Methods of charging objects Friction – rub an insulator to add or remove electrons Conduction – touch a charged object to an uncharged object Induction – bring a charged object near (but not touching) an uncharged object, ground the uncharged object, remove the ground and then remove the charged object ( acquires the same charge as the charging object ) ( acquires the opposite charge as the charging object )

F = 997 = p ∴ e = 803 2.0 nC A B (Q)2 QAQB k 4 x 10-5 = 9 x 109 1. -0.5 C 1 electron 3.125 x 1018 electrons x = -1.6 x 10-19 C 2. 1800 = e + p and 31 x 10-18 = e(-1.6 x 10-19) + p(+1.6 x 10-19) 1800 - p = e 31 x 10-18 = (1800 – p)(-1.6 x 10-19) + p(+1.6 x 10-19) 31 x 10-18 = -2.88 x 10-16 + 1.6 x 10-19p +1.6 x 10-19p 3.19 x 10-16 = 3.2 x 10-19p 997 = p ∴ e = 803 3. 2.0 nC A B (Q)2 QAQB F = k 4 x 10-5 = 9 x 109 (0.03)2 d2

F = A B QAQB (1.5 x 10-6) (0.5 x 10-6) k = 9 x 109 = d (1.3)2 4. A B QAQB (1.5 x 10-6) (0.5 x 10-6) F = k = 9 x 109 = d 2 (1.3)2 3.99 x 10-3 N attractive Thus ball B is pulled up with a force of 3.99 x 10-3 N So mass = 4.1 x 10-4 kg

F13 = F23 = Q1Q3 (3 x 10-6) (1 x 10-6) k = 9 x 109 = 8 N (repulsion) d 5. q3 +1μC .058 m q3 .03 m θ = 31° 20 N q1 q2 .05 m - 2μC +3μC Q1Q3 (3 x 10-6) (1 x 10-6) F13 = k = 9 x 109 = 8 N (repulsion) d 2 (.058)2 Q2Q3 (2 x 10-6) (1 x 10-6) F23 = k = 9 x 109 20 N (attraction) = d 2 (.03)2 X Y - 20 +6.9 +4.1 Add the forces 8 N X = 0 Y = - 20 X = + 6.9 Y = + 4.1 20 N 31° + 6.9 - 15.9

17.3 N @ 66.2° south of east 6.9 15.9