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(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of magnitude 1390 N/C just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

Short Answer

Expert verified
Approximately 6.93 billion excess electrons; electric field 10 cm outside is 1600 N/C.

Step by step solution

01

Understanding the Problem

We need to find the number of excess electrons needed to produce an electric field of 1390 N/C just outside the surface of a plastic sphere with a diameter of 30.0 cm. Additionally, we must find the electric field at a point 10.0 cm outside the sphere.
02

Calculate the Radius and Convert to Meters

The diameter of the sphere is 30.0 cm, so the radius is half of that: 15.0 cm. Convert the radius to meters: \[ r = \frac{30.0}{2} = 15.0 \text{ cm} = 0.15 \text{ m} \]
03

Use Gauss's Law to Find Charge

Gauss's Law states \( \Phi = E A = \frac{Q_{enc}}{\varepsilon_0} \), where \(E\) is the electric field, \(A\) is the area of the spherical surface, and \(\varepsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12}\, \text{C}^2/\text{N}\cdot \text{m}^2\)). Rearrange to solve for charge \(Q_{enc}\): \[ Q_{enc} = E \cdot A \cdot \varepsilon_0 \]The area \(A\) of the sphere is \(4\pi r^2\). Substituting known values:\[ Q_{enc} = 1390 \cdot 4 \pi (0.15)^2 \cdot 8.854 \times 10^{-12} \approx 1.111 \times 10^{-9} \text{ C} \]
04

Calculate Number of Electrons

Each electron has a charge of \(-e\), where \(e = 1.602 \times 10^{-19} \text{ C}\). Find the number of electrons by dividing the total charge by the charge of one electron:\[ n = \frac{1.111 \times 10^{-9}}{1.602 \times 10^{-19}} \approx 6.93 \times 10^9 \]
05

Calculate Electric Field 10.0 cm Outside the Sphere

The distance from the center of the sphere to the point 10.0 cm outside the surface is \(0.15 \text{ m} + 0.10 \text{ m} = 0.25 \text{ m}\). Use the formula for the electric field due to a point charge at a distance \(r\):\[ E = \frac{k \cdot Q_{enc}}{r^2} \]where \(k = 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2\). Substitute the values:\[ E = \frac{8.99 \times 10^9 \cdot 1.111 \times 10^{-9}}{(0.25)^2} \approx 1600 \text{ N/C} \]

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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 that connects an electric field to the charge enclosed within a surface. It states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. In formula terms, we write:
  • \( \Phi = E \times A = \frac{Q_{\text{enc}}}{\varepsilon_0} \)
Where:
  • \( \Phi \) is the electric flux
  • \( E \) is the electric field
  • \( A \) is the area of the closed surface
  • \( Q_{\text{enc}} \) is the enclosed charge
  • \( \varepsilon_0 \) is the permittivity of free space
This law is particularly useful for symmetrical shapes such as spheres, where it simplifies calculations. To find the electric field just outside a charged sphere, considering it as a closed surface, is straightforward using Gauss's Law.
Excess Electrons
Excess electrons refer to the additional electrons present on an object that give it a negative charge. When electrons are added to an object, it attains a negative electric charge, leading to an overall imbalance between the number of protons and electrons.
To determine how many excess electrons are needed to achieve a certain electric field around an object, one must first compute the required charge using concepts like Gauss's Law. Each electron contributes a charge of \(-1.602 \times 10^{-19} \text{ C}\). Therefore, by dividing the total charge by the charge of a single electron, we discover the number of excess electrons required. This calculation is fundamental in solving problems where electric fields of specified magnitudes are involved.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a physical constant that describes how an electric field affects and is affected by a vacuum. It is approximately equal to \( 8.854 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \). This constant is crucial in the equations governing electromagnetism, particularly in Gauss's Law and Coulomb's Law.
The permittivity of free space influences how strongly electric forces act between charges in a vacuum. It determines the electric flux per unit charge, facilitating the equations that predict behavior and enable calculations of interactions in physics. Understanding \( \varepsilon_0 \) is the key to linking observed electric fields to theoretical charge distributions.
Point Charge Formula
The point charge formula is essential for calculating the electric field produced by a point charge at a certain distance. The formula used is:
  • \( E = \frac{k \cdot Q}{r^2} \)
Where:
  • \( E \) is the electric field
  • \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2 \)
  • \( Q \) is the charge creating the field
  • \( r \) is the distance from the charge
This formula hinges on the idea of an isolated point charge, where the field's strength diminishes with the square of the distance. In practical scenarios involving spheres, by considering the charge on the sphere as residing at its center, this formula can simplify the calculation of fields at points outside the sphere.

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Most popular questions from this chapter

An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface charge density +2.90 \(\times\) 10\(^{-12}\) C/m2. (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 0.050 m from the sheet? (b) What is the speed of the electron when it is 0.050 m from the sheet?

Using Thomson's (outdated) model of the atom described in Problem 22.50, consider an atom consisting of two electrons, each of charge \(-e\), embedded in a sphere of charge \(+2e\) and radius \(R\). In equilibrium, each electron is a distance \(d\) from the center of the atom (\(\textbf{Fig. P22.51}\)). Find the distance \(d\) in terms of the other properties of the atom.

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes \(x = d\) and \(x = -d\). The \(y\)- and \(z\)-dimensions of the slab are very large compared to \(d\); treat them as essentially infinite. The slab has a uniform positive charge density \(\rho\). (a) Explain why the electric field due to the slab is zero at the center of the slab (\(x =\) 0). (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 90.0 N/C that is directed at 20\(^\circ\) from the plane of the sheet (\(\textbf{Fig. E22.2}\)). Find the magnitude of the electric flux through the sheet.

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located a distance (a) \(r > R\) from the center of the shell (outside the shell); (b) \(r < R\) from the center of the shell (inside the shell).

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