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Chapter 24: Q. 27 (page 684)

FIGURE EX24.27 shows a hollow cavity within a neutral conductor. A point charge Qis inside the cavity. What is the net electric flux through the closed surface that surrounds the conductor?

Short Answer

Expert verified

The net electric flux through the closed surface isΦe=Qε0

Step by step solution

01

Given information and Theory used 

Given figure :

Theory used :

The amount of electric field that travels through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is provided by equation :

Φe=∮E→·dA→=Qinε0 (1)

02

Calculating the net electric flux through the closed surface 

The cavity's positive charge Qinduces a negative charge-Qon the conductor's inner side.

As seen in the diagram below, the induced negative charge produces a positive charge Qat the conductor's surface.

Qin=Q+(-Q)+Q=Qis the total enclosed charge through the closed surface that surrounds the conductor.

To calculate the electric flux, enter this number into equation (1).

Φe=Qinε0=Qε0

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Most popular questions from this chapter

What is the net electric flux through the two cylinders shown inFIGURE EX24.16? Give your answer in terms of RandE

A proton is fired from far away toward the nucleus of an iron atom. Iron is element number 26, and the diameter of the nucleus is 9.0 fm. What initial speed does the proton need to just reach the surface of the nucleus? Assume the nucleus remains at rest.

Newton’s law of gravity and Coulomb’s law are both inversesquare laws. Consequently, there should be a “Gauss’s law for gravity.” a. The electric field was defined as E u = F u on q /q, and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g u of a point mass?

Write your answer using the unit vector nr, but be careful with signs; the gravitational force between two “like masses” is attractive, not repulsive. b. What is Gauss’s law for gravity, the gravitational equivalent of Equation 24.18? Use ΦG for the gravitational flux, g u for the gravitational field, and Min for the enclosed mass. c. A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as r = r011 - r/2R2, where r0 is the density at the center. Determine r0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses. d. Find an expression for the gravitational field strength inside the planet at distance r 6 R.

FIGUREP24.38shows a solid metal sphere at the center of a hollow metal sphere. What is the total charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow sphere, and (c) the exterior surface of the hollow sphere?

All examples of Gauss's law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we've claimed that the net Φe=Qin/ϵ0is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length Lcentered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral.

a. Consider the face parallel to the yz-plane. Define area dA→as a strip of width dyand height Lwith the vector pointing in the x-direction. One such strip is located at position localid="1648838849592" y. Use the known electric field of a wire to calculate the electric flux localid="1648838912266" dΦthrough this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

b. Now integrate dΦto find the total flux through this face.

c. Finally, show that the net flux through the cube is Φe=Qin/ϵ0.
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