The three parallel planes of charge shown in FIGURE P24.45have surface charge densities -12,h,handlocalid="1649410735638" -12,h- . Find the electric fields localid="1649410752965" Eu1to localid="1649410757308" Eu4in regions localid="1649410763257" 1to localid="1649410765846" 4.

Short Answer

Expert verified

The areas in Zones 1to localid="1649410567945" 4are irrigated.

E1=EA+EBECj^=0

E2=EA+EBECj^=η2ε0j^

E3=EAEBECj^=η2ε0j^

E4=EAEB+ECj^=0

Step by step solution

01

Step :1 Introduction 

To compute the sector, we will utilize the collocation method, which stipulates that the net horizontal component Ein a point equals the magnitude of the vector of the piezoelectric effect emitted by all diverse perspectives.

Distinct electrified orthogonal planes serve as the generators in this example. Each of these provides a horizontal to the flat usually focusing whose size is the zeta potential densities divided by 2ε0. The field points towards the plane if the plane is neutralized, and away from it if the plane is polarised.

02

Step :2 Magnitude of the field 

The amplitudes of the energies emitted by planes A,B, andlocalid="1649410440703" Care then calculated.

localid="1649410460704" EA=η4ε0

role="math" EB=η2ε0

EC=η4ε0

03

Step :3 Filed of planes 

The fields due to these planes in regions1,2,3and 4are

E1=EA+EBECj^=0

E2=EA+EBECj^=η2ε0j^

E3=EAEBECj^=η2ε0j^

E4=EAEB+ECj^=0

E3=EAEBECj^=η2ε0j^

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Most popular questions from this chapter

A sphere of radius Rhas total charge Q. The volume charge density C/m3within the sphere is

ρ=ρ01-rR

This charge density decreases linearly from \(\rho_{0}\) at the center to zero at the edge of the sphere.

a. Show that ρ0=3Q/πR3.

b. Show that the electric field inside the sphere points radially outward with magnitude

E=Qr4πϵ0R34-3rR

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?

A spark occurs at the tip of a metal needle if the electric field strength exceeds 3.0×106N/C, the field strength at which air breaks down. What is the minimum surface charge density for producing a spark?

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 dAas 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 earth has a vertical electric field at the surface, pointing down, that averages100N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?

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