/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 A straight, nonconducting plasti... [FREE SOLUTION] | 91Ó°ÊÓ

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A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

Short Answer

Expert verified
Straight wire: Apply integration, field is vertical upward. Circular wire: Symmetry results in vertical upward field.

Step by step solution

01

Understand the Problem

We have a charged wire of length 8.50 cm, with a linear charge density of +175 nC/m, lying straight on a table. We need to find the electric field at a point 6.00 cm above the midpoint of the wire. In the second part, after bending the wire into a circle, we need to find the electric field at the same point directly above the center of this circle.
02

Set Up for Straight Wire

Initially, the wire is straight and lying on the table. To find the electric field at a point along the perpendicular bisector of the wire, we can use the formula for the electric field due to an infinite line of charge and modify it for a finite line segment.The line charge density \(\lambda = +175 \text{ nC/m} = 175 \times 10^{-9} \text{ C/m}\). The wire length is 8.5 cm which is equivalent to 0.085 meters.The electric field at a distance \(d = 6 \text{ cm} = 0.06 \text{ m}\) directly above the midpoint due to a finite line charge is calculated using integration over the length of the wire.
03

Calculate Electric Field for Straight Wire

The formula for the electric field, E, due to a finite line of charge at its axis is given by:\[E = \frac{\lambda}{4\pi\varepsilon_0 d} \left(\frac{L/2}{\sqrt{(L/2)^2 + d^2}} + \frac{L/2}{\sqrt{(L/2)^2 + d^2}}\right),\]where \(\varepsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\).Substituting the values \( \lambda = 175 \times 10^{-9} \text{ C/m} \), \(L = 0.085 \text{ m}\), and \(d = 0.06 \text{ m}\), calculate \(E\). The direction is vertically upward since the point is directly above the wire.
04

Discuss the Direction for Straight Wire

Since the electric field is symmetrical about the midpoint of the wire and we are considering a point on the perpendicular bisector, the horizontal components from each infinitesimal charge segment cancel out. Thus, the electric field is vertically upward and along the vertical line above the midpoint of the wire.
05

Set Up for Circular Wire

When the wire is bent into a circle, the net electric field at a point directly above the center of the circle is due to the symmetry of the charge distribution. The circle can be thought of having a radius \(r = \frac{L}{2\pi}\), where \(L = 0.085 \text{ m} \). We have a point at a height \( h = 0.06 \text{ m} \) above the center.
06

Calculate Electric Field for Circular Wire

For a ring of charge, the vertical component of the electric field at an axial point is given by:\[E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q \cdot h}{(r^2 + h^2)^{3/2}}.\]The total charge \(q = \lambda L = 175 \times 10^{-9} \times 0.085 \). Using these, calculate \(E\). The field direction is still vertically upward along the axis of the circle.
07

Discuss the Direction for Circular Wire

The circular symmetry ensures that the electric field produced by each infinitesimal charge segment along the ring has a horizontal component that cancels out. Therefore, the entire contribution to the electric field is vertically upwards.

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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 charge is distributed along a line, such as a wire. It is represented by the symbol \( \lambda \), and it's defined as the charge per unit length of the wire. In the exercise, the linear charge density \( \lambda \) is given as \(+175 \text{ nC/m}\). This means that every meter of the wire has 175 nanoCoulombs of charge spread uniformly along its length.
Understanding linear charge density is crucial when determining the electric field produced by a charged line, as it directly influences the strength of the field. Since field calculations often require integration over the length, knowing \( \lambda \) situates the problem in a more calculable framework.
When dealing with finite lines or specific shapes (like a circle), it's important to adjust the setup appropriately in terms of linear charge density to model the electric field accurately.
Permittivity of Free Space
Permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental physical constant characterizing the ability of the vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2\).
This constant appears frequently in electrostatic equations, such as the ones used in finding the electric field due to different charge distributions. In the problem, \( \varepsilon_0 \) plays a key role in calculating the electric field generated by the wire. It occurs in the denominator of formulas to demonstrate its inverse relationship with force; namely, higher permittivity implies that a given charge produces a weaker field in the medium.
Understanding \( \varepsilon_0 \) helps in realizing how electric forces act in different environments, providing a baseline to compare dielectric materials.
Finite Line Charge
A finite line charge is a charge distribution along a limited-length straight line. Unlike an infinite line charge, which extends indefinitely, a finite line has clear boundaries. The exercise talks about how to compute the electric field at a point directly above the midpoint of a straight wire that represents a finite line charge.

When calculating the electric field from a finite line charge, the integration covers only the extent of the wire, differentiating it from calculations involving infinite lines where symmetry and simplification play larger roles. The resulting field not only depends on the distance from the line and the linear charge density but also requires considering the entire geometry due to the wire's limited length.

Knowing how a finite line charge operates allows you to solve problems involving real-world objects like wires and rods, giving a more practical perspective on electrical field distributions.

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Most popular questions from this chapter

CP A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P21. 104 ). 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 \(\vec{E} .\) Consider points both above and below the annulus in Fig. P21. \(104 .\) (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.)

Two positive point charges \(q\) are placed on the \(x\) -axis, one al \(x=a\) and one at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0 .\) (b) Derive an expression for the electric field at points on the \(x\) -axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x,\) for values of \(x\) between \(-4 a\) and \(+4 a .\)

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{Cis~placed}\) at the point \(x=+0.150 \mathrm{m}, y=0\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m}\)

CALC Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

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