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Chapter 22: Problem 1

A flat sheet of paper of area 0.250 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 N/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) Flux is 1.75 Nm虏/C. (b) No, it doesn't depend on shape. (c) Max when \(\theta = 0^\circ\); min when \(\theta = 90^\circ\).

Step by step solution

01

Write Down Known Values and Formulas

The area of the sheet is given as \(A = 0.250\, \text{m}^2\), the electric field magnitude is \(E = 14\, \text{N/C}\), and the angle between the normal to the sheet and the electric field is \(\theta = 60^\circ\). We can use the formula for electric flux, given by \(\Phi_E = EA \cos\theta\), where \(\Phi_E\) is the electric flux.
02

Calculate the Electric Flux

Substitute the known values into the formula. \[\Phi_E = (14\, \text{N/C}) \times (0.250\, \text{m}^2) \times \cos(60^\circ)\]Since \(\cos(60^\circ) = 0.5\), we have:\[\Phi_E = 14 \times 0.250 \times 0.5 = 1.75\, \text{Nm}^2/\text{C}\]Thus, the magnitude of the electric flux through the sheet is \(1.75\, \text{Nm}^2/\text{C}\).
03

Determine Dependency on Shape

The magnitude of the electric flux depends on the orientation of the surface relative to the electric field, not on the shape of the sheet. Since we use a uniform electric field and account for its interaction with the area, any shape with the same area and orientation would result in the same electric flux.
04

Determine Angles for Maximum and Minimum Flux

(i) The electric flux is largest when \(\cos\theta = 1\), i.e., \(\theta = 0^\circ\). This means the electric field is perpendicular to the surface.(ii) The electric flux is smallest when \(\cos\theta = 0\), i.e., \(\theta = 90^\circ\). In this case, the electric field is parallel to the 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 is a region around a charged object where other charges experience a force. Imagine it like an invisible force field created by electric charges. Electric fields are represented by vectors, meaning they have both a direction and a magnitude. The direction of the field is the direction a positive charge would move if placed in the field. The strength or magnitude of the electric field is measured in newtons per coulomb (N/C).
Key points about electric fields:
  • The electric field is stronger where the lines are closer together, and weaker where they are further apart.
  • The field lines never intersect.
  • A uniform electric field has parallel lines that are evenly spaced.
Understanding electric fields is crucial when calculating electric flux, as they determine the force experienced by surfaces placed in the field.
Orientation of Surfaces
The orientation of a surface in an electric field can significantly affect the amount of electric flux passing through it. Imagine a flat sheet of paper inside a steady breeze (the electric field). The angle at which you hold the paper against the breeze determines how much wind hits it, similar to how a surface's angle affects electric flux.
Orientation involves the angle between the surface's normal (a line perpendicular to the surface) and the electric field lines. This angle is crucial in calculating electric flux:
  • A small angle means the field lines hit the surface directly, increasing the flux.
  • A larger angle reduces the effective area exposed to the field, thus reducing the flux.
Understanding the orientation helps comprehend how surfaces interact differently with electric fields under various angles.
Flux Calculation
Electric flux quantifies how much of an electric field passes through a given surface. It is denoted by \( \Phi_E \), and its calculation is akin to measuring how much wind flows through a window. The formula for electric flux through a flat surface is given by \( \Phi_E = EA \cos\theta \), where:
  • \( E \) is the electric field strength
  • \( A \) is the area of the surface
  • \( \theta \) is the angle between the surface normal and the field direction
To calculate \( \Phi_E \), use the cosine of the angle to adjust for the effective area exposed to the field. A direct hit (\
Perpendicular and Parallel Alignment
When discussing electric flux, the terms perpendicular and parallel alignments are key concepts. These refer to the orientation of the surface relative to the electric field, which dramatically affects the resulting flux.- **Perpendicular Alignment**: When the surface is oriented so that the electric field lines hit it at a right angle (90 degrees to the surface's normal), this results in the maximum flux. This is because all field lines are passing directly through the surface, which is observed when \( \theta = 0^\circ \). Mathematically, this is when \( \cos\theta = 1 \), maximizing the flux equation.- **Parallel Alignment**: Here, the surface is in line with the electric field (\

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Most popular questions from this chapter

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) A small sphere of mass \(m =\) 8.00 \(\times\) 10\(^{-6}\) kg and charge \(q\) is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho\)(\(r\)) given as follows: $$\rho(r) = \rho_0 \bigg(1 - \frac{r}{R}\bigg) \space \space \space \mathrm{for} \space r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \mathrm{for} \space r \leq R$$ where \(\rho_0 = 3Q/{\pi}R^3\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q\). (b) Show that the electric field in the region \(r \geq R\) is identical to that produced by a point charge \(Q\) at \(r =\) 0. (c) Obtain an expression for the electric field in the region \(r \leq R\). (d) Graph the electric-field magnitude \(E\) as a function of \(r\). (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$\rho(r) = 3ar/2R \space \space \space \space \space \mathrm{for} \space r \leq R/2$$ $$\rho(r) = \alpha[1-(r/R)^2] \space \space \space \space \mathrm{for} \space R/2 \leq r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \mathrm{for} \space r \geq R$$ Here \(\alpha\) is a positive constant having units of C/m\(^3\). (a) Determine \(\alpha\) in terms of \(Q\) and \(R\). (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r\). Do this separately for all three regions. Express your answers in terms of \(Q\). (c) What fraction of the total charge is contained within the region \(R/2 \leq r \leq R\)? (d) What is the magnitude of \(\overrightarrow{E}\) at \(r = R/2\)? (e) If an electron with charge \(q' = -e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

Early in the 20th century, a leading model of the structure of the atom was that of 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 electron's equilibrium position 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 electron's motion. 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 SHM 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(s) in the atom. What radius would a Thomson-model atom need 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\(^{-10}\) m (see Appendix F). (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\(^{-15}\) m.)

In a region of space there is an electric field \(\overrightarrow{E}\) that is in the z-direction and that has magnitude \(E =\) [964 N/(C \(\cdot\) m)]\(x\). Find the flux for this field through a square in the \(xy\)-plane at \(z =\) 0 and with side length 0.350 m. One side of the square is along the \(+x\)-axis and another side is along the \(+y\)-axis.
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