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Chapter 24: Q. 4 (page 681)

InFIGURE Q24.4, where the field is uniform, is the magnitude of Φ1larger than, smaller than, or equal to the magnitude of Φ2? Explain.

Short Answer

Expert verified

Φ1=Φ2

Step by step solution

01

Given information and formula used  

Given :

Figure -

The field is uniform

Theory used :

Electric flux, property of an electric field that may be thought of as the number of electric lines of force (or electric field lines) that intersect a given area.

Electric field lines are considered to originate on positive electric charges and to terminate on negative charges.

02

Determining if the magnitude of Φ1 larger than, smaller than, or equal to the magnitude of Φ2

The electric field is consistent across both surfaces. The electric flow is constant in this scenario. Because the electric fields inner 'surface 1' and outward 'surface 2' are identical, the two surfaces are joined to form a single closed surface.

So,Φ1=Φ2

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Most popular questions from this chapter

What is the electric flux through the surface shown in FIGURE EX24.9?

FIGURE EX24.27 shows a hollow cavity within a neutral conductor. A point charge Qis inside the cavity. What is the net electric flux through the closed surface that surrounds the conductor?

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/ϵ1, where ϵis the permittivity of the material. (Technically, ϵ0is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density 250nC/mis covered with insulation whose permittivity is 2.5ϵ0. What is the electric field strength at a point inside the insulation that is 1.5mmfrom the axis of the wire?

The two spheres in FIGURE Q24.8 on the next page surround equal charges. Three students are discussing the situation.

Student 1: The fluxes through spheres A and B are equal because they enclose equal charges.

Student 2: But the electric field on sphere B is weaker than the electric field on sphere A. The flux depends on the electric field strength, so the flux through A is larger than the flux through B.

Student 3: I thought we learned that flux was about surface area. Sphere B is larger than sphere A, so I think the flux through B is larger than the flux through A.

Which of these students, if any, do you agree with? Explain.

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.
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