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Chapter 24: Q. 5 (page 682)

The electric field is constant over each face of the cube shown in FIGURE EX24.5. Does the box contain positive charge, negative charge, or no charge? Explain.

Short Answer

Expert verified

The charge of the box is postive

Step by step solution

01

Step 1. Given information

Over each face of the cube illustrated, the electric field is constant.

02

Step 2. Is there a positive charge, a negative charge, or no charge in the box?

According to Gauss's law , 鈭�E鈫�,dA鈫�=Q蔚0, where E is the electric field, A is the area, Q is the charge and 蔚0is a constant

Let apply Gauss's law,

Q20=鈭�E鈫�路dA鈫�=(-5-15-10+10+20+10)A=10A

So, the Charge Q=10A蔚0>0

The box is charged positively.

Gauss's law for electric flux is used to determine the charge of the box.

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Most popular questions from this chapter

FIGUREEX24.18shows 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) -q/系0, (b) q/系0, (c) 3q/系0, and (d) 4q/系0.

The charged balloon in FIGURE Q24.7 expands as it is blown up, increasing in size from the initial to final diameters shown. Do the electric field strengths at points 1, 2, and 3 increase, decrease, or stay the same? Explain your reasoning for each.

A sphere of radius Rhas total charge Q. The volume charge Calc density role="math" localid="1648722354966" Cm3within the sphere is 蚁r=Cr2, whereC is a constant to be determined.
a. The charge within a small volume dVis dq=蚁dV. The integral of 蚁dVover the entire volume of the sphere is the total chargeQ. Use this fact to determine the constant Cin terms of QandR .
Hint: Let dVbe a spherical shell of radiusr and thicknessdr. What is the volume of such a shell?
b. Use Gauss's law to find an expression for the electric field strengthE inside the sphere, ,r鈮�R in terms of QandR.
c. Does your expression have the expected value at the surface,r=R ? Explain.

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/系0, with Qin/系, where 系 is the permittivity of the material. (Technically,系0 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?

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.
See all solutions

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