Question

Chapter 21, Problem 008 Your answer is partially correct. Try again. In the figure three identical conducting spheres initially have the following charges: sphere A, 30; sphere B, -80; and sphere C, O. Spheres A and B are fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere C is touched to sphere A and then separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere C is touched to sphere B and then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1? Number 10,197 unit: 2 No units : the tolerance is +/-2% SHOW HINT

Answer 1

When two charged sphere (charge Q. , Q and radio 81, 82) touch, charge transfer occurs until their surface potential become equal Initially, Surface potentials via ka, V₂ =kQ₂ here R= coulomb's constant - finally, Surface potentials v = RQ! 82 31 32 : v=vý → RQ!=kQs' => Q = 21 If Spheres are identical (8,283), then TQi -Qu'll As per Principle of conservation of charge Q, +Q, = Qi +Q Let Qi=Q; = Q Qu+Q2 = Qi'+Q* = Q'+&! To=Q, +Q2 |

Experiment lia Contact of a anda final charge on sphere A , QA = QA +Qc - 3Q+031 =15Q final charge onspherec Q = Q = 30= 1.5Q Contact of B and C final charge on sphere B, Qs = Qs + Qć = -8Q+ 3 f Qe' = - 3.25Q force between A and B F = kQA Qe' k (1.5Q)(-3.250) d2 d2 Experiment 2: contact of B and c: final charge on B, Qs = Q3 + Qc = -80 to Qe = -4Q final charge onc, Qc = Qs = -4Q Contact of A and c final charge on A, QA = Qc + QA = - 4Q + 30 QA=-0.5Q

Force between A and B so Fz = RQA Q á =R (-0.5Q) (-4Q) d2 se negde acopection d² Thus (-3-25Q This = 2605QX***@) /(52)-3-250) d2 d2

(유)(-) - (-A-) (주) - ( ( (를) -) 주 - (75-)(5) . 나 (-) (50)