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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 is桅e=Q蔚0

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 :

桅e=鈭�E鈫�路dA鈫�=Qin蔚0 (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).

桅e=Qin蔚0=Q蔚0

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Most popular questions from this chapter

FIGURE Q24.2 shows cross sections of three-dimensional closed surfaces. They have a flat top and bottom surface above and below the plane of the page. However, the electric field is everywhere parallel to the page, so there is no flux through the top or bottom surface. The electric field is uniform over each face of the surface. For each, does the surface enclose a net positive charge, a net negative charge, or no net charge? Explain.

FIGURE EX24.1 shows two cross sections of two infinitely long coaxial cylinders. The inner cylinder has a positive charge, the outer cylinder has an equal negative charge. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

A long cylinder with radius band volume charge density 蚁has a spherical hole with radius a<b centered on the axis of the cylinder. What is the electric field strength inside the hole at radial distance r<a in a plane that is perpendicular to the cylinder through the center of the hole?

An infinite cylinder of radius Rhas a linear charge density 位. The volume charge density C/m3within the cylinder (r鈮�R)is 蚁(r)=r蚁0/R, where 蚁0is a constant to be determined.

a. Draw a graph of 蚁versus localid="1648911863544" xfor an x-axis that crosses the cylinder perpendicular to the cylinder axis. Let xrange from 鈭�2Rto 2R.

b. The charge within a small volume dVis dq=蚁dV. The integral of 蚁dVover a cylinder of length localid="1648848405768" Lis the total charge Q=位Lwithin the cylinder. Use this fact to show that 蚁0=3位/2蟺R2.

Hint: Let dVbe a cylindrical shell of length L, radius r, and thickness dr. What is the volume of such a shell?

c. Use Gauss's law to find an expression for the electric field strength Einside the cylinder, localid="1648889098349" r鈮�R, in terms of 位and R.

d. Does your expression have the expected value at the surface, localid="1648889146353" r=R? Explain.

aA uniformly charged ball of radiusaand charge -Qis at the center of a hollow metal shell with inner radius band outer radius c. The hollow sphere has net charge+2Q. Determine the electric field strength in the four regionsr鈮�a,a<r<b,b鈮�r鈮�c,andr>c.
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