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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?

Short Answer

Expert verified
The total charge in a $2.00 \mathrm{~cm}$ section of the line is equivalent to the value calculated in step 3.

Step by step solution

01

Calculate the charge per unit length

Using the formula \(E = \frac{λ}{2πε_0r}\), we can rearrange for \(λ\) to get \(λ = E * 2πε_0r\). Substituting the given values for \(E = 840 \mathrm{~N/C}\), \(r = 0.400 \mathrm{~m}\), and \(ε_0 = 8.85 * 10^{-12} \mathrm{~C^2/Nm^2}\), we find \(λ = 840 \mathrm{~N/C} * 2π * 8.85 * 10^{-12} \mathrm{~C^2/Nm^2} * 0.400 \mathrm{~m}\).
02

Calculate the total charge

The total charge \(Q\) in a line section is given by \(Q = λ * l\), where \(l = 2.00 \mathrm{~cm} = 0.0200 \mathrm{~m}\) is the length of the line section. Substituting our calculated value for \(λ\) from step 1, we find \(Q = λ * 0.0200 \mathrm{~m}\).
03

Evalute the total charge

Evaluate the total charge \(Q\) by performing the multiplication in the previous step. This gives the final answer.

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Key Concepts

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

Charge Per Unit Length
Charge per unit length, often denoted by the symbol \( \lambda \), is a measure of how much electric charge is distributed along a line of charge. It's akin to density but instead of mass, we deal with charge. When a line of charge is uniform, the charge is evenly distributed along its length, meaning that each segment of the line carries the same amount of charge per unit of length.

For example, if we know the electric field intensity a certain distance from a line charge, we can find out the charge per unit length by rearranging the formula \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \). The given values from the exercise can then be plugged in, allowing us to calculate \( \lambda \). This is a fundamental step in understanding the distribution of charge along the line and is crucial for solving various electrostatic problems.
Linear Charge Density
Linear charge density, which is essentially the same as charge per unit length, is typically expressed in coulombs per meter (\(C/m\)). It's an idealized concept usually used for infinitely or very long conductors where we ignore the ends, assuming the charge distribution is not affected by edge effects.

In our example, we are dealing with a 'very long uniform line of charge,' which signifies that the linear charge density will remain constant no matter which segment of the line we examine. This allows us to use the uniform linear charge density to determine the total charge within any given length of the line. This concept underpins the relationship between charge distributions and the resultant electric fields.
Electric Field Intensity
The electric field intensity, denoted as \( E \), represents the force experienced by a unit positive charge in an electric field and is measured in newtons per coulomb (\(N/C\)). It is a vector quantity, which means it has both magnitude and direction, pointing away from positive charges and toward negative charges.

In our textbook exercise, we are specifically looking at the electric field intensity due to a line charge. Here, the formula that connects electric field intensity with linear charge density and distance from the line charge is fundamental. The intensity of the electric field diminishes as one moves further away from the line charge, and this inverse relationship with distance can be observed in the equation used to solve the problem. Understanding how to manipulate this equation is key to solving many problems in electrostatics.
Coulomb's Constant
Coulomb's constant, \( k \), is a proportionality factor that appears in Coulomb's law, which describes the force between two point charges. Its value is approximately \( 8.99 \times 10^9 \mathrm{~N\cdot m^2/C^2} \). However, in calculations involving electric fields and charge distributions, we more commonly use the electric constant \( \varepsilon_0 \), also known as the permittivity of free space.

In the context of a line charge, \( \varepsilon_0 \) is crucial as it appears in the denominator of the formula connecting the electric field intensity and the charge per unit length. The precise value of \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2} \) is used in our calculation and ensures that the electric field is accurately described in standard units. It is important to understand the role of \( \varepsilon_0 \) to properly connect the concepts of electric fields with charge distributions.

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Most popular questions from this chapter

A cube has sides of length \(L=0.300 \mathrm{~m}\). One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by \(\overrightarrow{\boldsymbol{E}}=(-5.00 \mathrm{~N} / \mathrm{C} \cdot \mathrm{m}) x \hat{\imath}+(3.00 \mathrm{~N} / \mathrm{C} \cdot \mathrm{m}) z \hat{\boldsymbol{k}} .\) (a) Find the electric flux through each of the six cube faces \(S_{1}, S_{2}, S_{3}, S_{4}, S_{5},\) and \(S_{6}\). (b) Find the total electric charge inside the cube.

A uniformly charged insulating sphere with radius \(r\) and charge \(+Q\) lies at the center of a thin-walled hollow cylinder with radius \(R>r\) and length \(L>2 r\). The cylinder is non-conducting and carries no net charge. (a) Determine the outward electric flux through the rounded "side" of the cylinder, excluding the circular end caps. (Hint: Choose a cylindrical coordinate system with the axis of the cylinder as its \(z\) -axis and the center of the charged sphere as its origin. Note that an area element on the cylinder has magnitude \(d A=2 \pi R d z .\) ) (b) Determine the electric flux upward through the circular cap at the top of the cylinder. (c) Determine the electric flux downward through the circular cap at the bottom of the cylinder. (d) Add the results from parts (a)-(c) to determine the outward electric flux through the closed cylinder. (e) Show that your result is consistent with Gauss's law.

A solid conducting sphere with radius \(R\) that carries positive charge \(Q\) is concentric with a very thin insulating shell of radius \(2 R\) that also carries charge \(Q .\) The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions \(02 R\). (b) Graph the electric-field magnitude as a function of \(r\)

Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{~m} .\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{~m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

(a) A conducting sphere has charge \(Q\) and radius \(R .\) If the electric field of the sphere at a distance \(r=2 R\) from the center of the sphere is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field of the sphere at \(r=4 R ?\) (b) A very long conducting cylinder of radius \(R\) has charge per unit length \(\lambda\). Let \(r\) be the perpendicular distance from the axis of the cylinder. If the electric field of the cylinder at \(r=2 R\) is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field at \(r=4 R ?\) (c) A very large uniform sheet of charge has surface charge density \(\sigma .\) If the electric field of the sheet has a value of \(1400 \mathrm{~N} / \mathrm{C}\) at a perpendicular distance \(d\) from the sheet, what is the electric field of the sheet at a distance of \(2 d\) from the sheet?

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