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Chapter 23: Problem 4

In Fig. 23-32, a butterfly net is in a uniform electric field of magnitude \(E=3.0 \mathrm{mN} / \mathrm{C}\). The rim, a circle of radius \(a=11 \mathrm{~cm}\), is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.

Short Answer

Expert verified
Electric flux through the net is approximately \(1.14 \times 10^{-3} \text{ Nm}^2/\text{C}\).

Step by step solution

01

Understand the Concept of Electric Flux

Electric flux \( \Phi_E \) is a measure of the electric field passing through a given area. Mathematically, it can be calculated using the formula: \( \Phi_E = E \cdot A \cdot \cos(\theta) \), where \( E \) is the electric field, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the field lines and the normal (perpendicular) to the surface.
02

Calculate the Area of the Circular Rim

The net rim is a circle with radius \( a = 11 \) cm. The area \( A \) of the circle is given by the formula \( A = \pi a^2 \). First, convert the radius to meters: \( a = 0.11 \) m. Then, calculate \( A = \pi \times (0.11)^2 \) square meters.
03

Substitute Values into the Electric Flux Formula

Since the rim is aligned perpendicular to the electric field, \( \theta = 0 \), and \( \cos(\theta) = 1 \). Substitute the values into the electric flux formula: \( \Phi_E = E \times A \times \cos(\theta) = 3.0 \times 10^{-3} \times \pi \times (0.11)^2 \times 1 \).
04

Perform the Calculation

Calculate the expression: \( \Phi_E = 3.0 \times 10^{-3} \text{ mN/C} \times \pi \times 0.0121 \text{ m}^2 \). After performing the multiplication, \( \Phi_E \approx 1.14 \times 10^{-3} \text{ Nm}^2/\text{C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
An electric field is a vector field that represents the force exerted per unit charge at a given point in space. It is denoted by the letter \( E \) and is measured in newtons per coulomb (N/C).
The direction of the electric field is defined as the direction a positive test charge would move if placed in the field. Thus, electric field lines point away from positively charged objects and toward negatively charged ones.
Consider a uniform electric field, like that in the case of the butterfly net exercise. Here, the field has the same magnitude and direction at every point in space.
  • The given electric field magnitude is \( E = 3.0 \text{ mN/C} \), which can also be expressed as \( 3.0 \times 10^{-3} \text{ N/C} \).
  • Understanding the uniformity and direction of the field is crucial when calculating the electric flux through any surface aligned with it.
Circular Area
To calculate the electric flux through a surface, knowing the area of that surface is essential. In this exercise, the surface is a circular rim of a butterfly net.
The formula for the area \( A \) of a circle is \( A = \pi a^2 \), where \( a \) is the radius of the circle. The radius given is \( a = 11 \text{ cm} \), which needs to be converted to meters for consistency with the SI unit system.
  • Convert the radius: \( a = 11 \text{ cm} = 0.11 \text{ m} \).
  • Calculate the area: \( A = \pi \times (0.11)^2 \).
  • The result is approximately \( 0.038 \text{ m}^2 \).
Knowing the area allows you to proceed with finding the electric flux, as it directly factors into the electric flux formula with the electric field magnitude and the angle of incidence.
Angle of Incidence
The angle of incidence, in the context of electric flux, involves the angle \( \theta \) between the electric field lines and the normal (perpendicular) to the surface. This angle drastically affects the electric flux calculation because it influences how much field passes through the area.
In mathematical terms, it is denoted by \( \cos(\theta) \), where \( \theta \) impacts the formula: \( \Phi_E = E \cdot A \cdot \cos(\theta) \).
  • For a surface perpendicular to the electric field, as in this exercise, \( \theta = 0 \).
  • Thus, \( \cos(0) = 1 \), simplifying the formula to \( \Phi_E = E \cdot A \).
This simplification occurs because when the surface is fully perpendicular, all the field lines pass through, maximizing the flux. Therefore, understanding the angle's role helps predict how much of the electric field contributes to the flux, crucial for various applications in physics and engineering.

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Most popular questions from this chapter

A thin-walled metal spherical shell has radius \(25.0 \mathrm{~cm}\) and charge \(2.00 \times 10^{-7} \mathrm{C}\). Find \(E\) for a point (a) inside the shell, (b) just outside it, and (c) \(3.00 \mathrm{~m}\) from the center.

Charge of uniform volume density \(\rho=3.2 \mu \mathrm{C} / \mathrm{m}^{3}\) fills a nonconducting solid sphere of radius \(5.0 \mathrm{~cm} .\) What is the magnitude of the electric field (a) \(3.5\) \(\mathrm{cm}\) and (b) \(8.0 \mathrm{~cm}\) from the sphere's center?

SSM The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of \(300 \mathrm{~m}\) the field has magnitude \(60.0 \mathrm{~N} / \mathrm{C} ;\) at an altitude of \(200 \mathrm{~m}\), the magnitude is \(100 \mathrm{~N} / \mathrm{C}\). Find the net amount of charge contained in a cube \(100 \mathrm{~m}\) on edge, with horizontal faces at altitudes of 200 and \(300 \mathrm{~m}\).

Charge \(Q\) is uniformly distributed in a sphere of radius \(R\). (a) What fraction of the charge is contained within the radius \(r=R / 2.00 ?\) (b) What is the ratio of the electric field magnitude at \(r=R / 2.00\) to that on the surface of the sphere?

. The chocolate crumb mystery. Explosions ignited by electrostatic discharges (sparks) constitute a serious danger in facilities handling grain or powder. Such an explosion occurred in chocolate crumb powder at a biscuit factory in the 1970 s. Workers usually emptied newly delivered sacks of the powder into a loading bin, from which it was blown through electrically grounded plastic pipes to a silo for storage. Somewhere along this route, two conditions for an explosion were met: (1) The magnitude of an electric field became \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) or greater, so that electrical breakdown and thus sparking could occur. (2) The energy of a spark was \(150 \mathrm{~mJ}\) or greater so that it could ignite the powder explosively. Let us check for the first condition in the powder flow through the plastic pipes. Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius \(R=5.0 \mathrm{~cm}\). Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density \(\rho\). (a) Using Gauss' law, find an expression for the magnitude of the electric field \(\vec{E}\) in the pipe as a function of radial distance \(r\) from the pipe center. (b) Does \(E\) increase or decrease with increasing \(r ?\) (c) Is \(\vec{E}\) directed radially inward or outward? (d) For \(\rho=1.1 \times 10^{-3} \mathrm{C} / \mathrm{m}^{3}\) (a typical value at the factory), find the maximum \(E\) and determine where that maximum field occurs. (e) Could sparking occur, and if so, where? (The story continues with Problem 70 in Chapter 24.)
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