Suppose the spheres A and B in Exercise 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
The charge on A and B,q1=q2=6.5×107C. The distance between the sphere r=0.5m. Since the spheres are same size, they will possess equal charges on being brought in contact. When uncharged sphere c(q3=0) is bought in contact with A,m charge left on A. q1′=2q1+q3=26.5×10−7+0=3.25×10−7C and charge on sphere C,q31=3.25×10−7C Therefore, when the sphere C is brought in contact with B, charge left on sphere B, q2′=2q2+q3′=26.5×10−7+3.25×10−7 =4.875×10−7C Therefore, new force of repulsion between the sphere A and B, F=4π∈01×r2q11×q21 =9×109×(0.5)23.25×10−7×4.875×10−7 =5.704×10−3N.
