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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

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\). (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\). (b) What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R\). (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R\).

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.

The electric field \(0.400 \mathrm{~m}\) from a very long uniform line of charge is \(840 \mathrm{~N} / \mathrm{C}\). How much charge is contained in a \(2.00 \mathrm{~cm}\) section of the line?

Two very long uniform lines of charge are parallel and are separated by \(0.300 \mathrm{~m}\). Each line of charge has charge per unit length \(+5.20 \mu \mathrm{C} / \mathrm{m} .\) What magnitude of force does one line of charge exert on a \(0.0500 \mathrm{~m}\) section of the other line of charge?

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{~m}\). What is the total electric flux due to the two point charges through a spherical surface centered at the origin and with ra\(\operatorname{dius}(\mathrm{a}) 0.500 \mathrm{~m},(\mathrm{~b}) 1.50 \mathrm{~m},(\mathrm{c}) 2.50 \mathrm{~m} ?\)
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