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22.20. 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 charge contained in the 2.00 cm section is approximately 4.44 nC.

Step by step solution

01

Understanding the Problem

We are given the electric field strength at a certain distance from a uniformly charged line and need to calculate the charge contained in a specific length of that line.
02

Formula for Electric Field Due to an Infinite Line of Charge

The electric field due to an infinite line of charge is given by the formula \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where \( E \) is the electric field, \( \lambda \) is the linear charge density, \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \)), and \( r \) is the distance from the line of charge.
03

Rearranging the Formula for Linear Charge Density

We need to find \( \lambda \), the linear charge density. Rearrange the electric field formula to solve for \( \lambda \):\[ \lambda = E \cdot 2 \pi \varepsilon_0 r \]
04

Calculating Linear Charge Density

Substitute the given values into the rearranged formula:\[ \lambda = 840 \text{ N/C} \cdot 2 \pi \cdot 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \cdot 0.400 \text{ m} \]Calculate \( \lambda \).
05

Converting Linear Charge Density to Total Charge

The charge contained in a section of the line is given by \( Q = \lambda \cdot L \), where \( L \) is the length of the section. Convert \( L \) from cm to m by dividing by 100, so \( L = 0.020 \text{ m} \).
06

Calculating the Total Charge

Use the linear charge density \( \lambda \) calculated previously to find the total charge:\[ Q = \lambda \cdot 0.020 \text{ m} \]Compute \( Q \).
07

Final Result

Substitute the calculated value of \( \lambda \) to get the numeric result for the total charge \( Q \) contained in the 2.00 cm section.

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Key Concepts

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

Linear Charge Density
Linear charge density, denoted by \( \lambda \), is a measure that tells us how much charge is distributed along a line per unit length. This is particularly relevant when dealing with objects like wires or rods that have charge evenly spread along their length.
  • It is expressed in units of charge per length, commonly coulombs per meter (C/m).
  • The linear charge density allows us to easily understand the intensity of charge on an object of considerable length.
  • In problems involving electric fields, knowing the linear charge density is key to understanding how the field is influenced by an object.

To calculate the linear charge density, we rearrange the formula that relates the electric field \( E \) due to an infinite line of charge:\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \rightarrow \lambda = E \cdot 2 \pi \varepsilon_0 r \]This formula demonstrates the relationship between the electric field and how densely charge is distributed along the line. The rearranged formula lets us calculate \( \lambda \) when \( E \), \( r \), and \( \varepsilon_0 \) are known.
Permittivity of Free Space
The permittivity of free space, symbolized by \( \varepsilon_0 \), is a fundamental physical constant that describes how electric fields interact with the vacuum of space. Its influence is seen in Coulomb’s law and many electromagnetic theories.
  • The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).
  • It serves as a constant of proportionality in equations that describe electric fields, like the one for an infinite line of charge.
  • It describes how easily electric field lines can permeate space, essentially dictating the strength and reach of an electric field.

The presence of \( \varepsilon_0 \) in the formula for the electric field due to an infinite line of charge highlights its role in determining the field’s behavior and strength. In practice, \( \varepsilon_0 \) makes calculations possible by establishing the baseline relationship between electric field intensity, distance, and charge distribution.
Infinite Line of Charge
An infinite line of charge is a theoretical model used to illustrate electric fields produced by lines of charged particles that extend indefinitely. This model simplifies calculations by focusing on symmetry and uniformity.
  • Due to its infinite nature, the electric field at any point depends only on the perpendicular distance from the line.
  • It assumes uniform linear charge density, allowing the use of symmetry to simplify calculations.
  • In practice, real objects are finite, but for sufficiently long charged rods or wires, this model serves as an accurate approximation.

The formula derived from Gauss’s Law for the electric field due to an infinite line of charge is:\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]This indicates that the electric field intensity decreases as you move away from the line, inversely proportional to the radial distance \( r \). This model is particularly useful in understanding fundamental concepts in electrostatics, such as symmetry and the behavior of electric fields in uniform charge distributions.

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Most popular questions from this chapter

22.5. A hemispherical surface with radius \(r\) in a region of uniform electric field \(\vec{E}\) has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

\(22.66 .\) 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 $$ \begin{array}{ll}{\rho(r)=\alpha} & {\text { for } r \leq R / 2} \\\ {\rho(r)=2 \alpha(1-r / R)} & {\text { for } R / 2 \leq r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ 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 \(\vec{E}\) as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q\) . Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?\left(\text { d) If an electron with charge } q^{\prime}=-e \text { is }\right.\) oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2,\) show that the motion is simple harmonic. (Hint: Review the discussion of simple harmonic motion in Section 13.2. If, and only if, the net force on the electron is proportional to its displacement from equilibrium, then the motion is simple harmonic. \()(\mathrm{e})\) What is the period of the motion in part \((\mathrm{d}) ?(\mathrm{f})\) If the amplitude of the motion described in part (e) is not?

22\. 13. A \(9.60-\mu\) C point charge is at the center of a cube with sides of length 0.500 \(\mathrm{m}\) (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 \(\mathrm{m}\) long? Explain.

22.9. A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{cm},\) giving it a charge of \(-15.0 \mu \mathrm{C}\) . Find the electric field (a) just inside the paint layer; (b) just outside the paint layer: (c) 5.00 \(\mathrm{cm}\) outside the surface of the paint layer.

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