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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 electric flux is 1.75 N·m²/C. (b) No, flux does not depend on sheet shape. (c) Flux is largest at 0°, smallest at 90°.

Step by step solution

01

Understand the Concept

Electric flux is defined as the electric field multiplied by the area through which the electric field lines pass, and specifically, it is the component of the electric field perpendicular to the surface. This is expressed by the equation: \( \Phi = E \cdot A \cdot \cos(\theta) \), where \( E \) is the electric field, \( A \) is the area, and \( \theta \) is the angle between the electric field and the normal to the surface.
02

Substitute Known Values for Part (a)

We are given an electric field magnitude \( E = 14 \, \mathrm{N/C} \), an area \( A = 0.250 \, \mathrm{m^2} \), and an angle \( \theta = 60^{\circ} \). Substitute these values into the formula for electric flux: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times \cos(60^{\circ}) \].
03

Calculate the Electric Flux for Part (a)

Calculate \( \cos(60^{\circ}) \), which is 0.5. Substitute this into the equation to find the flux: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times 0.5 = 1.75 \, \mathrm{N \cdot m^2/C} \].
04

Analyze Part (b) Dependence on Shape

The answer to part (a) does not depend on the shape of the sheet. Electric flux depends only on the electric field's component perpendicular to the surface area, not on how the area is distributed.
05

Determine the Angle for Maximum Flux in Part (c)(i)

The flux is largest when \( \cos(\theta) \) is maximum, which occurs at \( \theta = 0^{\circ} \). This means the electric field is perpendicular to the sheet.
06

Determine the Angle for Minimum Flux in Part (c)(ii)

The flux is smallest when \( \cos(\theta) \) is minimum, which occurs at \( \theta = 90^{\circ} \). This means the electric field is parallel to the sheet's surface.

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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 represents the force per unit charge exerted on a charged object. It is symbolized by the letter \( E \) and measured in newtons per coulomb (\( ext{N/C} \)).
Electric fields can be visualized as lines of force extending outward in a uniform manner. These lines show how the electric field spreads out through space. The field's intensity indicates how strongly it can influence charged particles.
In situations like our exercise, where there's a flat sheet and a uniform electric field, understanding the interaction between the field and surfaces is key. The way electric field lines intersect with a surface plays a huge role in determining the electric flux.
Flux Calculation
Calculating electric flux involves assessing how many electric field lines pass through a given surface. To do this, you need to execute a simple but detailed calculation involving the surface area, electric field strength, and the angle of incidence.
Here’s the fundamental formula to use:
  • Electric flux, \( \Phi = E \cdot A \cdot \cos(\theta) \)
In this equation:
  • \( E \) is the magnitude of the electric field.
  • \( A \) is the area through which the field lines pass.
  • \( \theta \) is the angle between the electric field lines and the normal (perpendicular) to the surface.
By substituting in the known values, such as those in the exercise, you can calculate the electric flux. For instance, using \( E = 14 \, \mathrm{N/C} \), \( A = 0.250 \, \mathrm{m^2} \), and \( \theta = 60^{\circ} \) yields a flux value of 1.75 \( \mathrm{N \cdot m^2/C} \). Remember, the cosine function in the formula accounts for the direction of the field relative to the surface.
Angle Dependence of Flux
Electric flux is highly dependent on the angle \( \theta \) at which the electric field meets a surface. This relationship is described by the cosine function in the flux calculation formula.
Understanding this dependence is crucial:
  • When \( \theta = 0^{\circ} \), \( \cos(\theta) = 1 \). Here, the electric field is perfectly perpendicular to the surface, maximizing the flux.
  • As \( \theta \) increases towards \( 90^{\circ} \), \( \cos(\theta) \) approaches zero. Thus, when the field is parallel to the surface, the flux is minimized.
The varying angle directly affects how effective the electric field lines are in penetrating the surface. This concept shows why the orientation of a surface in an electric field is key to analyzing electric phenomena.
Perpendicular and Parallel Components
Components of the electric field with respect to a surface divide into perpendicular and parallel parts. These components help us understand how the electric field interacts with surfaces.
Here's how each component behaves:
  • The **perpendicular component** influences electric flux. This part of the field is effective in passing through the surface, as it aligns with the normal vector of the surface.
  • The **parallel component** runs along the surface and does not contribute to the electric flux. These field lines skim the surface instead of passing through it.
In practical terms, the electric flux is entirely due to the perpendicular component. When calculating flux, we break down the electric field into parts and focus on the perpendicular aspect to determine the interactions with any given surface. This decomposition aids in visualizing and solving complex field problems.

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Most popular questions from this chapter

The gravitational force between two point masses separated by a distance \(r\) is proportional to \(1 / r^{2},\) just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss's law for gravitation. (a) Let \(\vec{g}\) be the acceleration due to gravity caused by a point mass \(m\) at the origin, so that \(\vec{g}=-\left(G m / r^{2}\right) \hat{r}\) . Consider a spherical Gaussian surface with radius \(r\) centered on this point mass, and show that the flux of \(\vec{g}\) through this surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G m$$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show that the flux of \(\vec{g}\) through any closed surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G M_{\mathrm{encl}}$$ where \(M_{\text { encl }}\) is the total mass enclosed within the closed surface.

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length \(+5.20 \mu \mathrm{C} / \mathrm{m} .\) What magnitude of force does one line of charge exert on a \(0.0500-\mathrm{m}\) section of the other line of charge?

A hemispherical surface with radius \(r\) in a region of uniform electric field \(\vec{E}\) has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

Thomson's Model of the Atom. In the early years of the 20 th century, a leading model of the structure of the atom was that of the English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e,\) which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\) (a) Explain why the equilibrium position of the electron is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the motion of the electron. If the electron is displaced from equilibrium by a distance less than \(R,\) show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron or electrons in the atom. What would the radius of a Thomson-model atom have to be for it to produce red light of frequency \(4.57 \times 10^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of \(10^{-10} \mathrm{m}\) (see Appendix For data about the electron). (d) If the electron were displaced from equilibrium by a distance greater than \(R,\) would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (Historical note: In \(1910,\) the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume as Thomson supposed, but is concentrated in the tiny nucleus of radius \(10^{-14}\) to \(10^{-15} \mathrm{m} . )\)

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

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