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Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?

Short Answer

Expert verified
Answer: The ratio of the magnitude of the force on either sphere after they are touched to that before they were touched is -0.8.

Step by step solution

01

Calculate the initial force between the spheres before they touched

As given, the charges on the spheres are: \(Q_1 = +5.0 \mu \mathrm{C}\) and \(Q_2 = -1.0 \mu \mathrm{C}\). The magnitude of the force between the two spheres can be calculated using Coulomb's Law: \(F_1 = k \frac{Q_1 Q_2}{L^2}\), where \(k = 8.99 \times 10^9 \ Nm^2/C^2\) is Coulomb's constant.
02

Calculate the charges on the spheres after they touched

After touching, the total charge on the two spheres will be equal to the initial sum of charges (\(Q_{total} = Q_1 + Q_2\)). Since they are conducting spheres, the charges will be equally shared between them when the spheres are touched. Therefore, the charges on the spheres after touching will be: \(Q' = \frac{Q_{total}}{2} = \frac{(+5.0 - 1.0) \mu\mathrm{C}}{2} = +2.0 \mu\mathrm{C}\)
03

Calculate the force between the spheres after they touched and moved back to distance L

Now that the spheres have been moved back to their original distance L, we can calculate the force between them using Coulomb's Law again: \(F_2 = k\frac{Q' Q'}{L^2}\)
04

Calculate the ratio of the force magnitudes

Finally, we find the ratio between the magnitude of the force on either sphere after they are touched to that before they were touched: \(\frac{F_2}{F_1} = \frac{k \frac{Q' Q'}{L^2}}{k \frac{Q_1 Q_2}{L^2}} = \frac{Q' Q'}{Q_1 Q_2} = \frac{(2.0 \mu \mathrm{C})^2}{(5.0 \mu \mathrm{C})(-1.0 \mu \mathrm{C})} = \frac{4.0}{-5.0} = -0.8\) Therefore, the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched is -0.8. The negative sign indicates that the force between the spheres changed direction after they touched.

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Most popular questions from this chapter

Two small metal spheres are \(25.0 \mathrm{cm}\) apart. The spheres have equal amounts of negative charge and repel each other with a force of \(0.036 \mathrm{N}\). What is the charge on each sphere?

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