Question

Q1 Q2 Q3 d1 d2 Q1 a2 Q3 11 12 Question 5 (20 points) In the diagram above, the charge Q1 and Q3 are fixed in place. The charge Q2 is free to move and is initially at rest in the position depicted in the diagram (a). Upon release, the charge Q2 will move toward Q3, as shown in the diagram (b). The charges Q1, Q2, and Q3, are 3nc, Inc and 2nc, respectively. The initial distance between the charge (diagram (a)) are d1 = 1m and d2 - 3m. Given that r1 - 2.5m, and the mass of Q2 is 10 10kg, find the velocity of Q2 in the diagram (b). Hint: d1 + d2 - 11+12 and the mechanical energy is constant.

Answer 1

SI initially 92 o * + 14 di dz - initial potential energy of system 91837 I 328 YITO h3 uitho R † ㅗ 8, Q₂ To r2 Si= charge of a = 3nc B2 = charge of 92= 11 83 z charge of 9 3= distance sin 193=dirdz=1+3 = 4m 122 = distance bin 243= odz= 3m R2² distama bin 142 = di = lm 2 2nc. · 9x109 aho 9 initial potential eversy 1 "ITE -9 ( 3X10 x 10 im 9 10 X 2X10 3m + 3x10'x xx10? 42 41% [ 3+3+3)xio's 2 3 9X10 (3+1)x1D" kxid C) xide 2 46.5X TO J

0 As all particles are at rest initially initially kineti energy finally (6) 14 " 7:2.5m 1 ft2 = di Edz = n2 di td-r, =1t3-2-521-5m. final potential energy of system = 1 oleh UITO {878 18 & zgisi 72 El 22 8, = 3nc ₂2 inc 83 222 42 = 2.5m riz= um 2 = 1.5m 2 final potential eversy ㅗ sxiong XID ។ 01X2X 0178 чп%C + 2.5 + LXD X10 115 4 | -18 + YITO (of to 3+ * xio axið Gf 3x10 9x109( +.03) X10 -18 -18 As no, 49, are + 2!*&+9€ 2 sur twytlum habean munt rulf (futtatt (orthu Trust at rest ":V;=0) N

enersy initial total energy = initial potential energy & Initial kineter 46.5x1dJ+o = 46-5x10'J eversy final total enersy Anal potential everary & final kinetic 36-3x105 + 1 m v? (m2 = 10 kg) of everry initial total enerary final to fleversy from law of conservation 2) 46.5X109 2 36-3x10 + 1 mzu² 10.2xjo? m_v² 2 - 10 10 - 20.4x10 2. = 20.4x10 V = √204 = 14.28ml quarhiza 2