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Chapter 24: Q. 3 (page 682)

FIGURE EX24.3 shows a cross section of two infinite parallel planes of charge. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

Short Answer

Expert verified

The direction of the electric field for two infinite parallel planes of charge is discovered

Step by step solution

01

Given information

Cross section of two infinite parallel planes of charge

02

Step 2. On your paper, draw this figure, then draw electric field vectors to depict the shape of the electric field.

The cross section of two infinite parallel positive charge planes is depicted in the diagram.

Because similar charges repel each other. As a result, the electric fields of the positive charges in the two infinite planes migrate away from each other (that is, they repel each other). The electric field is in an outward direction. As a result, the electric field between the two positive charge parallel planes is zero.

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Most popular questions from this chapter

FIGURE P24.48shows two very large slabs of metal that are parallel and distance lapart. The top and bottom of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1has total charge localid="1648838411434" Q1=Qand metal 2has total charge localid="1648838418523" Q2=2Q. Assume Qis positive. In terms of Qand localid="1648838434998" A, determine a. The electric field strengths localid="1648838424778" E1to localid="1648838441501" E5in regions 1to 5. b. The surface charge densities localid="1648838447660" ηuto localid="1648838454086" ηdon the four surfaces ato d.

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 spherical shell has inner radius Rinand outer radius Rout. The shell contains total charge Q, uniformly distributed. The interior of the shell is empty of charge and matter.

a. Find the electric field strength outside the shell,r≥Rout .

b. Find the electric field strength in the interior of the shell, r≤Rin.

c. Find the electric field strength within the shell, Rin≤r≤Rout.

d. Show that your solutions match at both the inner and outer boundaries

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.

The conducting box in FIGURE EX24.26 has been given an excess negative charge. The surface density of excess electrons at the center of the top surface is 5.0×1010electrons/m2 . What are the electric field strengths E1to E3at points 1 to 3?

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