have
F1 = (-F12
- F13cos45°)i + (-F13sin45°
+ F14)jj
= [- (2kQ2/l2)
- (3kQ2/2l2)(Ö2/2)]i
+ [ -(3kQ2/2l2)(Ö2/2)
+ (4kQ2/l2)]j
= (kQ2/l2)[-2
- 3Ö2/4)i + (4 - 3Ö2/4)]j
For Charge 2Q we have
F2 = (F12
+ F24cos45°)i + (F24sin45°
- F23)j
= [(2kQ2/l2)
+ (4kQ2/l2)(Ö2/2)]i
+ [(4kQ2/l2)(Ö2/2)
- (6kQ2/l2)]j
=
(kQ2/l2)[(2
+ 2Ö2)i + (-6 +2Ö2)j]
For Charge 2Q we have
F3 = (F34
+ F13cos45°)i + (F13sin45° + F23)j
= [(-12kQ2/l2)
- (3kQ2/2l2)(Ö2)]i
+ [(3kQ2/2l2)(Ö2/2)
+ (6kQ2/l2)]j
= (kQ2/l2)[(-12
- 3Ö2/4)i + (6+ 3Ö2/4)j].
For charge 4Q we have
F4 = (- F34
- F24cos45°)i + (- F24sin45°
- F14)j
= [(12kQ2/l2)
- (4kQ2/l2)(Ö2/2)]i
+ [- (4kQ2/l2)(Ö2/2)
- (4kQ2/l2)]j
= (kQ2/l2)[(12 - 2Ö2)i + (-4 - 2Ö2)j].
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| 4. The directions of the individual fields will be along the
diagonals of the square, as shown.
We find the magnitudes of the individual fields:
E1 = kQ1/(L/Ö2)2
= 2kQ1/L2
= 2(9.0 x 109N.m2/C2)(45.0 x 10-6C)(0.525m)2
= 2.94 x106N/C.
E2 = E3
= E4 = kQ2/L/Ö2)2
= 2kQ2/L2
=
2(9.0 x 109N.m2/C2)(27.0 x 10-6C)/(0.525m)2
=
1.76 x 106N/C.
From the symmetry, we see that the resultant field will be
along the diagonal shown as the x-axis. For the net field, we have
E = E1
+ E3 = 2.94 x 106N/C + 1.76 x 106N/C
= 4.70 x 106N/C.
Thus the field at the center is 4.70
x 106N/C away from the positive charge.
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| 5. In Example 21-9, the field produced on the axis of a
single ring is given in terms of the distance from the center of the ring
We use two expressions with the origin shifted to the position between the
two rings: E =
(Q/4peo)({(x
+ 1/2l)2/[(x + 1/2l)2 + R2]3/2}
+{(x - 1/2l)/[(x -
1/2l)2 + R2]3/2})i
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6. (a) We find the acceleration produced by the electric field:
qE = ma
(--1.60 x 10-19C)[(2.0 x 104N/C)i + (8.0 x
104N/C)j] = (9.11 x 10-31kg)a
which gives a
= -(3.5 x 1015m/s2)i - (1.41 x 1016m/s2)j
Because the field is
constant, the acceleration is constant.
(b) We find the velocity from
v
= vo + at
= (8.0 x10m/s)i + [ -(3.5 x 1015m/s2i - (1.41 x 1016m/s2)j](1.0 x 10-9s)
= (-3.43 x 106m/s)i - (1.41 x 107m/s)j.
The direction of the
electron is the direction of its velocity:
tanq =
vy/vx = (-1.41 x 107m/s)/(-3.43 x 106m/s) = 4.11, or
q = -104°
7. (a) We find the net charge from
p = Ql;
3.4 x 10-30C.m = Q(1.0 x10-10m), which gives Q
= 3.4 x 10-20C.
(b) No,
this is not an integral multiple of e. The covalent bonding means the
electron is shared between the H and Cl atoms, so the effective net charge is
less than e.
(c) The maximum torque on the dipole is
t = pEsinq = (3.4
x 10-30C.m)(2.5 x 104N/c)sin90° = 8.5
x10-26m.N.
(d) The lowest potential energy is when the dipole
and electric field are parallel. Thus the energy needed to change the potential
energy is
W = DU = ( -p×E)f
- (-p&×E)i
= (-pEcos45° ) - (-pE) = pE(1 - cos45°)
= (3.4 x 10-30C.m)(2.5 x 104N/C)(1 - 0.707) = 2.5
x 10-26J.
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8*. (a) We assume the angles are small enough
that the Coulomb force between the charges can be treated as being
horizontal.
From the force diagram, we apply SF
= 0 on each charge:
Q1 : FT1sinq1
= F = kQ1Q2/r2
= 2kQ2/r2;
FT1sinq1= m1g,
or tanq1 = 2kQ2/m1gr2.
Q2 : FT2sinq2
= F = kQ1Q2/r2
= 2kQ2/r2;
FT2sinq2 = m2g,
or tanq2 = 2kQ2/m2gr2.
If we divide the two results and use tanq
»q for small angles, we get
tanq1/tanq2=
q1/q2
= m2/m1=1.
(b) The analysis of forces is the same, so we have
tanq1/ tanq2=
q1/q2
= m2/m1=2.
(c) The distance between the charges is
r = lsinq1 + lsinq2
= lq1 + lq2.
For the conditions in part (a), we have
ra = 2lq1 =
2l(2kQ2/mgra2),
which gives = (4klQ2/mg)1/3.
For the conditions in part (a), we have
rb = lq1
+ lq2 = 3lq2
= 3l(2kQ2/m2grb2),
which gives = (3klQ2/mg)1/3.
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| 9*. From the symmetry, we see that there are
only three
magnitudes for the seven forces from the other charges:
3 adjacent corners: F1
= kQQ/l2
= kQ2/l2;
3 diagonal corners: F2
= kQQ/(lÖ2)2
= kQ2/2l2
;
1 opposite corner:
F3
= kQQ/(lÖ3)2
= kQ2/3l2
.
The directions of the forces are determined from the signs
of the charges and are indicated on the diagram.
We could add the seven forces, but we can use the
symmetry to reduce the process.
Each of the components
will have the same magnitude and will be in the
corresponding negative direction.
Thus we find one of them:
Fx = – F1
– F2/Ö2
– F2/Ö2
– F3/Ö3
= – [(kQ2/l2)
+ 2(kQ2/2l2)/Ö2
+ (kQ2/3l2)/Ö3]
= – (kQ2/l2)[1
+ (1/Ö2) + (1/3Ö3)] = – 1.90kQ2/l2.
Thus the resultant force is
F
= – (1.90kQ2/Ö2)(i + j + k).
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| 10.* We choose a differential element of the rod dx'
a distance x' from the origin of the coordinate system, as shown in
the diagram. Because the positive direction of x' is to the left,
the limits for x' are 0 to l. The charge of the element is dq
= (Q/i)dx'. we find the field by integrating along the rod :
E = [1/4pe0ò 0ldq/(x
+ x')2]i = (Q/4pe0l)[ò 0ldx'/(x
+ x')2]i
= (Q/4pe0l)[-1/(x
+ x')]0li
= (-Q/4pe0l)[1/(x
+ l) - 1/x]i
= [Q/4pe0x(x
+ l)]i
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11* (a) The field along the axis of the
ring is
E = (-Q/4pe0)[x/(x2
+ R2)3/2]i,
so the force on the charge is
F = qE = (-qQ/4pe0)[x/(x2
+ R2)3/2]i
= (-qQx/4pe0R3)/[1
+ (x/R)2)]3/2i.
If x << R, we can use the approximation (1 +
u)-n = 1 -nu :
F » -qQx/4pe0R3)[1
+ 3/2(x/R)2] » -qQx/4pe0R3.
We see that the force is a restoring force proportional to the
displacement, so the motion will be simple harmonic.
(b) The effective spring constant is
k = qQ/4pe0R3,
So the period is
T = 2p(m/k)1/2 = 2p(4pe0mR3/Qq)1/2.
| 12* If the charges of the dipole are
separated by dx, the dipole moment is p = Qdxi.
If the negative charge is at x, where the electric field is E(x),
the electric field at the positive charge is
E(x + dx) = E(x) + (dE/dx)dx.
The net force on the dipole is
F = F+ - F- = [QE(x
+ dx) - QE(x)]i
= Q(dE/dx)dxi
= p(dE/dx)i = [p · (dE/dx)]i.
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