/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A flat sheet of paper of area \(... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 1

A flat sheet of paper of area \(0.250 \mathrm{~m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude \(14 \mathrm{~N} / \mathrm{C}\). (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

Short Answer

Expert verified
a) The magnitude of the electric flux through the sheet is \(1.75 \mathrm{~N \cdot m^{2}/C}\). b) The shape of the sheet does not affect this value because the electric flux depends only on the electric field, the area, and the angle between them. c) (i) The magnitude of the flux is largest when the normal to the sheet and the electric field are aligned (\(胃 = 0\)), and (ii) smallest when they are perpendicular (\(胃 = 90^{\circ}\)).

Step by step solution

01

Calculate electric flux

First, we need to calculate the magnitude of the electric flux through the sheet of paper. According to the formula \(桅 = E \cdot A \cdot cos胃\), plug in the given values: \(桅 = 14 \mathrm{~N/C} \cdot 0.250 \mathrm{~m}^{2} \cdot cos60^{\circ} = 1.75 \mathrm{~N \cdot m^{2}/C}\).
02

Consider the impact of the sheet's shape

The shape of the sheet would not affect the calculation of the electric flux because the electric flux is the product of the electric field, the area through which the field is passing, and the cosine of the angle between them. So, as long as the area and the angle stay the same, the shape of the area does not matter.
03

Determine the angles for maximum and minimum flux

The magnitude of the flux through the sheet is largest when the normal to the sheet is aligned with the electric field, i.e., when \(胃 = 0\). This is because cos胃 is 1 at \(胃 = 0\), giving the maximum possible product in the flux formula. On the other hand, the flux is smallest when \(胃 = 90^{\circ}\) because cos胃 is 0 at \(胃 = 90^{\circ}\), making the product in the flux formula zero.

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Key Concepts

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

Uniform Electric Field
Imagine a vast space where, regardless of the position you choose, the strength and direction of the electric field remain constant - this is what we call a uniform electric field. It's like the gravity inside a room: everywhere you go, it pulls you down with the same force. Now, why is this concept so important?
In our exercise, the electric field is uniform, denoting a consistent force acting upon charges in a particular region. For a charge placed anywhere within this field, the experience would be the same; a uniform push or pull in one direction. This simplifies calculations and concepts such as the electric flux, which leads us to an understanding of how the field interacts with surfaces and objects, represented, in our case, by the sheet of paper.
Gauss's Law
Gauss's Law is a fundamental principle linking the electric field and the amount of charge within a given space. It states that the net electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space.
\[\begin{equation}\Phi = \frac{Q_{\text{enc}}}{\epsilon_0}\end{equation}\]This might sound technical, but think of it like a net catching fish in the ocean 鈥� the amount of fish (charge) inside the net (closed surface) can be determined by how tightly packed the netting is (permittivity of free space) and how widely you extend it. Gauss's Law is crucial for understanding scenarios like our exercise, as it provides a mathematical way to relate the field outside, represented by flux, to the charge within.
Cosine Angle of Incidence
The 'cosine angle of incidence' sounds like a mouthful, but it's just a fancy way of expressing how aligned a surface is relative to an electric field. The angle of incidence is the angle between the normal (an imaginary line perpendicular to the surface) and the direction of the field. The cosine of this angle scales the electric field to give us the component that is 'effective' in producing electric flux through the surface.
In other words, it鈥檚 like sunbathing at different times of the day. When the sun is directly overhead, you get the full warmth (flux is maximized), which is akin to an angle of incidence of 0 degrees and thus a cosine of 1. As the sun sets, the warmth lessens (flux decreases) because it hits you at an angle. This reflects the idea that when the electric field is parallel to the normal, the flux is greatest, and when it is perpendicular, the flux is zero, as showcased in our exercise.

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Most popular questions from this chapter

A solid conducting sphere carrying charge \(q\) has radius \(a\). It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c .\) The hollow sphere has no net charge. (a) Derive expressions for the electric-field magnitude in terms of the distance \(r\) from the center for the regions \(rc .\) (b) Graph the magnitude of the electric field as a function of \(r\) from \(r=0\) to \(r=2 c\). (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius \(2 c\).

(a) A conducting sphere has charge \(Q\) and radius \(R .\) If the electric field of the sphere at a distance \(r=2 R\) from the center of the sphere is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field of the sphere at \(r=4 R ?\) (b) A very long conducting cylinder of radius \(R\) has charge per unit length \(\lambda\). Let \(r\) be the perpendicular distance from the axis of the cylinder. If the electric field of the cylinder at \(r=2 R\) is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field at \(r=4 R ?\) (c) A very large uniform sheet of charge has surface charge density \(\sigma .\) If the electric field of the sheet has a value of \(1400 \mathrm{~N} / \mathrm{C}\) at a perpendicular distance \(d\) from the sheet, what is the electric field of the sheet at a distance of \(2 d\) from the sheet?

An electron is released from rest at a distance of \(0.300 \mathrm{~m}\) from a large insulating sheet of charge that has uniform surface charge density \(+2.90 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}\). (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point \(0.050 \mathrm{~m}\) from the sheet? (b) What is the speed of the electron when it is \(0.050 \mathrm{~m}\) from the sheet?

Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field \(\vec{E}\) of magnitude \(1 \times 10^{6} \mathrm{~N} / \mathrm{C}\) at \(25 \mathrm{~m}\) from the center of the sphere. What net charge on each sphere is needed? (a) \(-0.07 \mathrm{C} ;\) (b) \(-8 \mathrm{mC}\) (c) \(-80 \mu \mathrm{C}\) (d) \(-1 \times 10^{-20} \mathrm{C}\)

A cube has sides of length \(L=0.300 \mathrm{~m}\). One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by \(\overrightarrow{\boldsymbol{E}}=(-5.00 \mathrm{~N} / \mathrm{C} \cdot \mathrm{m}) x \hat{\imath}+(3.00 \mathrm{~N} / \mathrm{C} \cdot \mathrm{m}) z \hat{\boldsymbol{k}} .\) (a) Find the electric flux through each of the six cube faces \(S_{1}, S_{2}, S_{3}, S_{4}, S_{5},\) and \(S_{6}\). (b) Find the total electric charge inside the cube.
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