Chapter 3: Problem 1

Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin, \(Q_{1}\), and with hemispheres of radius a. A second charge \(Q_{2}\) (this time a point charge) is located at distance \(R\) from \(Q_{1}\), where \(R>>a .(a)\) What is the force on the point charge before the hemispheres are assembled around \(Q_{1} ?\) (b) What is the force on the point charge after the hemispheres are assembled but before they are discharged? ( \(c\) ) What is the force on the point charge after the hemispheres are assembled and after they are discharged? ( \(d\) ) Qualitatively, describe what happens as \(Q_{2}\) is moved toward the sphere assembly to the extent that the condition \(R>>a\) is no longer valid.

Short Answer

Expert verified

Answer: The force on Q2 remains constant and can be calculated using Coulomb's force formula (F = k(Q1Q2)/R^2) in the scenarios before the assembly, after the assembly, and after the discharging of the hemispheres. However, when Q2 moves closer to the sphere assembly, the net force on Q2 changes due to the influence of the charged hemispheres and its position relative to the assembly.

Step by step solution

The force on the point charge Q2 can be calculated using Coulomb's force formula: \(F = k\frac{Q_1Q_2}{R^2}\). Since the conditions given in the problem are \(Q_1\), \(Q_2\), and \(R\), we can use the given values to compute the force on Q2 before the hemispheres are assembled. #b) After the hemispheres are assembled but before they are discharged:

Since the hemispheres do not have charges on them before they are discharged, and they are assembled symmetrically around Q1, the force on Q2 remains unchanged. This is because the electric field created by Q1 on Q2 would not be altered by the presence of the hemispheres. Hence, the force on Q2 in this case is the same as in part (a): \(F = k\frac{Q_1Q_2}{R^2}\). #c) After the hemispheres are discharged:

When the hemispheres are discharged, they obtain charges. However, since they are symmetrically placed around Q1 and at a greater distance than Q2, the electric field produced by these charges would cancel each other out. Thus, even after discharging the hemispheres, the force on Q2 remains the same: \(F = k\frac{Q_1Q_2}{R^2}\). #d) Describing what happens as Q2 moves closer to the sphere assembly:

When Q2 moves closer to the sphere assembly, and the condition \(R >> a\) is no longer valid, the impact of the charged hemispheres on Q2 becomes significant. As Q2 approaches the sphere assembly, the force between Q2 and each charged hemisphere would not cancel each other out anymore. This will result in a different net force acting on Q2. Additionally, the shape of the electric field lines near the sphere assembly will change due to their influence, causing a change in the force experienced by Q2. Overall, the force on Q2 will depend on its position relative to the sphere assembly and the charges on the hemispheres.