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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

FIGURE EX24.2 shows a cross section of two concentric spheres. The inner sphere has a negative charge. The outer sphere has a positive charge larger in magnitude than the charge on the inner sphere. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

FIGUREP24.38shows a solid metal sphere at the center of a hollow metal sphere. What is the total charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow sphere, and (c) the exterior surface of the hollow sphere?

FIGURE EX24.17shows 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) 2q/ϵ0, (b) q/ϵ0, (c) 0,and (d) 5q/ϵ0.

FIGURE shows three Gaussian surfaces and the electric flux through each. What are the three charges q1,q2andq3?

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ϵ0, with Qin/ϵ, where ϵ is the permittivity of the material. (Technically,ϵ0 is called the vacuum permittivity.) Suppose that a 50nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is2500N/C . What is the permittivity of rubber?

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