Homework 1: HRW Chapter 22, Problem # 8, 9, 27, 30, 37, 52, 54, 83 8. 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 charges. The individual magnitudes | |, | |, | |,E E E1 2 3 and | |E4 are figured from Eq. 22-3, where the absolute value signs for q1, q2, and q3 are unnecessary since those charges are positive (assuming q > 0). We note that the contribution from q1 cancels that of q2 (that is, | | | |E E1 2= ), and the net field (if there is any) should be along the y axis, with magnitude equal to ( )43net22 2 20 01 1 12 3ˆ ˆj j4 4 42qqq qEd d ddπε πε  = − = −      which is seen to be zero. A rough sketch of the field lines is shown next: 9. (a) The vertical components of the individual fields (due to the two charges) cancel, by symmetry. Using d = 3.00 m and y = 4.00 m, the horizontal components (both pointing to the –x direction) add to give a magnitude of 9 2 2 19,net2 2 3/ 2 2 2 3/ 20102 | | 2(8.99 10 N m C )(3.20 10 C)(3.00 m)4 ( ) [(3.00 m) (4.00 m) ]1.38 10 N/C .xq dEd yπε−−× ⋅ ×= =+ += × . (b) The net electric field points in the –x direction, or 180° counterclockwise from the +x axis. 27. From symmetry, we see that the net field at P is twice the field caused by the upper semicircular charge )q Rλ π+ = ((and that it points downward). Adapting the steps leading to Eq. 22-21, we find ( )90net2 2900 0ˆ ˆ2 j sin j.4qER Rθε ε°− ° = − = − π λπ (a) With R = 8.50 × 10− 2 m and q = 1.50 × 10−8 C, net| | 23.8 N/C.E = (b) The net electric field netE points in the ˆj−direction, or 90− °counterclockwise from the +x axis.30. We use Eq. 22-16, with “q” denoting the charge on the larger ring: 3/ 22 2 3/ 2 2 2 3/ 20 0130 4.194 ( ) 4 [ (3 ) ] 5qz qzq Q Qz R z Rπε πε + = ⇒ = − = − + + . Note: We set z = 2R in the above calculation. 37. We use Eq. 22-26, noting that the disk in Figure 22-57(b) is effectively equivalent to the disk in Figure 22-57(a) plus a concentric smaller disk (of radius R/2) with the opposite value of σ. That is, E(b) = E(a) – σ2εo 1 − 2R(2R)2 + (R/2)2 where E(a) = σ2εo 1 − 2R(2R)2 + R2 . We find the relative difference and simplify: E(a) – E(b) E(a) = 1 2 / 4 1/ 4 1 2 / 17 / 4 0.02990.2830.10561 2 / 4 1 1 2 / 5− + −= = =− + − or approximately 28%. 52. (a) Due to the fact that the electron is negatively charged, then (as a consequence of Eq. 22-28 and Newton’s second law) the field E → pointing in the same direction as the velocity leads to deceleration. Thus, with t = 1.5 × 10− 9 s, we find 194 90 0314(1.6 10 C)(50 N/C)| | 4.0 10 m/s (1.5 10 s)9.11 10 kg2.7 10 m/s .eEv v a t v tm−−−×= − = − = × − ××= × (b) The displacement is equal to the distance since the electron does not change its direction of motion. The field is uniform, which implies the acceleration is constant. Thus, 505.0 10 m.2v vd t−+= = × 54. Due to the fact that the electron is negatively charged, then (as a consequence of Eq. 22-28 and Newton’s second law) the field E → pointing in the +y direction (which we will call “upward”) leads to a downward acceleration. This is exactly like a projectile motion problem as treated in Chapter 4 (but with g replaced with a = eE/m = 8.78 × 1011 m/s2). Thus, Eq. 4-21 gives 660 03.00 m1.96 10 scos (2.00 10 m/s)cos40.0xtvθ−= = = ×× °.This leads (using Eq. 4-23) to 6 11 2 60 05sin (2.00 10 m/s)sin40.0 (8.78 10 m/s )(1.96 10 s)4.34 10 m/s .yv v atθ−= − = × °− × ×= − × Since the x component of velocity does not change, then the final velocity is v → = (1.53 × 106 m/s) i^ − (4.34 × 105 m/s) j^ . 83. The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field. The field causes a torque that tends to align the dipole with the field. When placed in an electric field ,E the potential energy of the dipole pis given by Eq. 22-38: ( ) cosU p E pEθ θ= − ⋅ = −. The torque caused by the electric field is (see Eq. 22-34) .p Eτ= ×  (a) From Eq. 22-38 (and the facts that ɵɵii=1⋅ and ɵɵj i = 0⋅ ), the potential energy is ()()( )3026ˆ ˆ ˆ3.00i 4.00j 1.24 10 C m 4000 N C i1.49 10 J.U p E−−  = − ⋅ = − + × ⋅ ⋅  = − × (b) From Eq. 22-34 (and the facts that ɵɵii0×= and ɵɵɵj i = k× − ), the torque is ()()( )()30 26ˆ ˆ ˆ ˆ3.00i 4.00j 1.24 10 C m 4000 N C i 1.98 10 N m k.p Eτ− −  = × = + × ⋅ × = − × ⋅    (c) The work done is W U p E p p Ei f= = − ⋅ = − ⋅= + − − + × ⋅ ⋅= ×−−∆ ∆didie j e jc hb g3 00 4 00 124 10 40003 47 103026.ɵ ɵ.ɵ ɵ.ɵ.i 4.00j i 3.00j C m N C iJ. The work done by the agent is equal to the change in the potential energy of the