/*! 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 20 A flat surface with area \(2.0 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 20

A flat surface with area \(2.0 \mathrm{m}^{2}\) is in a uniform \(850-\mathrm{N} / \mathrm{C}\) electric field. Find the electric flux through the surface when it's (a) at right angles to the field, (b) at \(45^{\circ}\) to the field, and (c) parallel to the field.

Short Answer

Expert verified
The electric flux through the surface is (a) 1700 N.m^2/C when the surface is at right angles to the field, (b) 1202 N.m^2/C when the surface is at 45 degrees to the field, and (c) 0 N.m^2/C when the surface is parallel to the field.

Step by step solution

01

Calculate the Electric Flux at Right Angles

When the surface is at right angles to the field, the angle \(\Theta\) is 0. Substituting into the formula gives \(\Phi = E \cdot A \cdot cos(0) = (850 \, \mathrm{N/C})(2.0 \, \mathrm{m}^{2}) \cdot 1 = 1700 \, \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\)
02

Calculate the Electric Flux at 45 Degrees

When the surface is at \(45^{\circ}\) to the field, the angle \(\Theta\) is \(45^\circ\). Using the same formula gives \(\Phi = E \cdot A \cdot cos(45^{\circ}) = (850 \, \mathrm{N/C})(2.0 \, \mathrm{m}^{2}) \cdot 0.7071 = 1202 \, \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\). We used the fact that \(cos(45^{\circ}) = 0.7071\)
03

Calculate the Electric Flux Parallel to the Field

When the surface is parallel to the field, the angle \(\Theta\) is \(90^\circ\). Substituting this into the formula gives \(\Phi = E \cdot A \cdot cos(90^{\circ}) = (850 \, \mathrm{N/C})(2.0 \, \mathrm{m}^{2}) \cdot 0 = 0\). The electric flux is zero because the electric field lines don't pass through the surface when it's parallel to the field

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You're sitting inside an uncharged, hollow spherical shell. Suddenly someone dumps a billion coulombs of charge on the shell, distributed uniformly. What happens to the electric field at your location?

A coaxial cable in electrostatic equilibrium carries charge \(-Q\) on its inner conductor and \(+Q\) on its shield. If the charge on the shield only is doubled, a. the magnitude of the electric field between the conductors will double. b. the magnitude of the electric field outside the shield will double. c. the magnitude of the electric field at the outer surface of the shield will become twice the magnitude of the field at the shield's inner surface. d. the magnitude of the electric field at the outer surface of the shield will equal the magnitude of the field at the shield's inner surface.

A \(6.8-\mu \mathrm{C}\) charge and a-4.7- \(\mu \mathrm{C}\) charge are inside an uncharged sphere. What's the electric flux through the sphere?

The charge density within a charged sphere of radius \(R\) is given by \(\rho=\rho_{0}-a r^{2},\) where \(\rho_{0}\) and \(a\) are constants and \(r\) is the distance from the center. Find an expression for \(a\) such that the electric field outside the sphere is zero.

A point charge \(-q\) is at the center of a spherical shell carrying charge \(+2 q .\) That shell, in turn, is concentric with a larger shell carrying \(-\frac{3}{2} q .\) Draw a cross section of this structure, and sketch the electric field lines using the convention that eight lines correspond to a charge of magnitude \(q.\)
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Linear Momentum

Read Explanation

Scientific Method Physics

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.