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Chapter 21: Problem 52

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 electric- field magnitude equal to 2.50 \(\mathrm{N} / \mathrm{C}\)?

Short Answer

Expert verified
The distance is approximately 1.08 cm.

Step by step solution

01

Understand the Problem

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

Recognize the Electric Field Formula for a Line Charge

The electric field due to a long straight wire at a distance \(r\) is given by: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where \(\varepsilon_0\) is the permittivity of free space and its value is \(8.85 \times 10^{-12} \, \mathrm{C^2/(N\cdot m^2)}\).
03

Plug in the Values and Rearrange for r

Substitute \(E = 2.50 \, \mathrm{N/C}\) and \(\lambda = 1.50 \times 10^{-10} \, \mathrm{C/m}\) into the formula: \[ 2.50 = \frac{1.50 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times r} \] Rearranging for \(r\), we have: \[ r = \frac{1.50 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times 2.50} \]
04

Calculate r

Perform the calculation: \[ r = \frac{1.50 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times 2.50} \] Using a calculator: \[ r \approx 1.08 \times 10^{-2} \, \mathrm{m} \]
05

Conclusion

The distance from the wire where the electric field magnitude is \(2.50 \, \mathrm{N/C}\) is approximately \(1.08 \times 10^{-2} \, \mathrm{m}\), 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, represented as \( \lambda \), is a measure of the amount of electric charge per unit length along a line, such as a wire.
It is an important concept when dealing with problems related to electric fields generated by line charges.
  • It's defined in terms of coulombs per meter (C/m).
  • In the given problem, the wire has a linear charge density \( \lambda = 1.50 \times 10^{-10} \mathrm{C/m} \).
Understanding linear charge density helps determine how charge is distributed along a wire, which directly affects the electric field's magnitude and direction around the wire.
Long Straight Wire
A long straight wire is a common model used in physics to simplify the calculation of electric fields.
This model assumes the wire is infinitely long, which allows certain symmetries to simplify mathematical calculations.
  • By modeling a very long wire, the electric field can be treated as uniform in a plane perpendicular to the wire.
  • This assumption allows us to ignore edge effects, making calculations easier.
In the context of the given problem, the simplification of an infinite wire helps in applying Gauss's law easily to find the electric field.
Permittivity of Free Space
The permittivity of free space, \( \varepsilon_0 \), is a fundamental constant that characterizes how electric fields interact with the vacuum.
Its value is \( 8.85 \times 10^{-12} \mathrm{C^2/(N\cdot m^2)} \), making it crucial for calculating electric fields in a vacuum or air.
  • It acts as a proportionality factor in the formulas for electric fields, linking electric field intensity to charge density.
  • In the electric field equation \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \), \( \varepsilon_0 \) directly influences the field strength.
Understanding \( \varepsilon_0 \) is key to solving problems related to electric fields in spaces where matter is negligible.
Electric Field Formula
The electric field formula for a long straight wire provides a way to compute the electric field created by the wire at a certain distance.
This formula is derived using Gauss's law, which relates the electric field to the charge enclosed by a Gaussian surface.
  • For a linear charge, the formula is \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \).
  • The variables \( \lambda \) and \( r \) denote linear charge density and the distance from the wire, respectively.
  • \( \varepsilon_0 \) is the permittivity of free space and acts as a scaling factor.
By applying this formula, we can determine the strength of the electric field at various points around the wire, showing how distance and charge density affect the field.

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Most popular questions from this chapter

Two very large parallel sheets are 5.00 \(\mathrm{cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2},\) and sheet \(B,\) which is to the right of \(A,\) carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}\) . Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 \(\mathrm{cm}\) to the right of sheet \(A\) (b) 4.00 \(\mathrm{cm}\) to the left of sheet \(A ;(c) 4.00 \mathrm{cm}\) to the right of sheet \(B\) .

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L .\) Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and \((b)\) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

Two small plastic spheres are given positive electrical charges. When they are 15.0 \(\mathrm{cm}\) apart, the repulsive force between them has magnitude 0.220 \(\mathrm{N} .\) What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about \(20,000 \mathrm{C} / \mathrm{s} ;\) this lasts for 100\(\mu\) or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?

\(A+2.00-n C \quad\) point charge is at the origin, and a second \(-5.00\) -n \(\mathrm{C}\) point charge is on the \(x\) -axis at \(x=0.800 \mathrm{m}\) . (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\) -axis: (i) \(x=\) \(0.200 \mathrm{m} ;\) (ii) \(x=1.20 \mathrm{m} ;\) (iii) \(x=-0.200 \mathrm{m} .\) (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).
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