What Is the Ratio of Electrostatic Forces After Introducing a Third Sphere?

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When a third identical sphere is introduced and touches the first sphere, it takes half of sphere 1's charge, leaving it with half its original charge. Upon touching sphere 2, the third sphere takes a quarter of the charge from sphere 2, resulting in sphere 2 having three-quarters of its original charge. According to Coulomb's law, the forces between the spheres should be equal in magnitude but opposite in direction, leading to the conclusion that the ratio of forces F' to F would be -1. The calculations appear to be correct, confirming that the forces are indeed equal and opposite. The discussion emphasizes the application of Coulomb's law in determining electrostatic forces between charged spheres.
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Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters. The electrostatic force acting on sphere 2 due to sphere 1 is F and the force acting on sphere 1 due to sphere 2 is -F'. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1, then to sphere 2, and finally removed. What is the ratio F'/F?

So when the thrid sphere touches the first sphere it takes with it half the total charge of sphere 1. Then touching it to sphere two it takes only 1/4 of the charge.

So now sphere 1 has 1/2 the original charge and sphere 2 has 3/4 of the original charge. According to Coulomb's law the force of sphere 1 on sphere 2 is going to be k|q_1q_2|/r^2.

So wouldn't the force of sphere 1 on sphere 2 be the same as the force of sphere 2 on sphere 1? And if that is the case, then the ratio would be -1, right?
 
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