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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.

E→1=−EA+EB−ECj^=0→

E→2=EA+EB−ECj^=η2ε0j^

E→3=EA−EB−ECj^=−η2ε0j^

E→4=EA−EB+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 E→in 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

E→1=−EA+EB−ECj^=0→

E→2=EA+EB−ECj^=η2ε0j^

E→3=EA−EB−ECj^=−η2ε0j^

E→4=EA−EB+ECj^=0→

E→3=EA−EB−ECj^=−η2ε0j^

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Most popular questions from this chapter

What is the net electric flux through the torus (i.e., doughnut shape) of FIGURE ?

A 3.0-cm-diameter circle lies in the xz-plane in a region where the electric field isE→=(1500i^+1500j^-1500k^)N/C. What is the electric flux through the circle?

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.

| A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball 1r … R2 is E1r2 = r4 Emax /R4 . a. What is Emax in terms of Q and R? b. Find an expression for the volume charge density r1r2 inside the ball as a function ofr.c. Verify that your charge density gives the total charge Q when integrated over the volume of the ball

A positive point chargeq sits at the center of a hollow spherical shell. The shell, with radius R and negligible thickness, has net charge -2q. Find an expression for the electric field strength (a) inside the sphere, r<K, and (b) outside the sphere, r>K. In what direction does the electric field point in each case?

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