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.
 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].
 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/v = (-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

WDU = ( -p×E)f - (-pE)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.

 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.
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    = – F1F2/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).

 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

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.