10CHAPTER 15 (b) When the spheres are connected by a conducting wire, the net charge will divide equally between the two identical spheres. Thus, the force is now 9126.010Cnetqqq−=+= −×()()()229292226.010C2N m8.9910 C40.30 menetkqFr−−×⋅==×or 79.010N (repulsion)F−=×15.10The forces are as shown in the sketch at the right. ()()()6629121222-2126.0010C1.5010CN m8.9910 89.9 NC3.0010mek q qFr−−××⋅==×=×()()()6621392222-2136.0010C2.0010CN m8.9910 43.2C5.0010mek q qFr−−××⋅==×=×N
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