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

Short Answer

Expert verified
The surface charge density is \( 2.03 \times 10^{-6} \text{ C/m}^2 \).

Step by step solution

01

Recall the Relation Between Electric Field and Surface Charge Density

The electric field just outside the surface of a charged conductor is given by the equation \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma \) is the surface charge density and \( \varepsilon_0 \) is the permittivity of free space, which is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).
02

Rearrange the Formula to Solve for Surface Charge Density \(\sigma\)

Rearrange the formula from Step 1 to solve for \(\sigma\). The equation becomes \( \sigma = E \times \varepsilon_0 \).
03

Substitute the Known Values into the Equation

Substitute the given electric field \( E = 2.3 \times 10^{5} \text{ N/C} \) and the value of \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \) into the equation \( \sigma = E \times \varepsilon_0 \):\[\sigma = (2.3 \times 10^{5} \text{ N/C}) \times (8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2)\]
04

Perform the Calculation

Multiply the values together to find \(\sigma\):\[\sigma = 2.3 \times 8.85 \times 10^{5 - 12} \text{ C/m}^2 = 2.03 \times 10^{-6} \text{ C/m}^2\]
05

Conclude with the Surface Charge Density Value

The surface charge density \( \sigma \) on the drum is therefore \( 2.03 \times 10^{-6} \text{ C/m}^2 \).

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Key Concepts

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

Surface Charge Density
Surface charge density is a measure of how much electric charge resides on a surface per unit area. It is denoted by the symbol \( \sigma \). To understand it simply, consider it as the concentration of electric charge spread over a surface such as that of a conductor. Surface charge density is crucial because it influences the strength of the electric field emanating from a charged surface.

In many practical applications, such as in the photocopying machine drum in the exercise, it helps in determining the effectiveness of charge distribution over a surface. The formula connecting surface charge density to the electric field, \( E = \frac{\sigma}{\varepsilon_0} \), shows that a higher surface charge density on a conductor will result in a stronger electric field right next to it.
  • The unit used for surface charge density is \( \text{C/m}^2 \).
  • It provides insight into how a charged object will interact with other charged particles around it.
  • Determining \( \sigma \) helps in practical calculations to avoid electrical breakdowns or discharges in devices.
Permittivity of Free Space
Permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental constant in physics that characterizes the ability of the vacuum to permit electric field lines. Its value is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

This constant is vital in understanding how electric fields interact in a vacuum and it plays a central role in Coulomb's law, which describes the electric force between two point charges. In the exercise, \( \varepsilon_0 \) helps link the electric field \( E \) to the surface charge density \( \sigma \) through the equation \( E = \frac{\sigma}{\varepsilon_0} \).
  • \( \varepsilon_0 \) is critical for calculating forces in electromagnetic fields.
  • It determines how effective a material or vacuum is at "holding" electric field lines.
  • A lower permittivity implies stronger electric forces for the same amount of surface charge.
Charged Conductor
A charged conductor is any conductor (e.g., metal) that holds an electric charge. When an electric charge is placed on a conductor, the charge distributes itself evenly on the surface because like charges repel each other and try to get as far apart as possible.

Conductors like metals allow charges to move freely over their surface until they reach equilibrium, where the electric field inside the conductor is zero. This property makes conductors crucial in managing and manipulating electric charges in a controlled way, such as the charged drum in a photocopier.
  • In conductors, charges reside only on the surface, not inside.
  • Electric fields are strongest near sharp points or edges of the conductor.
  • Understanding charged conductors helps in designing devices that need uniform charge distribution.
  • Charged conductors are vital in many electrical devices for shielding or directing electric fields.

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Most popular questions from this chapter

An isolated conductor has net charge \(+10 \times 10^{-6} \mathrm{C}\) and a cavity with a particle of charge \(q=+3.0 \times 10^{-6} \mathrm{C}\). What is the charge on (a) the cavity wall and (b) the outer surface?

ILW The volume charge denrigure \(20=\) oblem \(52 .\) sity of a solid nonconducting sphere of radius \(R=5.60 \mathrm{~cm}\) varies with radial distance \(r\) as given by \(\rho=\) \(\left(14.1 \mathrm{pC} / \mathrm{m}^{3}\right) r / R .\) (a) What is the sphere's total charge? What is the field magnitude \(E\) at (b) \(r=0\), (c) \(r=R / 2.00\), and (d) \(r=R ?\) (e) Graph \(E\) versus

. The chocolate crumb mystery. Explosions ignited by electrostatic discharges (sparks) constitute a serious danger in facilities handling grain or powder. Such an explosion occurred in chocolate crumb powder at a biscuit factory in the 1970 s. Workers usually emptied newly delivered sacks of the powder into a loading bin, from which it was blown through electrically grounded plastic pipes to a silo for storage. Somewhere along this route, two conditions for an explosion were met: (1) The magnitude of an electric field became \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) or greater, so that electrical breakdown and thus sparking could occur. (2) The energy of a spark was \(150 \mathrm{~mJ}\) or greater so that it could ignite the powder explosively. Let us check for the first condition in the powder flow through the plastic pipes. Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius \(R=5.0 \mathrm{~cm}\). Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density \(\rho\). (a) Using Gauss' law, find an expression for the magnitude of the electric field \(\vec{E}\) in the pipe as a function of radial distance \(r\) from the pipe center. (b) Does \(E\) increase or decrease with increasing \(r ?\) (c) Is \(\vec{E}\) directed radially inward or outward? (d) For \(\rho=1.1 \times 10^{-3} \mathrm{C} / \mathrm{m}^{3}\) (a typical value at the factory), find the maximum \(E\) and determine where that maximum field occurs. (e) Could sparking occur, and if so, where? (The story continues with Problem 70 in Chapter 24.)

Charge of uniform volume density \(\rho=3.2 \mu \mathrm{C} / \mathrm{m}^{3}\) fills a nonconducting solid sphere of radius \(5.0 \mathrm{~cm} .\) What is the magnitude of the electric field (a) \(3.5\) \(\mathrm{cm}\) and (b) \(8.0 \mathrm{~cm}\) from the sphere's center?

A charged particle 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 charge of the particle?

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