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A very long, straight wire has charge per unit length \(1.50 \times 10^{-10} \mathrm{C} / \mathrm{m}\). At what distance from the wire is the electricfield magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)

Short Answer

Expert verified
The distance from the wire where the electric field is 2.50 N/C is approximately 1.08 cm.

Step by step solution

01

Understand the Problem

We have a long, straight wire with a given linear charge density \( \lambda = 1.50 \times 10^{-10} \, \mathrm{C/m} \). We need to find the distance \( r \) from the wire at which the electric field magnitude is \( E = 2.50 \, \mathrm{N/C} \).
02

Recall the Formula for Electric Field due to a Line Charge

The electric field \( E \) at a distance \( r \) from an infinitely long straight wire with linear charge density \( \lambda \) is given by the formula:\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \) is the permittivity of free space.
03

Rearrange the Formula to Solve for \( r \)

Solve the electric field formula for \( r \):\[ \begin{align*} E &= \frac{\lambda}{2 \pi \varepsilon_0 r} \2 \pi \varepsilon_0 r E &= \lambda \r &= \frac{\lambda}{2 \pi \varepsilon_0 E} \end{align*} \]
04

Substitute Known Values

Substitute \( \lambda = 1.50 \times 10^{-10} \, \mathrm{C/m} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \), and \( E = 2.50 \, \mathrm{N/C} \) into the equation:\[ r = \frac{1.50 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times 2.50} \]
05

Calculate the Value of \( r \)

Perform the calculation:\[ r = \frac{1.50 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times 2.50} \approx 1.08 \times 10^{-2} \, \mathrm{m} \]
06

Interpret the Result

The distance \( r \) from the wire where the electric field magnitude is \( 2.50 \, \mathrm{N/C} \) is approximately \( 0.0108 \, \mathrm{m} \) or \( 1.08 \, \mathrm{cm} \).

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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 is a measure of how much electric charge is distributed along a line, such as a wire. It is denoted by the symbol \( \lambda \) and is defined as the amount of charge per unit length of the line. This concept is crucial when calculating the electric fields generated by line charges since it provides a straightforward way to express the distribution of charge.

- For example, if you have a wire carrying a uniform charge, and its linear charge density is given as \( \lambda = 1.50 \times 10^{-10} \, \mathrm{C/m} \), this means there is \( 1.50 \times 10^{-10} \) Coulombs of electric charge for every meter of the wire.

- Understanding linear charge density helps in simplifying the calculation of electric fields around such objects, which are otherwise continuous and potentially complex distributions. It ultimately serves as one of the fundamental building blocks in electromagnetic theory.
Permittivity of Free Space
Permittivity of free space, also known as the vacuum permittivity, is a constant that appears in the equations of electromagnetism, indicating the ability of free space to permit electric field lines. This is symbolized by \( \varepsilon_0 \) and has a value of \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \).

- It plays a critical role in Coulomb's Law, which calculates the force between two point charges, and in the formula for the electric field produced by a continuous charge distribution like a line charge.

- In practical terms, the lower the permittivity, the stronger an electric field might be for a certain amount of charge. Understanding \( \varepsilon_0 \) is vital for determining the force interactions between charges in free space, and it is a key parameter in many applications, such as capacitors, where it governs the relationship between the charge, voltage, and the configuration of the system.
Electric Field due to a Line Charge
The electric field due to a line charge is an important concept in electrostatics, especially when dealing with scenarios like the exercise, where one calculates the field generated by a long, straight wire. In this context, the electric field \( E \) at a distance \( r \) from a wire with a linear charge density \( \lambda \) is given by the formula:\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]

- Here, \( \varepsilon_0 \) is the permittivity of free space, and \( r \) is the radial distance from the wire to the point where the electric field is measured.

- This relationship indicates that the electric field is directly proportional to the linear charge density and inversely proportional to the distance from the wire. As the distance from the wire increases, the electric field decreases, which is expected as the influence of the charge becomes weaker further away.

- Understanding this formula allows for calculating the electric field at any point around a charged wire, which is essential for various real-world applications such as designing electrical circuits and understanding natural phenomena.

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Most popular questions from this chapter

Positive charge \(Q\) is distributed uniformly along the positive \(y\)-axis between \(y=0\) and \(y=a\). A negative point charge \(-q\) lies on the positive \(x\)-axis, a distance \(x\) from the origin (Fig. P1.48). (a) Calculate the \(x\) - and \(y\)-components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\)-axis. (b) Calculate the \(x\) - and \(y\)-components of the force that the charge distribution \(Q\) exerts on \(q\). (c) Show that if \(x \gg a, F_{x} \cong-Q q / 4 \pi \epsilon_{0} x^{2}\) and \(F_{y} \cong+Q q a / 8 \pi \epsilon_{0} x^{3} .\) Explain why this result is obtained.

A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P1.57). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\)-plane, with its center at the origin. For an arbitrary point on the \(x\)-axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus in Fig. P1.56. (c) Show that at points on the \(x\)-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\)-axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(2.25 \mathrm{~cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

\(\cdots\) A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\)-coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\)-axis at \(x=4.00 \mathrm{~cm}\). If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{~cm}, y=3.00 \mathrm{~cm}\), find the \(x\) - and \(y\)-components of the total force exerted on this charge by the other two.

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{~N} ?\)

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