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Chapter 21: Problem 54

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
(a) Electric field E_z = 561 N/C vertically upward. (b) Electric field is zero due to symmetry.

Step by step solution

01

Understand the Line Charge

The wire has a uniform linear charge density \(\lambda = +175 \, \mathrm{nC/m}\), or \( \lambda = 175 \times 10^{-9} \, \mathrm{C/m} \). The length of the wire is \( L = 8.5 \, \mathrm{cm} = 0.085 \, \mathrm{m} \). We want to calculate the electric field due to this line charge at a point directly above its midpoint.
02

Setup for the Uniform Line Charge

Consider the wire to lie along the x-axis from \(-L/2\) to \(L/2\). The point is 6 cm or 0.06 m above the midpoint, at \( (0, 0, h) \) where \( h = 0.06 \, \mathrm{m} \). By symmetry, the electric field will only have a vertical component \(E_z\) at point \(P\), due to horizontal components canceling out.
03

Electric Field Due to Element of Charge

For a small element of the wire \(dx\) at position \(x\), \(dE_z\), the vertical component of the electric field produced at \(P\), is given by\[ dE_z = \frac{k_e \lambda \, dx}{(x^2 + h^2)} \cdot \frac{h}{\sqrt{x^2 + h^2}} \]where \( k_e = \frac{1}{4\pi\epsilon_0} \approx 8.99 \times 10^9 \, \mathrm{N\,m^2/C^2} \) is the electrostatic constant.
04

Integrate Across the Wire

Integrate the expression for \(dE_z\) over the line charge from \(-L/2\) to \(L/2\):\[ E_z = \int_{-L/2}^{L/2} \frac{k_e \lambda h}{(x^2 + h^2)^{3/2}} \, dx \]Evaluating this integral gives:\[ E_z = \frac{k_e \lambda}{h} \left[ \frac{x}{\sqrt{x^2 + h^2}} \right]_{-L/2}^{L/2} \]After substituting the limits, calculate \(E_z\).
05

Calculate Numerical Solution for Line Charge

Substitute values into the expression:\[ E_z = \frac{(8.99 \times 10^9)(175 \times 10^{-9})}{0.06} \left[ \frac{0.0425}{\sqrt{0.0425^2 + 0.06^2}} - \frac{-0.0425}{\sqrt{0.0425^2 + 0.06^2}} \right] \]Calculate the result to find \(E_z\).
06

Wire Bent into a Circle

When the wire is bent into a circle (radius \(r\)), the electric field at the center due to equal charge distribution is zero as the components cancel each other due to symmetry.
07

Calculate Circle Electric Field

The circle's radius is \(r = \frac{L}{2\pi}\). For point directly above center of circle, since all components of electric field cancel out by symmetry:\[ \vec{E}_{\text{circle}} = 0 \text{ at } h = 0.06 \, \mathrm{m} \] The direction of this field is null.

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Key Concepts

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

Line Charge
In the context of electrostatics, a line charge is a configuration where charge is distributed along a line, such as a wire or rod. This is represented by a linear charge density, denoted as \( \lambda \), which indicates how much charge exists per unit length of the wire or rod.

  • The linear charge density in our example is \(+175 \, \mathrm{nC/m}\) which translates to \(+175 \times 10^{-9} \, \mathrm{C/m}\).
  • The given wire is 8.5 cm long, which in meters is 0.085 meters.
  • We aim to calculate the electric field generated by this line charge at a point situated 6 cm above the wire's midpoint.

Understanding a line charge involves recognizing how this charge distribution influences and creates an electric field in the surrounding space. It considers the cumulative effect of these charges as they manifest in the vertical component of the electric field at a specific point.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. The concepts here are crucial to understanding how electric fields are produced by line charges or other charge distributions.

  • In electrostatics, the electric field \( \mathbf{E} \) due to a point charge can be calculated using Coulomb's Law, but for continuous charge distributions, we must integrate over the distribution.
  • The electric field due to a line charge at a point is calculated by considering the contributions from infinitesimally small charge elements of the line.
  • The principle of superposition applies; hence, the total electric field is the vector sum of the individual fields produced by these differential charges.

In the problem provided, we analyze the electrostatic field at a point above a straight line of charge. This involves understanding both the computation of fields from small elements and their integration over a continuous charge distribution.
Charge Distribution
Charge distribution refers to how electric charges are spatially arranged. In the given problem, the charge distribution involves two scenarios: a straight wire and a circular wire.

  • Firstly, when the wire is straight, the charge is linearly distributed over the length of the wire, exerting an electric field directly above its midpoint which requires integrating the expressions for vertical electric fields from each segment.
  • Secondly, when the wire is bent into a circle, each segment's field contributes but cancels out entirely at the center due to perfect symmetry, leading to zero electric field at that point.
  • Understanding various charge distributions helps us apply theoretical frameworks to predict how they will produce electric fields in space, crucial when modeling and solving physical scenarios.

Each arrangement of charge distributions profoundly impacts the character and behavior of the generated electric field, illustrating the importance of symmetry in simplifying complex calculations.

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Most popular questions from this chapter

(a) What must the charge (sign and magnitude) of a \(1.45-\mathrm{g}\) particle be for it to remain stationary when placed in a downward- directed electric field of magnitude 650 \(\mathrm{N} / \mathrm{C} 2\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitnde to its weight?

Estimate how many electrons there are in your body. Make any assumptions you feel are necessary, but clearly state what they are. (Hint: Most of the atoms in your body have equal numbers of electrons, protons, and neutrons.) What is the combined charge of all these electrons?

A proton is placed in a uniform electric field of \(2.75 \times\) \(10^{3} \mathrm{N} / \mathrm{C}\) . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00\(\mu \mathrm{s}\) in the field, assuming it starts from rest.

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\) . The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\text { max }}\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. ( \(b\) ) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\text { max }}\) and \(d\) if \(E=500 \mathrm{N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{m} / \mathrm{s},\) and \(\alpha=30.0^{\circ} .\)

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu \mathrm{C}\) is at the origin, and charge \(q_{2}=-5.00 \mu \mathrm{C}\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-\mathrm{x}\) -direction?
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