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An infinitely long charged wire produces an electric field of magnitude \(1.23 \cdot 10^{3} \mathrm{~N} / \mathrm{C}\) at a distance of \(50.0 \mathrm{~cm}\) perpendicular to the wire. The direction of the electric field is toward the wire. a) What is the charge distribution? b) How many electrons per unit length are on the wire?

Short Answer

Expert verified
Question: Determine the charge distribution per unit length and the number of electrons per unit length on an infinitely long charged wire, given that the electric field magnitude at a distance of 50 cm from the wire is 1.23 x 10^3 N/C. Answer: The charge distribution per unit length on the wire is approximately 3.43 x 10^-9 C/m, and there are approximately 2.14 x 10^10 electrons/m present on the wire.

Step by step solution

01

Review Gauss's Law for a cylindrical symmetry situation

Using Gauss's law, we can calculate the charge distribution on the wire (charge per unit length). Gauss's law states that: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}\) Where \(\vec{E}\) is the electric field, \(d\vec{A}\) is differential vector area, \(Q_{enc}\) is the enclosed charge, and \(\epsilon_0\) is the electric constant (approximately \(8.85 \times 10^{-12} \mathrm{C^2/N\cdot m^2}\)). In a cylindrical symmetry situation, the electric field will be constant on the cylinder's surface, and the direction of the electric field will be parallel to dA. Thus, we can simplify the integral into: \(E \cdot A = \frac{Q_{enc}}{\epsilon_0}\)
02

Calculate the charge per unit length, λ

First, we can determine the charge per unit length using Gauss's law. We know that \(E = 1.23 \cdot 10^{3} \mathrm{~N/C}\) and \(r = 50.0~\mathrm{cm} = 0.5~\mathrm{m}\). For a cylindrical Gaussian surface around the wire with radius r, the area is \(A = 2\pi r L\), where L is the unit length of the wire. Plugging in the values, we get: \(E\cdot A = \frac{\lambda L}{\epsilon_0} \Rightarrow 1.23 \cdot 10^{3}\mathrm{~N/C} \cdot 2\pi \cdot 0.5\mathrm{~m} \cdot L = \frac{\lambda L}{8.85 \times 10^{-12}\mathrm{C^2/N\cdot m^2}}\) Solving for λ, we obtain: \(\lambda = 1.23 \cdot 10^{3}\mathrm{~N/C} \cdot 2\pi \cdot 0.5\mathrm{~m} \cdot 8.85 \times 10^{-12}\mathrm{C^2/N\cdot m^2} \approx 3.43 \times 10^{-9}\mathrm{C/m}\)
03

Calculate the number of electrons per unit length

Now that we have the charge per unit length, we can determine the number of electrons per unit length. We know that the charge of an electron is \(e = -1.6 \times 10^{-19}\mathrm{C}\). To find the number of electrons (\(n\)) per unit length, we can divide the charge per unit length by the charge of an electron: \(n = \frac{\lambda}{-e} = \frac{3.43 \times 10^{-9}\mathrm{C/m}}{-1.6 \times 10^{-19}\mathrm{C}} \approx 2.14 \times 10^{10}\mathrm{electrons/m}\) Since the electric field is towards the wire, the wire must have a negative charge, which is consistent with our result. The charge per unit length on the wire is \(\lambda \approx 3.43 \times 10^{-9}\mathrm{C/m}\), and there are approximately \(2.14 \times 10^{10}\mathrm{electrons/m}\) present on the wire.

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Key Concepts

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

Understanding Electric Field
The electric field is a fundamental concept in electromagnetism, representing the force per unit charge exerted on a small positive test charge placed at a point in space. An electric field is produced by electric charges or varying magnetic fields. In our problem, an infinitely long charged wire creates an electric field that has a specific magnitude at a given distance from the wire. It is vectorial, having both magnitude and direction, and in this case, the field is directed toward the wire indicating the wire carries a negative charge. The strength of the electric field diminishes with distance and can be calculated using the formula
\[ E = \frac{F}{q} \]
where \(E\) represents the electric field strength, \(F\) is the force experienced by the test charge, and \(q\) is the magnitude of that test charge.
Charge Distribution on the Wire
Charge distribution refers to how electric charge is spread along, over, or throughout an object. For example, a uniform charge distribution would be equally spaced charges along the length of our wire. In contrast, non-uniform distribution might have more charges accumulated at one end of the wire than the other. In the given problem, we are required to determine the charge distribution of an infinitely long wire, which we quantify as charge per unit length. Uniform charge distribution is often assumed for mathematical simplicity, especially in infinitely long or large conductors.
Cylindrical Symmetry and Its Implications for Gauss's Law
Cylindrical symmetry means that a system's properties are invariant under rotations about its axis and translations along the axis. When applying Gauss's Law, cylindrical symmetry simplifies the calculations drastically, as the electric field's strength will be the same at all points equidistant from the axis of symmetry—the long wire in this scenario. Imagining a cylindrical 'Gaussian surface' surrounding the wire, we exploit this symmetry to state that the electric field at any point on the surface only depends on the distance from the wire and not on the angle around it or the position along the length of the wire. Thus, we can treat the problem in a more straightforward manner.
The Role of the Electric Constant (Permittivity of Free Space)
The electric constant, also known as the permittivity of free space (\(\epsilon_0\)), is a physical constant that describes how easily an electric field can permeate the vacuum of free space. It is a measure of the resistance encountered when forming an electric field, featuring in Coulomb's law as well as Gauss's law for electricity. In the case of Gauss's law, \(\epsilon_0\) appears in the denominator, demonstrating that the electric field strength is inversely proportional to the permittivity of free space. The precise value of the electric constant is key to solving many electromagnetic problems as it relates the electric field to the quantity of charge responsible for its formation.
Determining Charge Per Unit Length
Charge per unit length (\(\lambda\)) is a measurement indicating the amount of charge located along a line—here, along our infinite wire. We calculate it by dividing the total enclosed charge by the length over which this charge extends. With cylindrical symmetry, the enclosed charge can be equated to the product of charge per unit length and the length of the cylinder. In the exercise, Gauss's law helps us find the value for \(\lambda\) by taking the electric field and multiplying it by the area of our Gaussian surface. Dividing this product by the electric constant gives us the total enclosed charge along the length of the wire.

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Most popular questions from this chapter

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer).

A point charge, \(+Q,\) is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. c) less than zero. b) greater than zero. d) none of the above.

At which of the following locations is the electric field the strongest? a) a point \(1 \mathrm{~m}\) from a \(1-\mathrm{C}\) point charge b) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a 1 -m-long wire with \(1 \mathrm{C}\) of charge distributed on it c) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}^{2}\) sheet of charge with \(1 \mathrm{C}\) of charge distributed on it d) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell with a radius of \(1 \mathrm{~m}\) e) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell with a radius of \(0.5 \mathrm{~m}\) and a charge of \(1 \mathrm{C}\)

A thin, hollow, metal cylinder of radius \(R\) has a surface charge distribution \(\sigma\). A long, thin wire with a linear charge density \(\lambda / 2\) runs through the center of the cylinder. Find an expression for the electric field and determine the direction of the field at each of the following locations: a) \(r \leq R\) b) \(r \geq R\)

In order to use Gauss's Law to calculate the electric field created by a known distribution of charge, which of the following must be true? a) The charge distribution must be in a nonconducting medium. b) The charge distribution must be in a conducting medium. c) The charge distribution must have spherical or cylindrical symmetry. d) The charge distribution must be uniform. e) The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made.

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