ANSWERS TO CHAPTER 21 HOMEWORK
1.(P.569 Ex.4) The magnitude of the Coulomb force is
F = kQ_{1}Q_{2}/r^{2 }= (9.0 x 10^{9}N.m^{2}/C^{2})(1.60 x 10 ^{19}C)(1.60 x 10^{19}C)/(5.0 x 10^{12}m)^{2} = 9.2N
2. Using the symbols in the figure, we find the magnitudes of
the three individual forces.
F_{12 }= F_{21} = kQ_{1}Q_{2}/r_{12}^{2} = kQ_{1}Q_{2}/L^{2} ^{ }= (9.0 x 10^{9}N.m^{2}/C^{2})(70 x 10^{6}C)(48 x 10^{6}C)/(0.35m)^{2} = 2.47 x 10^{2}N F_{13} = F_{31} = kQ_{1}Q_{2}/r_{13}^{2} = kQ_{1}Q_{3}/(2L)^{2} = (9.0 x 10^{9}N.m^{2}/C^{2})(70 x 10^{6}C)(80 x 10^{6}C)/[2(0.35m)]^{2} = 1.03 x 10^{2}N F_{23} = F_{32} = kQ_{2}Q_{3}/r_{23} = kQ_{2}Q_{3}/L^{2} =(9.0 x 10^{9}N.m^{2}/C^{2})(48 x 10^{6}C)(80 x 10^{6}C)/(0.35m)^{2} = 2.82 x 10^{2}N. 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 F_{1} = F_{13}  F_{12} = 1.03 x 10^{2}N  2.47 x 10^{2}N = 1.4 x10^{2}N (left) F_{2} = F_{21} + F_{23} = 2.47 x 10^{2}N + 2.82 x 10^{2}N = +5.3 x 10^{2}N (right) F_{3} = F_{31}  F_{32} = 1.03 x10^{2}N  2.82 x 10^{2}N = 3.9 x 10^{2}N (left) Note that the sum for the three charges is zero.


3. The magnitudes of the individual forces on the charges are
F_{12} = kQ2Q/l^{2 }= 2kQ^{2}/l^{2}; F_{13 }= kQ3Q/(lÖ2)^{2} = 3kQ^{2}/2l^{2}; F_{14} = kQ4Q/l^{2} = 4kQ^{2}/l^{2};
F_{23} = k2Q3Q/l^{2 }= 6kQ^{2}/l^{2}; F_{24} = k2Q4Q/(lÖ2)^{2} = 4kQ^{2}/l^{2} F_{34} = k3Q4Q/l^{2} = 12kQ^{2}/l^{2}. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For charge Q we have F_{1} = (F_{12 } F_{13}cos45°)i + (F_{13}sin45° + F_{14})jj = [ (2kQ^{2}/l^{2})  (3kQ^{2}/2l^{2})(Ö2/2)]i + [ (3kQ^{2}/2l^{2})(Ö2/2) + (4kQ^{2}/l^{2})]j = (kQ^{2}/l^{2})[2  3Ö2/4)i + (4  3Ö2/4)]j For Charge 2Q we have F_{2} = (F_{12 }+ F_{24}cos45°)i + (F_{24}sin45°  F_{23})j = [(2kQ^{2}/l^{2}) + (4kQ^{2}/l^{2})(Ö2/2)]i + [(4kQ^{2}/l^{2})(Ö2/2)  (6kQ^{2}/l^{2})]j = (kQ^{2}/l^{2})[(2 + 2Ö2)i + (6 +2Ö2)j] For Charge 2Q we have F_{3} = (F_{34} + F_{13}cos45°)i + (F_{13}sin45° + F_{23})j = [(12kQ^{2}/l^{2})  (3kQ^{2}/2l^{2})(Ö2)]i + [(3kQ^{2}/2l^{2})(Ö2/2) + (6kQ^{2}/l^{2})]j = (kQ^{2}/l^{2})[(12  3Ö2/4)i + (6+ 3Ö2/4)j]. For charge 4Q we have F_{4} = ( F_{34}  F_{24}cos45°)i + ( F_{24}sin45°  F_{14})j = [(12kQ^{2}/l^{2})  (4kQ^{2}/l^{2})(Ö2/2)]i + [ (4kQ^{2}/l^{2})(Ö2/2)  (4kQ^{2}/l^{2})]j = (kQ^{2}/l^{2})[(12  2Ö2)i + (4  2Ö2)j].

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: E_{1} = kQ_{1}/(L/Ö2)^{2} = 2kQ_{1}/L^{2} = 2(9.0 x 10^{9}N.m^{2}/C^{2})(45.0 x 10^{6}C)(0.525m)^{2} = 2.94 x10^{6}N/C. E_{2} = E_{3} = E_{4} = kQ_{2}/L/Ö2)^{2} = 2kQ_{2}/L^{2} = 2(9.0 x 10^{9}N.m^{2}/C^{2})(27.0 x 10^{6}C)/(0.525m)^{2} = 1.76 x 10^{6}N/C. From the symmetry, we see that the resultant field will be along the diagonal shown as the xaxis. For the net field, we have E = E_{1} + E_{3 } = 2.94 x 10^{6}N/C + 1.76 x 10^{6}N/C = 4.70 x 10^{6}N/C. Thus the field at the center is 4.70 x 10^{6}N/C away from the positive charge. 
5. In Example 219, 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/4pe_{o})({(x + 1/2l)^{2}/[(x + 1/2l)^{2} + R^{2}]^{3/2}} +{(x  1/2l)/[(x  1/2l)^{2} + R^{2}]^{3/2}})i 
6. (a) We find the acceleration produced by the electric field:
qE = ma
(1.60 x 10^{19}C)[(2.0 x 10^{4}N/C)i + (8.0 x 10^{4}N/C)j] = (9.11 x 10^{31}kg)a
which gives a = (3.5 x 10^{15}m/s^{2})i  (1.41 x 10^{16}m/s^{2})j
Because the field is constant, the acceleration is constant.
(b) We find the velocity from
v = v_{o + }at
= (8.0 x10m/s)i + [ (3.5 x 10^{15}m/s^{2}i  (1.41 x 10^{16}m/s^{2})j](1.0 x 10^{9}s)
= (3.43 x 10^{6}m/s)i  (1.41 x 10^{7}m/s)j.
The direction of the electron is the direction of its velocity:
tanq = v_{y}/v_{x } = (1.41 x 10^{7}m/s)/(3.43 x 10^{6}m/s) = 4.11, or q = 104°
7. (a) We find the net charge from
p = Ql;
3.4 x 10^{30}C.m = Q(1.0 x10^{10}m), which gives Q = 3.4 x 10^{20}C.
(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^{30}C.m)(2.5 x 10^{4}N/c)sin90° = 8.5 x10^{26}m.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^{30}C.m)(2.5 x 10^{4}N/C)(1  0.707) = 2.5 x 10^{26}J.
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 = [1/4pe_{0}ò _{0}^{l}dq/(x + x')^{2}]i = (Q/4pe_{0}l)[ò _{0}^{l}dx'/(x + x')^{2}]i = (Q/4pe_{0}l)[1/(x + x')]_{0}^{l}i = (Q/4pe_{0}l)[1/(x + l)  1/x]i = [Q/4pe_{0}x(x + l)]i 

11* (a) The field along the axis of the ring is
E = (Q/4pe_{0})[x/(x^{2} + R^{2})^{3/2}]i,
so the force on the charge is
F = qE = (qQ/4pe_{0})[x/(x^{2} + R^{2})^{3/2}]i
= (qQx/4pe_{0}R^{3})/[1 + (x/R)^{2})]^{3/2}i.
If x << R, we can use the approximation (1 + u)^{}^{n} = 1 nu :
F » qQx/4pe_{0}R^{3})[1 + 3/2(x/R)^{2}] » qQx/4pe_{0}R^{3}.
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/4pe_{0}R^{3},
So the period is
T = 2p(m/k)^{1/2} = 2p(4pe_{0}mR^{3}/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. 
