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Chapter 24: Q. 18 (page 683)

FIGUREEX24.18shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) -q/系0, (b) q/系0, (c) 3q/系0, and (d) 4q/系0.

Short Answer

Expert verified

a.Diagram for closed surface with charge-qis .

b.Diagram for closed surface with chargeqis.

c.Diagram for closed surface with charge3qis.

d.Diagram for closed surface with charge4qis.

Step by step solution

01

Figure for closed surface with electric flux -qε0 (part a)

(a).

The amount of electric field that travels through a closed surface is referred to as the electric flux.

桅e=鈭�E鈫�路dA鈫�=Qin系o

Because any flux owing to charges outside the closed surface is zero, we construct a closed surface around the number of charges the flux equals to obtain the flux.

Negative charge is surrounded by a closed surfacelocalid="1649268816933" -q.

02

 Figure for closed surface with electric flux qε0(part b)

(b).

A closed surface is drawn around net charges ofq. If the two charges are combined, this might happen-qand2q.

03

 Figure for closed surface with electric flux 3qε0(part c)

(c).

A closed surface is drawn around net charges of3q.

This could occur if we combine the three charges -q,2qand2q.

04

Figure for closed surface with electric flux 4qε0 (part d)

(d).

A closed surface is drawn around net charges of 4q.

If the two charges 2qand2qare combined, this might happen.

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Most popular questions from this chapter

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 桅e=Qin/系0is 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 位. 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鈫�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" d桅through 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 d桅to find the total flux through this face.

c. Finally, show that the net flux through the cube is 桅e=Qin/系0.

Figure 24.32bshowed a conducting box inside a parallel-plate capacitor. The electric field inside the box is E鈫�=0鈫�. Suppose the surface charge on the exterior of the box could be frozen. Draw a picture of the electric field inside the box after the box, with its frozen charge, is removed from the capacitor.

Hint: Superposition.

Charges q1=-4Qandq2=+2Qare located atx=-aandx=+a, respectively. What is the net electric flux through a sphere of radius 2acentered

(a) at the origin and

(b) at x=2a?

A hollow metal sphere has6cmand 10cminner and outer radii, respectively. The surface charge density on the inside surface is -100nC/m2. The surface charge density on the exterior surface is +100nC/m2. What are the strength and direction of the electric field at points 4,8and12cm from the center?

A sphere of radius Rhas total charge Q. The volume charge Calc density role="math" localid="1648722354966" Cm3within the sphere is 蚁r=Cr2, whereC is a constant to be determined.
a. The charge within a small volume dVis dq=蚁dV. The integral of 蚁dVover the entire volume of the sphere is the total chargeQ. Use this fact to determine the constant Cin terms of QandR .
Hint: Let dVbe a spherical shell of radiusr and thicknessdr. What is the volume of such a shell?
b. Use Gauss's law to find an expression for the electric field strengthE inside the sphere, ,r鈮�R in terms of QandR.
c. Does your expression have the expected value at the surface,r=R ? Explain.
See all solutions

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