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. |
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 |
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.
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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