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If the flux of the gravitational field through a closed surface is zero, what can you conclude about the region interior to the surface?

Short Answer

Expert verified
The region interior to the surface contains no materials (objects or masses) that produce a gravitational field.

Step by step solution

01

Understanding Gauss' law for gravity

The Gauss' law for gravitation states that the total of the gravitational field leaving a volume is proportional to the total mass of matter within the volume. For a closed surface, the surface integral for the gravitational field equals the enclosed mass (if any) times the gravitational constant \( G \). Hence, if the flux through the surface is 0, there must be no enclosed mass.
02

Conclusion

From the provided details, with the flux through a closed surface being 0, it can be deducted that there are no massive objects within the surface boundary. Hence, the gravitational field within this volume is zero.

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Most popular questions from this chapter

A coaxial cable carries equal but opposite charges on its two conductors. In electrostatic equilibrium, charge on the shield a. lies entirely on its outer surface. b. is divided evenly between inner and outer surfaces. c. lies entirely on its inner surface. d. distributes itself differently depending on the magnitude of the charge.

A coaxial cable in electrostatic equilibrium carries charge \(-Q\) on its inner conductor and \(+Q\) on its shield. If the charge on the shield only is doubled, a. the magnitude of the electric field between the conductors will double. b. the magnitude of the electric field outside the shield will double. c. the magnitude of the electric field at the outer surface of the shield will become twice the magnitude of the field at the shield's inner surface. d. the magnitude of the electric field at the outer surface of the shield will equal the magnitude of the field at the shield's inner surface.

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A 250-nC point charge is placed at the center of an uncharged spherical conducting shell \(20 \mathrm{cm}\) in radius. Find (a) the surface charge density on the outer surface of the shell and (b) the electric field strength at the shell's outer surface.

A spherical shell of radius \(15 \mathrm{cm}\) carries \(4.8 \mu \mathrm{C}\) distributed uniformly over its surface. At the center of the shell is a point charge. If the electric field at the sphere's surface is \(750 \mathrm{kN} / \mathrm{C}\) and points outward, what are (a) the point charge and (b) the field just inside the shell?

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