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Chapter 22: Problem 1

22.1. 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) 1.75 Nm虏/C. (b) No, it doesn't. (c) Largest at 0掳, smallest at 90掳.

Step by step solution

01

Understanding the Concepts

Electric flux through a surface is defined by the equation \( \Phi = E \cdot A \cdot \cos(\theta) \), where \( E \) is the electric field strength, \( A \) is the area of the surface, and \( \theta \) is the angle between the normal to the surface and the electric field. Our goal is to apply this formula to solve the problem.
02

Calculate Electric Flux

Given \( E = 14 \, \mathrm{N/C} \), \( A = 0.250 \, \mathrm{m^2} \), and \( \theta = 60^{\circ} \), we plug these values into the flux formula: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times \cos(60^{\circ}) \]\( \cos(60^{\circ}) = 0.5 \), therefore: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times 0.5 = 1.75 \, \mathrm{N} \cdot \mathrm{m^2/C} \].
03

Does Shape Affect Flux?

The answer to part (a) does not depend on the shape of the sheet. Electric flux depends only on the area, the electric field, and the angle between the field and the normal; it does not matter how the area is distributed.
04

Determine Angles for Maximum and Minimum Flux

(i) The electric flux is largest when \( \theta = 0^{\circ} \). This is because \( \cos(0^{\circ}) = 1 \), so \( \Phi = E \cdot A \cdot 1 = E \cdot A \), giving the maximum possible flux.(ii) The electric flux is smallest when \( \theta = 90^{\circ} \) because \( \cos(90^{\circ}) = 0 \), resulting in zero flux through the sheet.

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Key Concepts

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

Electric Field
The electric field is a vector field that represents the force that a positively charged particle would feel at any given point in space. This force is expressed in Newtons per Coulomb (N/C).
In exercises like the one mentioned, the electric field (\( E = 14 \, \mathrm{N/C} \)) influences everything concerning the flux through a surface.
  • The stronger the field, the greater the force exerted on charges within it.
  • The direction of the electric field is crucial, affecting how it interacts with surfaces like the paper described.
Recognizing the characteristics of the electric field can help you predict and calculate interactions with different materials and objects in its path.
Surface Area
The surface area is a measure of the size of an object's surface. With regards to the electric flux, the surface area\( A = 0.250 \, \mathrm{m^2} \)) specifically determines how much of the field passes through.
* Area only affects the amount of field interacting with the surface, not its shape.* It proves crucial in the formula for electric flux,\( \Phi = E \cdot A \cdot \cos(\theta) \).In application, surface area acts as a gateway for the field lines. The greater the area, the more field lines can traverse.
Angle of Incidence
The angle of incidence, denoted as \( \theta \), represents the inclination between the normal to the surface and the electric field. In this example, the angle is \(60^{\circ}\). Through the \( \cos(\theta) \) term, this angle impacts the volume of field lines penetrating the surface.
  • A smaller angle signifies more field lines entering, maximizing flux.
  • When \( \theta = 90^{\circ} \), no field lines pass through, making flux zero.
Recognizing these effects helps optimize the orientation of materials for specific exposure to electric fields.
Flux Calculation
Flux calculation ties all these factors together. By formula, \( \Phi = E \cdot A \cdot \cos(\theta) \), where each term - electric field, surface area, and angle - plays a distinct role.
* The given parameters, \( E = 14 \, \mathrm{N/C} \), \( A = 0.250 \, \mathrm{m^2} \), and \( \theta = 60^{\circ} \) lead to the flux \( \Phi = 1.75 \, \mathrm{N} \cdot \mathrm{m^2/C} \).* Changes in any parameter directly alter the flux.Understanding how to calculate flux enables you to predict electric field behavior relative to surfaces, which is critical in many fields such as electronics and physics.

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Most popular questions from this chapter

22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.

22.10. A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m}\) . What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radins (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

22.43. A solid conducting sphere with radius \(R\) that carries poositive charge \(Q\) is concentric with a very thin insulating shell of radius 2\(R\) that also carries charge \(Q\) . The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions \(02 R\) . (b) Graph the electric- field magnitude as a function of \(r .\)

What Aflat, square surface with sides of length \(L\) is described by the equations $$ x-L \quad(0 \leq y \leq L, 0 \leq z \leq L) $$ (a) Draw this square and show the \(x\) - \(y\) - and \(z\) -axes. (b) Find the electric flux through the square due to a positive point charge \(q\) located at the origin \((x=0, y=0, z=0)\) . (Hint: Think of the square as part of a cube centered on the origin.)

22.23. A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric fiux through a spherical surface just inside the inner surface of the sphere?
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