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Charge at Center of Cube A point charge of \(1.8 \mu \mathrm{C}\) is at the center of a cubical Gaussian surface \(55 \mathrm{~cm}\) on edge. What is the net electric flux through the surface?

Short Answer

Expert verified
The net electric flux is \(2.03 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^2/\text{C}\).

Step by step solution

01

Understand the Problem

Determine the net electric flux through a cubical Gaussian surface with a point charge at its center.
02

Use Gauss's Law

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. The formula is given by \[ \text{Flux} = \frac{Q_{\text{enc}}}{\text{É›}_0} \], where \( Q_{\text{enc}} \) is the charge enclosed and \( \text{É›}_0 \) is the permittivity of free space (\( \text{É›}_0 \approx 8.85 \times 10^{-12} \mathrm{F/m} \)).
03

Identify Given Values

Identify the point charge \( Q_{\text{enc}} = 1.8 \mu \mathrm{C} \) and convert it to Coulombs. \[ Q_{\text{enc}} = 1.8 \times 10^{-6} \mathrm{C} \]
04

Calculate the Electric Flux

Substitute the given values into Gauss's law formula. \[ \text{Flux} = \frac{1.8 \times 10^{-6} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{F/m}} \; \]
05

Simplify the Calculation

Simplify the result to find the net electric flux. \[ \text{Flux} = 2.03 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^2/\text{C} \]

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism. It relates the electric flux passing through a closed surface to the charge enclosed within that surface. Gauss's Law can be mathematically expressed as \[ \text{Flux} = \frac{Q_{\text{enc}}}{\text{É›}_0} \] where \( Q_{\text{enc}} \) is the enclosed charge and \( \text{É›}_0 \) is the permittivity of free space.
Gauss's Law is especially useful in situations where the symmetry of the system allows for easy calculation of electric fields. This law helps us understand how charges generate electric fields and simplifies the process of calculating electric flux in complex situations involving symmetrical shapes.
Point Charge
A point charge refers to an idealized model of a charged object where all its charge is concentrated at a single point in space. This simplification helps in solving many electrostatics problems.
In many practical problems, an object can be approximated as a point charge if its size is very small compared to the distances involved. In the given exercise, the \( 1.8 \, \mu \text{C} \) charge is considered a point charge located at the center of a cubic Gaussian surface. Understanding point charges helps in applying Gauss's Law effectively because it simplifies complex charge distributions.
Electric Flux Calculation
Electric flux measures the quantity of the electric field passing through a given area. It is defined as \[ \text{Flux} = \mathbf{E} \cdot \mathbf{A} \] where \( \mathbf{E} \) is the electric field and \( \mathbf{A} \) is the area vector.
In this exercise, the electric flux is calculated using Gauss's Law because the point charge is enclosed by a Gaussian surface. Given the charge at the center of the cube, applying Gauss's Law simplifies the complex field calculations to a simple division: \[ \text{Flux} = \frac{Q_{\text{enc}}}{\text{É›}_0} \]. After substituting the given values, we have \( Q_{\text{enc}} = 1.8 \times 10^{-6} \text{C} \) and \( \text{É›}_0 = 8.85 \times 10^{-12} \text{F/m} \).
Permittivity of Free Space
The permittivity of free space, denoted by \( \text{É›}_0 \) , is a physical constant that describes how electric fields interact with the vacuum. Its value is approximately \( 8.85 \times 10^{-12} \text{F/m} \).
This constant appears in many fundamental equations of electromagnetism, like Gauss's Law and Coulomb’s Law.
To calculate the electric flux through a Gaussian surface, knowing \( \text{É›}_0 \) is essential. It helps relate the enclosed charge to the resulting electric flux. In our exercise, replacing \( \text{É›}_0 \) with its value gives us a straightforward way to find the net electric flux: \( \text{Flux} = \frac{1.8 \times 10^{-6} \text{C}}{8.85 \times 10^{-12} \text{F/m}} \approx 2.03 \times 10^5 \text{N} \cdot \text{m}^2/\text{C} \).

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Most popular questions from this chapter

Large Metal Plates Two large metal plates of area \(1.0 \mathrm{~m}^{2}\) face each other. They are \(5.0 \mathrm{~cm}\) apart and have equal but opposite charges on their inner surfaces. If the magnitude \(|\vec{E}|\) of the electric field between the plates is \(55 \mathrm{~N} / \mathrm{C}\), what is the amount of charge on each plate? Neglect edge effects.

Uniform Volume Charge Density Figure 24-29a shows a spherical shell of charge with uniform volume charge density \(\rho .\) Plot \(E\) due to the shell for distances \(r\) from the center of the shell ranging from zero to \(30 \mathrm{~cm}\). Assume that \(\rho=1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{3}, a=10 \mathrm{~cm}\), and \(b=20 \mathrm{~cm}\).

Sphere of Radius \(a\) In Fig \(24-28\) an insulating sphere, of radius \(a\) and charge \(+q\) uniformly distributed throughout its volume, is concentric with a spherical conducting shell of inner radius \(b\) and outer radius \(c\). This shell has a net charge of \(-q\). Find expressions for the electric field, as a function of the radius \(r,(a)\) within the sphere \((rc) .(\mathrm{e}) \mathrm{What}\) are the charges on the inner and outer surfaces of the shell?

Square Metal Plate A square metal plate of edge length \(8.0 \mathrm{~cm}\) and negligible thickness has a total charge of \(6.0 \times 10^{-6} \mathrm{C}\) (a) Estimate the magnitude \(E\) of the electric field just off the center of the plate (at, say, a distance of \(0.50 \mathrm{~mm}\) ) by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate \(E\) at a distance of \(30 \mathrm{~m}\) (large relative to the plate size) by assuming that the plate is a point charge.

Earth's Atmosphere It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of \(300 \mathrm{~m}\) the field has magnitude \(60.0 \mathrm{~N} / \mathrm{C}\), at an altitude of \(200 \mathrm{~m}\), the magnitude is \(100 \mathrm{~N} / \mathrm{C}\). Find the net amount of charge contained in a cube \(100 \mathrm{~m}\) on edge, with horizontal faces at altitudes of 200 and \(300 \mathrm{~m}\). Neglect the curvature of Earth.

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