/*! 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 39 Photocopying Machine The electri... [FREE SOLUTION] | 91Ó°ÊÓ

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Photocopying Machine The electric field just above the surface of the charged drum of a photocopying machine has a magnitude \(|\vec{E}|\) of \(2.3 \times 10^{5} \mathrm{~N} / \mathrm{C}\). What is the surface charge density on the drum, assuming that the drum is a conductor?

Short Answer

Expert verified
Surface charge density is \(2.0 \times 10^{-6} \, \mathrm{C}/\mathrm{m}^2\).

Step by step solution

01

Identify the Relationship

According to Gauss's Law, the surface charge density \(\sigma\) on a conductor is related to the electric field \(\vec{E}\) by the equation \[|\vec{E}| = \frac{\sigma}{\epsilon_0} \]where \(\epsilon_0\) is the permittivity of free space and has a value of \(8.85 \times 10^{-12} \, \mathrm{C}^{2}/(\mathrm{N} \, \mathrm{m}^{2})\).
02

Rearrange the Equation

Rearrange Gauss's Law equation to solve for the surface charge density \(\sigma\): \[\sigma = |\vec{E}| \cdot \epsilon_0 \]
03

Insert Known Values

Substitute the given values into the rearranged equation. The electric field is provided as \(2.3 \times 10^5 \, \mathrm{N}/\mathrm{C}\) and the permittivity of free space is \(8.85 \times 10^{-12} \, \mathrm{C}^{2}/(\mathrm{N} \, \mathrm{m}^{2})\): \[\sigma = (2.3 \times 10^5 \, \mathrm{N}/\mathrm{C}) \cdot (8.85 \times 10^{-12} \, \mathrm{C}^{2}/(\mathrm{N} \, \mathrm{m}^{2})) \]
04

Calculate the Surface Charge Density

Carry out the multiplication: \[ \sigma = 2.0355 \times 10^{-6} \, \mathrm{C}/\mathrm{m}^2 \]which simplifies to \[\sigma \approx 2.0 \times 10^{-6} \, \mathrm{C}/\mathrm{m}^2\]

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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 a force would be exerted on other charged objects. Think of it like the invisible force field around a charged particle. In this particular exercise, we are given the electric field's magnitude just above the surface of the charged drum in a photocopying machine. The electric field strength here is noted as \(|\vec{E}| = 2.3 \times 10^5 \mathrm{N}/\mathrm{C}\). The electric field is crucial because it directly affects how charges interact in space. In equations, we denote it with the symbol \(|\vec{E}|\). If you place a small positive test charge in this field, the force acting on it tells you about the direction and strength of the field. Understanding the electric field helps us determine other quantities like surface charge density using Gauss's Law. This law provides a connection between the field and the charge distribution on a surface.
surface charge density
Surface charge density, which we represent as \(\sigma\), is the amount of electric charge per unit area on a surface. In simpler terms, it's a measure of how much charge is packed into a given area. It's like measuring how crowded a surface is with charges. The units of surface charge density are \mathrm{C/m^2}\ (Coulombs per square meter). To find the surface charge density on the drum of the photocopying machine, we use the relationship given by Gauss's Law: \[ \sigma = |\vec{E}| \cdot \epsilon_0 \] This equation tells us that the surface charge density is the product of the electric field \(|\vec{E}|\) and the permittivity of free space \(\epsilon_0\). By plugging in the given electric field and the value of \(\epsilon_0\), we can calculate \(\sigma\). This helps in understanding how charged the surface is by looking at how strong the surrounding electric field is.
permittivity of free space
The permittivity of free space, denoted as \(\epsilon_0\), is a fundamental constant that quantifies how an electric field interacts with the vacuum. Its value is \(\epsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \ m^2)}\). This constant appears in many fundamental equations in electromagnetism, including Gauss's Law. It essentially measures the ability of the vacuum to permit electric field lines. Imagine you're in a vacuum trying to create an electric field; \(\epsilon_0\) tells you how much space resists or allows that field to form. In our exercise, \(\epsilon_0\) plays a crucial role in connecting the electric field \(|\vec{E}|\) to the surface charge density \(\sigma\). When we perform the calculation \[ \sigma = |\vec{E}| \cdot \epsilon_0 \], \(\epsilon_0\) allows us to transition from the field strength to surface charge density accurately. It's a bridge that translates how the presence of electric fields can define the amount of charge per unit area.

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Most popular questions from this chapter

Charge Causes Flux A point charge causes an electric flux of \(-750 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\) to pass through a spherical Gaussian surface of \(10.0 \mathrm{~cm}\) radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Arbitrary Shape Conductor An isolated conductor of arbitrary shape has a net charge of \(+10 \times 10^{-6} \mathrm{C}\). Inside the conductor is a cavity within which is a point charge \(q=+3.0 \times 10^{-6} \mathrm{C}\). What is the charge (a) on the cavity wall and (b) on the outer surface of the conductor?

Planar Slab A planar slab of thickness \(d\) has a uniform volume charge density \(\rho\). Find the magnitude of the electric field at all points in space both (a) within and (b) outside the slab, in terms of \(x\), the distance measured from the central plane of the slab.

Charge at Center of Shell A point charge \(+q\) is placed at the center of an electrically neutral, spherical conducting shell with inner radius \(a\) and outer radius \(b .\) What charge appears on (a) the inner surface of the shell and (b) the outer surface? What is the net electric field at a distance \(r\) from the center of the shell if (c) \(rr>a\), and (e) \(r>b\) ? Sketch field lines for those three regions. For \(r>b\), what is the net electric field due to (f) the central point charge plus the inner surface charge and (g) the outer surface charge? A point charge \(-q\) is now placed outside the shell. Does this point charge change the charge distribution on (h) the outer surface and (i) the inner surface? Sketch the field lines now. (j) Is there an electrostatic force on the second point charge? (k) Is there a net electrostatic force on the first point charge? (1) Does this situation violate Newton's Third Law?

Geiger Counter Figure \(24-33\) shows a Geiger counter, a device used to detect ionizing radiation (radiation that causes ionization of atoms). The counter consists of a thin, positively charged central wire surrounded by a concentric, circular, conducting cylinder with an equal negative charge. Thus, a strong radial electric field is set up inside the cylinder. The cylinder contains a low-pressure inert gas. When a particle of radiation enters the device through the cylinder wall, it ionizes a few of the gas atoms The resulting free electrons (labelled e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with other gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting "avalanche" of electrons is collected by the wire generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is \(25 \mu \mathrm{m}\), the radius of the cylinder \(1.4 \mathrm{~cm}\), and the length of the tube \(16 \mathrm{~cm}\). If the electric field component \(E_{r}\) at the cylinder's inner wall is \(+2.9 \times\) \(10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

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