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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 , ∮E→,dA→=Qε0, where E is the electric field, A is the area, Q is the charge and ε0is a constant

Let apply Gauss's law,

Q20=∮E→·dA→=(-5-15-10+10+20+10)A=10A

So, the Charge Q=10Aε0>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

The square and circle in FIGURE Q24.3 are in the same uniform field. The diameter of the circle equals the edge length of the square. Is Φsquarelarger than, smaller than, or equal to Φcircle? Explain.

II An infinite slab of charge of thickness 2z0lies in the XYplane between z=−z0andz=+z0. The volume charge density ρC/m3is a constant.

a. Use Gauss's law to find an expression for the electric field strength inside the slab −z0≤z≤z0.

b. Find an expression for the electric field strength above the slab z≥z0.

c. Draw a graph of Efrom z=0toz=3z0.

What is the net electric flux through the cylinder of FIGURE?

FIGURE EX24.3 shows a cross section of two infinite parallel planes of charge. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

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
See all solutions

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