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Chapter 24: Q. 4 (page 682)

The electric field is constant over each face of the cube shown in FIGURE EX24.4. Does the box contain positive charge, negative charge, or no charge? Explain.

Short Answer

Expert verified

The box has a negative charge.

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 $鈭� \overset{鈫�}{E} 路 \overset{鈫�}{d A} = \frac{Q}{蔚_{0}}$ ,

Here E is the electric field, A is the area, Q is the charge and $蔚_{0}$ is a constant

Now apply Gauss's law, to get

$\begin{array}{rcl} \frac{Q}{蔚_{0}} & = & 鈭� \overset{鈫�}{E} 路 \overset{鈫�}{d A} \\ = & (- 15 - 20 - 15 + 20 + 15 + 10) A \\ = & - 5 A \end{array}$

So, let have charge $Q = 5 A 蔚_{0} < 0$

$鈬�$ The box has negative charge

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

$55.3$ million excess electrons are inside a closed surface. What is the net electric flux through the surface?

The electric flux through the surface shown in FIGURE EX24.11 is $25 N m^{2} / C$ . What is the electric field strength?

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, $Q_{in} / 系_{0,}$ with $Q_{in} / 系_{1}$ , where $系$ is the permittivity of the material. (Technically, $系_{0}$ is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density $250 nC / m$ is covered with insulation whose permittivity is $2.5 系_{0}$ . What is the electric field strength at a point inside the insulation that is $1.5$ $mm$ from the axis of the wire?

A sphere of radius $R$ has total charge $Q$ . The volume charge Calc density role="math" localid="1648722354966" $(\frac{C}{m^{3}})$ within the sphere is $蚁 (r) = \frac{C}{r^{2}}$ , where $C$ is a constant to be determined.
a. The charge within a small volume $d V$ is $d q = 蚁 d V$ . The integral of $蚁 d V$ over the entire volume of the sphere is the total charge $Q$ . Use this fact to determine the constant $C$ in terms of $Q$ and $R$ .
Hint: Let $d V$ be a spherical shell of radius $r$ and thickness $d r$ . What is the volume of such a shell?
b. Use Gauss's law to find an expression for the electric field strength $E$ inside the sphere, , $r 鈮� R$ in terms of $Q$ and $R$ .
c. Does your expression have the expected value at the surface, $r = R$ ? Explain.

A sphere of radius $R$ has total charge $Q$ . The volume charge density $(C / m^{3})$ within the sphere is

$蚁 = 蚁_{0} (1 - \frac{r}{R})$

This charge density decreases linearly from \(\rho_{0}\) at the center to zero at the edge of the sphere.

a. Show that $蚁_{0} = 3 Q / 蟺 R^{3}$ .

b. Show that the electric field inside the sphere points radially outward with magnitude

$E = \frac{Q r}{4 蟺系_{0} R^{3}} (4 - 3 \frac{r}{R})$

See all solutions

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