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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

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

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?

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 sphere and ellipsoid in FIGURE Q24.9 surround equal charges. Four students are discussing the situation. Student 1: The fluxes through A and B are equal because the average radius is the same. Student 2: I agree that the fluxes are equal, but that’s because they enclose equal charges. Student 3: The electric field is not perpendicular to the surface for B, and that makes the flux through B less than the flux through A. Student 4: I don’t think that Gauss’s law even applies to a situation like B, so we can’t compare the fluxes through A and B. Which of these students, if any, do you agree with? Explain
See all solutions

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