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Chapter 16: Problem 83

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

A flat conducting sheet of area \(A\) has a charge \(q\) on each surface. (a) What is the electric field inside the sheet? (b) Use Gauss's law to show that the electric field just outside the sheet is \(E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} .\) (c) Does this contradict the result of Problem \(69 ?\) Compare the field line diagrams for the two situations.
Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.
What are the magnitude and direction of the electric field midway between two point charges, \(-15 \mu \mathrm{C}\) and \(+12 \mu \mathrm{C},\) that are $8.0 \mathrm{cm}$ apart?
A small sphere with a charge of \(-0.60 \mu \mathrm{C}\) is placed in a uniform electric field of magnitude \(1.2 \times 10^{6} \mathrm{N} / \mathrm{C}\) pointing to the west. What is the magnitude and direction of the force on the sphere due to the electric field?
A raindrop inside a thundercloud has charge - \(8 e\). What is the electric force on the raindrop if the electric field at its location (due to other charges in the cloud) has magnitude $2.0 \times 10^{6} \mathrm{N} / \mathrm{C}$ and is directed upward?
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