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(II) Estimate the electric field at a point 2.40\(\mathrm { cm }\) perpendicular to the midpoint of a uniformly charged 2.00 -m-long thin wire carrying a total charge of 4.75\(\mu \mathrm { C }\) .

Short Answer

Expert verified
The electric field is approximately \(3.00 \times 10^5 \mathrm{N/C}\).

Step by step solution

01

Understanding the Problem

You need to find the electric field at a point 2.40 cm (0.024 m) from the midpoint of a 2.00-m-long thin wire that is uniformly charged with a total charge of 4.75 μC. This involves calculating the electric field due to a line charge using integration.
02

Calculate the Linear Charge Density

The linear charge density, \( \lambda \), is the charge per unit length of the wire. It can be calculated using the formula \( \lambda = \frac{Q}{L} \), where \( Q = 4.75 \times 10^{-6} \mathrm{C} \) is the total charge, and \( L = 2.00 \mathrm{m} \) is the length of the wire. Thus, \( \lambda = \frac{4.75 \times 10^{-6}}{2.00} = 2.375 \times 10^{-6} \mathrm{C/m} \).
03

Setup the Electric Field Integral

Consider an infinitesimally small charge element \( dq = \lambda dx \) at a distance \( x \) along the wire. The electric field due to this element at the point located 2.40 cm away, perpendicular to the line, is given by \( dE = \frac{k_e dq}{(x^2 + y^2)} \times \frac{y}{\sqrt{x^2 + y^2}}\), where \( y = 0.024 \mathrm{m} \), and \( k_e \) is the Coulomb's constant \( 8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2 \).
04

Integrate to Find the Total Electric Field

Integrate \( dE \) from \(-L/2\) to \(L/2\) (or from \(-1.00 m\) to \(1.00 m\), as the wire's midpoint is at zero). The integral is \( E = \int_{-1}^{1} \frac{k_e \lambda y}{(x^2 + y^2)^{3/2}} \, dx \). Substitute \( k_e = 8.99 \times 10^9 \), \( \lambda = 2.375 \times 10^{-6} \), and \( y = 0.024 \) then integrate to find \( E \).
05

Solve the Integral

Perform the integration mathematically or numerically using appropriate methods (like numerical integration software) to evaluate the definite integral. Calculate to get \( E \approx 3.00 \times 10^5 \mathrm{N/C} \).
06

Interpret the Results

The electric field at the point 2.40 cm away from the midpoint of the wire is approximately \( 3.00 \times 10^5 \mathrm{N/C} \). This result indicates the strength and direction of the field produced by this thin uniformly charged wire at the given point.

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Key Concepts

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

Understanding Linear Charge Density
Linear charge density is a measure of how much charge is distributed over a certain length of a conductor, such as a wire. When dealing with problems involving continuous charge distributions, it's important to know the linear charge density, as this will tell us how charge spreads out along an object.
Linear charge density is denoted by the Greek letter \( \lambda \) and is defined as the total charge \( Q \) divided by the total length \( L \) of the object. In mathematical terms, \( \lambda = \frac{Q}{L} \).
Knowing \( \lambda \) allows us to calculate the electric field generated by the wire at a point outside of it. For instance, if you have a 2-meter long wire with a total charge of 4.75 μC, the linear charge density would be \( 2.375 \times 10^{-6} \) C/m. This value essentially tells us that every meter of the wire has roughly 2.375 microcoulombs of charge distributed evenly over it.
Coulomb's Constant in Electric Field Calculations
Coulomb's constant, represented by \( k_e \), plays a crucial role in electrostatic calculations. It is a constant that quantifies the amount of force between two charges. The value of Coulomb's constant is \( 8.99 \times 10^9 \) N·m²/C². This number forms a part of Coulomb's law, which describes how the force between two point charges depends on their sizes and the distance between them.
When calculating the electric field produced by a line charge, like a uniformly charged wire, Coulomb's constant is used to determine the strength of the field at a specific point around the wire. The concept is straightforward: the electric field due to a small charge element along the wire is recalculated using this constant. Then, we sum these tiny fields to estimate the whole field using integration.
Integrating Electric Fields to Find Total Effect
Integration is a powerful mathematical tool that allows us to calculate the total electric field generated by a continuous charge distribution. In the context of this exercise, it's used to sum the electric fields produced by each infinitesimally small charge segment along the wire.
The formula \( dE = \frac{k_e dq}{(x^2 + y^2)} \times \frac{y}{\sqrt{x^2 + y^2}} \) represents the tiny electric field contribution from a small segment with charge \( dq = \lambda dx \). By integrating this expression from one end of the wire to the other, we find the total electric field.
This process involves setting up definite integrals that cover the entire length of the wire, typically from \(-L/2\) to \(L/2\). By carefully evaluating this integral, either analytically or using computational software, we can precisely determine the strength and direction of the electric field at a point away from the charge distribution. In our example, integration results in an electric field magnitude of approximately \( 3.00 \times 10^5 \) N/C at a given distance from the wire.

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Most popular questions from this chapter

(II) An electron with speed \(v_{0}=27.5 \times 10^{6} \mathrm{~m} / \mathrm{s}\) is traveling parallel to a uniform electric field of magnitude \(E=11.4 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) (a) How far will the electron travel before it stops? (b) How much time will elapse before it returns to its starting point?

A one-dimensional row of positive ions, each with charge \(+Q\) and separated from its neighbors by a distance \(d\) occupies the right-hand half of the \(x\) axis. That is, there is a \(+Q\) charge at \(x=0, x=+d, x=+2 d, x=+3 d,\) and so on out to \(\infty .(a)\) If an electron is placed at the position \(x=-d,\) determine \(F,\) the magnitude of force that this row of charges exerts on the electron. \((b)\) If the electron is instead placed at \(x=-3 d,\) what is the value of \(F ?\) [Hint: The infinite sum \(\sum_{n=1}^{n=\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6},\) where \(n\) is a positive integer.]

(II) Estimate the electric field at a point \(2.40 \mathrm{~cm}\) perpendicular to the midpoint of a uniformly charged 2.00 -m-long thin wire carrying a total charge of \(4.75 \mu \mathrm{C}\).

(I) How many electrons make up a charge of \(-38.0 \mu \mathrm{C} ?\)

(a) The electric field near the Earth's surface has magnitude of about \(150 \mathrm{~N} / \mathrm{C} .\) What is the acceleration experienced by an electron near the surface of the Earth? (b) What about a proton? (c) Calculate the ratio of each acceleration to \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

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