/*! 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 48 You've hung two very large sheet... [FREE SOLUTION] | 91影视

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Chapter 27: Problem 48

You've hung two very large sheets of plastic facing each other with distance \(d\) between them, as shown in By rubbing them with wool and silk, you've managed to give one sheet a uniform surface charge density \(\eta_{1}=-\eta_{0}\) and the other a uniform surface charge density \(\eta_{2}=+3 \eta_{0} .\) What is the electric field vector at points \(1,2,\) and 37

Short Answer

Expert verified
The electric field vector at point 1 is \(\frac{\eta_{0}}{\epsilon_{0}}\), at point 2 is \(\frac{3\eta_{0}}{2\epsilon_{0}}\), and at point 3 is \(\frac{-\eta_{0}}{2\epsilon_{0}}\).

Step by step solution

01

Getting the expression for the electric field due to a sheet of charge

The electric field \(E\) due to an infinite sheet of charge is given by \(E = \frac{\eta}{2\epsilon_{0}}\), where \(\eta\) is the surface charge density and \(\epsilon_{0}\) is the permittivity of free space. This electric field is directed perpendicularly outward from the surface of the sheet if \(\eta > 0\), and perpendicularly inward if \(\eta < 0\).
02

Calculation of Electric Field at Point 1

Point 1 is in the region between the sheets, so the electric field at this point is the vector sum of the electric fields due to both sheets, i.e., \(E_{1} = E_{1,2} + E_{1,1}\), where \(E_{1,2} = \frac{3\eta_{0}}{2\epsilon_{0}}\) is the electric field due to sheet 2 and \(E_{1,1} = \frac{-\eta_{0}}{2\epsilon_{0}}\) is the electric field due to sheet 1. Hence \(E_{1} = \frac{3\eta_{0} - \eta_{0}}{2\epsilon_{0}} = \frac{2\eta_{0}}{2\epsilon_{0}} = \frac{\eta_{0}}{\epsilon_{0}}\).
03

Calculation of Electric Field at Point 2

Point 2 is beyond sheet 2, so the electric field at this point is the vector sum of the electric fields due to both sheets, i.e., \(E_{2} = E_{2,2} + E_{2,1}\), where \(E_{2,2} = \frac{3\eta_{0}}{2\epsilon_{0}}\) is the electric field due to sheet 2 and \(E_{2,1} = 0\) (because the electric field due to an infinite sheet of charge decreases with distance from the sheet). Hence \(E_{2} = \frac{3\eta_{0}}{2\epsilon_{0}} + 0 = \frac{3\eta_{0}}{2\epsilon_{0}}\).
04

Calculation of Electric Field at Point 3

Point 3 is beyond sheet 1, so the electric field at this point is the vector sum of the electric fields due to both sheets, i.e., \(E_{3} = E_{3,2} + E_{3,1}\), where \(E_{3,2} = 0\) (because the electric field due to an infinite sheet of charge decreases with distance from the sheet) and \(E_{3,1} = \frac{-\eta_{0}}{2\epsilon_{0}}\) is the electric field due to sheet 1. Hence \(E_{3} = 0 + \frac{-\eta_{0}}{2\epsilon_{0}} = \frac{-\eta_{0}}{2\epsilon_{0}}\).

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Key Concepts

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

Surface Charge Density
When studying electric fields, the term surface charge density often comes up. It's denoted by Greek letter \( \eta \) and represents the amount of electric charge per unit area on a surface. It's crucial because it helps determine the electric field produced by a charged object. In a way, you can think of it kind of like the density of a crowd at a concert鈥攖he greater the number of people (charge) in a given area, the higher the density. \
\

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Most popular questions from this chapter

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms \(30-\mu\) m-diameter drops of ink, charges them by spraying 800,000 electrons on the surface. and shoots them toward the page at a speed of \(20 \mathrm{m} / \mathrm{s}\). Along the way, the drops pass through two parallel electrodes that are \(6.0 \mathrm{mm}\) long, \(4.0 \mathrm{mm}\) wide, and spaced \(1.0 \mathrm{mm}\) apart. The distance from the center of the plates to the paper is \(2.0 \mathrm{cm} .\) To form the letters, which have a maximum height of \(6.0 \mathrm{mm}\), the drops need to be deflected up or down a maximum of \(3.0 \mathrm{mm}\). Ink, which consists of dye particles suspended in alcohol, has a density of \(800 \mathrm{kg} / \mathrm{m}^{3}\) a. Estimate the maximum electric field strength needed in the space between the electrodes. b. What amount of charge is needed on each electrode to produce this electric field?

Two 10 -cm-diameter charged disks face each other, \(20 \mathrm{cm}\) apart. Both disks are charged to \(+50 \mathrm{nC} .\) What is the electric ficld strength at (a) the midpoint between the two disks and (b) a point on the axis \(5.0 \mathrm{cm}\) from one disk?

The surface charge density on an infinite charged plane is \(-2.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A proton is shot straight away from the plane at \(2.0 \times 10^{6} \mathrm{m} / \mathrm{s} .\) How far does the proton travel before reaching its turning point?

The permanent electric dipole moment of the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) is \(6.2 \times 10^{-30} \mathrm{Cm} .\) What is the maximum possible torque on a water molecule in a \(5.0 \times 10^{8}\) N/C electric field?

Two 10 -cm-diameter charged rings face each other, \(20 \mathrm{cm}\) apart. The left ring is charged to -20 nC and the right ring is charged to \(+20 \mathrm{nC}\). a. What is the electric field \(\vec{E}\), both magnitude and direction, at the midpoint between the two disks? b. What is the force \(\vec{F}\) on \(a-1.0\) nC charge placed at the midpoint?
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