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Chapter 22: Problem 48

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

Short Answer

Expert verified
The number of excess electrons in the sphere is \(8.15 脳 10^{11}\). The magnitude of the electric field at the point 10 cm outside the sphere is \(447 N/C\).

Step by step solution

01

Calculate Charge

First, calculate the charge needed to produce the electric field just outside the sphere. The magnitude of the electric field \(E\) at a distance \(r\) from the center of a sphere with charge \(q\) is given by Gauss's law, \(E = \frac{q}{4\pi\epsilon_0 r^2}\). Solve this formula for \(q\). The field \(E\) and radius \(r\) are given, and the permittivity of free space \(\epsilon_0 = 8.85 脳 10^{-12}\) C^2/(N 鈰� m^2).
02

Calculate Excess Electrons

Then, use the formula of charge to calculate the number of excess electrons. The charge \(q\) of an electron is \(1.6 脳 10^{-19}\) C. Divide the total charge \(q\) by the charge per electron.
03

Calculate Electric Field at a Point

Next, calculate the magnitude of the electric field at a point 10 cm away from the surface of the sphere using the Gauss's law formula. You need to add the radius of the sphere with 10 cm to get the new radius \(r\) and the charge \(q\) was calculated in step 1.

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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 other charged objects would feel a force. It's like an invisible force field that can push or pull charges. In our exercise, we calculate the electric field just outside a plastic sphere. According to physics, the electric field, denoted as \(E\), is measured in newtons per coulomb (N/C). It tells us how strong the electric force would be on a positive test charge placed in the field.

Gauss's Law helps in finding the electric field created by symmetrical charge distributions like our sphere. 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 \(\epsilon_0\). Here, it simplifies to \(E = \frac{q}{4\pi\epsilon_0 r^2}\), where \(q\) is the charge. This formula is our go-to tool for solving electric field problems involving spheres.
Excess Electrons
Excess electrons are the additional electrons on an object, making it negatively charged. In a neutral object, the number of protons equals the number of electrons. Adding electrons will mean it becomes negatively charged since electrons bear a negative charge.

To find how many excess electrons are needed to create a given electric field, we first calculate the total negative charge required using the formula derived from Gauss's Law. The charge of a single electron is a tiny amount, precisely \(1.6 \times 10^{-19}\) coulombs. Thus, to find the number of electrons, we divide the total charge by the charge of one electron. This yields the number of additional electrons needed to create the observed electric field.
Permittivity of Free Space
The permittivity of free space, denoted \(\epsilon_0\), is a constant that characterizes the ability of a vacuum to permit electric field lines. It is a fundamental constant in electromagnetism and has a value of \(8.85 \times 10^{-12}\) C虏/(N鈰卪虏).

In equations like the one derived using Gauss's Law, \(\epsilon_0\) helps relate the electric field to the amount of charge producing it. It's essential in calculating how electric fields propagate through space. Whether you're working with charged spheres, plates, or any other charged objects, \(\epsilon_0\) is a key player in the equations that describe their electric fields.

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Most popular questions from this chapter

In a region of space there is an electric field \(\overrightarrow{\boldsymbol{E}}\) that is in the \(z\) -direction and that has magnitude \(E=[964 \mathrm{~N} /(\mathrm{C} \cdot \mathrm{m})] x\). Find the flux for this field through a square in the \(x y\) -plane at \(z=0\) and with side length \(0.350 \mathrm{~m}\). One side of the square is along the \(+x\) -axis and another side is along the \(+y\) -axis.

A charge of \(87.6 \mathrm{pC}\) is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of \(29.2 \mathrm{~cm}^{2} .\) A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C},\) what area of the sheet is enclosed by the Gaussian surface?

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$ \begin{array}{ll} \rho(r)=3 \alpha r / 2 R & \text { for } r \leq R / 2 \\ \rho(r)=\alpha\left[1-(r / R)^{2}\right] & \text { for } R / 2 \leq r \leq R \\\ \rho(r)=0 & \text { for } r \geq R \end{array} $$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3}\). (a) Determine \(\alpha\) in terms of \(Q\) and \(R\). (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of \(Q .\) (c) What fraction of the total charge is contained within the region \(R / 2 \leq r \leq R ?\) (d) What is the magnitude of \(\overrightarrow{\boldsymbol{E}}\) at \(r=R / 2 ?\) (e) If an electron with charge \(q^{\prime}=-e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

It was shown in Example 21.10 (Section 21.5 ) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E=\lambda / 2 \pi \epsilon_{0} r .\) Consider an imaginary cylinder with radius \(r=0.250 \mathrm{~m}\) and length \(l=0.400 \mathrm{~m}\) that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda=3.00 \mu \mathrm{C} / \mathrm{m} .\) (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r=0.500 \mathrm{~m} ?\) (c) What is the flux through the cylinder if its length is increased to \(l=0.800 \mathrm{~m} ?\)

Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field \(\vec{E}\) of magnitude \(1 \times 10^{6} \mathrm{~N} / \mathrm{C}\) at \(25 \mathrm{~m}\) from the center of the sphere. What net charge on each sphere is needed? (a) \(-0.07 \mathrm{C} ;\) (b) \(-8 \mathrm{mC}\) (c) \(-80 \mu \mathrm{C}\) (d) \(-1 \times 10^{-20} \mathrm{C}\)
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