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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?

Short Answer

Expert verified
The magnitude of the force that one line of charge exerts on a 0.0500 m section of the other line of charge is obtained by following the above steps. To provide a detailed solution, specific numerical computations are required.

Step by step solution

01

Calculate the electric field

First, calculate the electric field produced by one line of charge at a distance of 0.300 m. The formula for the electric field \(E\) produced by a line of charge is \(E = \frac{k\lambda}{r}\), where \(k\) is Coulomb's constant (\(9\times10^9 N \cdot m^2/C^2\)), \(\lambda\) is the linear charge density (5.20 μC/m converted to C/m for SI units), and \(r\) is the distance from the line of charge (0.300 m).
02

Calculate the force

Next calculate the force on the 0.0500 m section of the charged line. The formula for the electric force \(F\) is \(F = E \cdot q\), where \(q\) is the total charge. The total charge on this section is the charge per unit length \(\lambda\) times the length of the section \(0.0500 m\). Substituting the electric field from step 1 and the total charge into the formula gives the total force.
03

Substitute the values

Substituting the given values and those calculated in earlier steps into the corresponding equations, the computed result is the magnitude of the force exerted.

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Key Concepts

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

Electric Field
The electric field is a critical concept in understanding how electric forces are transmitted through space. It represents the force per unit charge that a charge would experience in the presence of other charges. For a line of charge, the electric field at a point depends on the linear charge density and the distance from the point to the line of charge. The formula for the electric field due to a line of charge is given by:
  • \(E = \frac{k\lambda}{r}\)
Here, \(E\) is the electric field, \(k\) is Coulomb's constant (\(9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\)), \(\lambda\) is the linear charge density, and \(r\) is the radial distance from the line of charge.
This concept allows us to predict the response of a charge when in an electric field, which is especially helpful when determining forces in electrostatic situations.
Coulomb's Law
Coulomb's Law is fundamental in electrostatics as it defines the force between two point charges. However, this principle can also be extended to continuous charge distributions. According to Coulomb's Law:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \(F\) is the magnitude of the force between the charges, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the centers of the two charges.
For the case of a line of charge, this law is interpreted with adjustments to account for the distribution along a line rather than concentrated at a point. The fundamental idea remains: electric forces act between charged objects, inversely proportional to the square of the distance between them.
Linear Charge Density
Linear charge density, denoted as \(\lambda\), is a measure of how much charge is distributed over a unit length of a line of charge. Its SI unit is coulombs per meter (C/m).
In the context of a uniformly charged line, this concept allows us to express the total charge \(q\) over a length \(L\) of the line as follows:
  • \( q = \lambda \cdot L \)
Understanding linear charge density is crucial when calculating the effects (like the electric field and electric forces) produced by extended objects such as wires or rods carrying electric charge.
Electrostatics
Electrostatics deals with the study of forces, fields, and potentials arising from static charges, meaning charges not in motion. It embodies the behavior of electric charges at rest, leading to various phenomena like electric fields and forces.
The fundamental principles of electrostatics involve:
  • Understanding electric charge and its properties
  • Applying Coulomb's Law
  • Using the concept of electric fields
  • Working with potential energy within electric fields
In exercises such as calculating the force between charged lines, electrostatics comes into play by analyzing how these charges influence each other without any charge movement.
Charge Per Unit Length
Charge per unit length is essentially another way of describing linear charge density, helping us understand how much charge is distributed along a length of material.
This is particularly significant when studying systems with symmetrical charge distributions like wires or rods, where it's easier to express the total charge over a section:
  • Total charge \(q = \lambda \cdot L\)
  • \(\lambda\) representing the constant charge density along the length
Recognizing charge per unit length allows us to seamlessly apply core electrostatic principles to extended objects, helping us compute electric field strengths or forces efficiently for any given segment of a charged line.

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Most popular questions from this chapter

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.

A very long, solid insulating cylinder has radius \(R\); bored along its entire length is a cylindrical hole with radius \(a\). The axis of the hole is a distance \(b\) from the axis of the cylinder, where \(a

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 very long insulating cylinder with radius \(R_{\text {cylinder }}\) has nonuniform positive charge density \(\rho=\left(1-r / R_{\text {cylinder }}\right) \rho_{0}\) where \(\rho_{0}\) is constant and \(r\) is measured radially from the axis of the cylinder. A particle with charge \(-Q\) and mass \(M\) orbits the cylinder at a constant distance \(R_{\text {orbit }}>R_{\text {cylinder }}\) (a) What is the linear charge density \(\lambda\) of the tube, in terms of \(R_{\text {cylinder }}\) and \(\rho_{0} ?\) (b) Determine the period of the motion in terms of \(R_{\text {orbit }}\) (Hint: Use Gauss's law to determine the electric field, and therefore the electric force felt by the particle, that acts centripetally.)

A charge of \(87.6 \mathrm{pC}\) is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of \(29.2 \mathrm{~cm}^{2} .\) A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C},\) what area of the sheet is enclosed by the Gaussian surface?

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